Mixing
How can I loop through a Python list and perform math calculations on elements of the list?
I am attempting to create a contract bridge match point scoring system. In the list below the 1st, 3rd, etc. numbers are the pair numbers (players) and the 2nd, 4th etc. numbers are the scores achieved by each pair. So pair 2 scored 430, pair 3 scored 420 and so on.
I want to loop through the list and score as follows:
for each pair score that pair 2 beats they receive 2 points, for each they tie 1 point and where they don't beat they get 0 points. The loop then continues and compares each pair's score in the same way. In the example below, pair 2 gets 7 points (beating 3 other pairs and a tie with 1), pair 7 gets 0 points, pair 6 gets 12 points beating every other pair.
My list (generated from an elasticsearch json object) is:
['2', '430', '3', '420', '4', '460', '5', '400', '7', '0', '1', '430', '6', '480']
The python code I have tried (after multiple variations) is:
nsp_mp = 0
ewp_mp = 0
ns_list =
for row in arr["hits"]["hits"]:
nsp = row["_source"]["nsp"]
nsscore = row["_source"]["nsscore"]
ns_list.append(nsp)
ns_list.append(nsscore)
print(ns_list)
x = ns_list[1]
for i in range(6): #number of competing pairs
if x > ns_list[1::2][i]:
nsp_mp = nsp_mp + 2
elif x == ns_list[1::2][i]:
nsp_mp = nsp_mp
else:
nsp_mp = nsp_mp + 1
print(nsp_mp)
which produces:
['2', '430', '3', '420', '4', '460', '5', '400', '7', '0', '1', '430', '6', '480']
7
which as per calculation above is correct. But when I try to execute a loop it does not return the correct results.
Maybe the approach is wrong. What is the correct way to do this?
The elasticsearch json object is:
arr = {'took': 0, 'timed_out': False, '_shards': {'total': 5, 'successful': 5, 'skipped': 0, 'failed': 0}, 'hits': {'total': 7, 'max_score': 1.0, 'hits': [{'_index': 'match', '_type': 'score', '_id': 'L_L122cBjpp4O0gQG0qd', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '2', 'ewp': '9', 'contract': '3NT', 'by': 'S', 'tricks': '10', 'nsscore': '430', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:32.896151'}}, {'_index': 'match', '_type': 'score', '_id': 'MPL122cBjpp4O0gQHEog', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '3', 'ewp': '10', 'contract': '4S', 'by': 'N', 'tricks': '10', 'nsscore': '420', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:33.027631'}}, {'_index': 'match', '_type': 'score', '_id': 'MfL122cBjpp4O0gQHEqk', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '4', 'ewp': '11', 'contract': '3NT', 'by': 'N', 'tricks': '11', 'nsscore': '460', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:33.158060'}}, {'_index': 'match', '_type': 'score', '_id': 'MvL122cBjpp4O0gQHUoj', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '5', 'ewp': '12', 'contract': '3NT', 'by': 'S', 'tricks': '10', 'nsscore': '400', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:33.285460'}}, {'_index': 'match', '_type': 'score', '_id': 'NPL122cBjpp4O0gQHkof', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '7', 'ewp': '14', 'contract': '3NT', 'by': 'S', 'tricks': '8', 'nsscore': '0', 'ewscore': '50', 'timestamp': '2018-12-23T16:45:33.538710'}}, {'_index': 'match', '_type': 'score', '_id': 'LvL122cBjpp4O0gQGkqt', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '1', 'ewp': '8', 'contract': '3NT', 'by': 'N', 'tricks': '10', 'nsscore': '430', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:32.405998'}}, {'_index': 'match', '_type': 'score', '_id': 'M_L122cBjpp4O0gQHUqg', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '6', 'ewp': '13', 'contract': '4S', 'by': 'S', 'tricks': '11', 'nsscore': '480', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:33.411104'}}]}}
python
edited Jan 2 at 12:07
Tagc
5,11553071
asked Jan 2 at 11:55
user1903663user1903663
5241931
add a comment |
I am attempting to create a contract bridge match point scoring system. In the list below the 1st, 3rd, etc. numbers are the pair numbers (players) and the 2nd, 4th etc. numbers are the scores achieved by each pair. So pair 2 scored 430, pair 3 scored 420 and so on.
I want to loop through the list and score as follows:
for each pair score that pair 2 beats they receive 2 points, for each they tie 1 point and where they don't beat they get 0 points. The loop then continues and compares each pair's score in the same way. In the example below, pair 2 gets 7 points (beating 3 other pairs and a tie with 1), pair 7 gets 0 points, pair 6 gets 12 points beating every other pair.
My list (generated from an elasticsearch json object) is:
['2', '430', '3', '420', '4', '460', '5', '400', '7', '0', '1', '430', '6', '480']
The python code I have tried (after multiple variations) is:
nsp_mp = 0
ewp_mp = 0
ns_list =
for row in arr["hits"]["hits"]:
nsp = row["_source"]["nsp"]
nsscore = row["_source"]["nsscore"]
ns_list.append(nsp)
ns_list.append(nsscore)
print(ns_list)
x = ns_list[1]
for i in range(6): #number of competing pairs
if x > ns_list[1::2][i]:
nsp_mp = nsp_mp + 2
elif x == ns_list[1::2][i]:
nsp_mp = nsp_mp
else:
nsp_mp = nsp_mp + 1
print(nsp_mp)
which produces:
['2', '430', '3', '420', '4', '460', '5', '400', '7', '0', '1', '430', '6', '480']
7
which as per calculation above is correct. But when I try to execute a loop it does not return the correct results.
Maybe the approach is wrong. What is the correct way to do this?
The elasticsearch json object is:
arr = {'took': 0, 'timed_out': False, '_shards': {'total': 5, 'successful': 5, 'skipped': 0, 'failed': 0}, 'hits': {'total': 7, 'max_score': 1.0, 'hits': [{'_index': 'match', '_type': 'score', '_id': 'L_L122cBjpp4O0gQG0qd', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '2', 'ewp': '9', 'contract': '3NT', 'by': 'S', 'tricks': '10', 'nsscore': '430', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:32.896151'}}, {'_index': 'match', '_type': 'score', '_id': 'MPL122cBjpp4O0gQHEog', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '3', 'ewp': '10', 'contract': '4S', 'by': 'N', 'tricks': '10', 'nsscore': '420', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:33.027631'}}, {'_index': 'match', '_type': 'score', '_id': 'MfL122cBjpp4O0gQHEqk', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '4', 'ewp': '11', 'contract': '3NT', 'by': 'N', 'tricks': '11', 'nsscore': '460', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:33.158060'}}, {'_index': 'match', '_type': 'score', '_id': 'MvL122cBjpp4O0gQHUoj', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '5', 'ewp': '12', 'contract': '3NT', 'by': 'S', 'tricks': '10', 'nsscore': '400', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:33.285460'}}, {'_index': 'match', '_type': 'score', '_id': 'NPL122cBjpp4O0gQHkof', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '7', 'ewp': '14', 'contract': '3NT', 'by': 'S', 'tricks': '8', 'nsscore': '0', 'ewscore': '50', 'timestamp': '2018-12-23T16:45:33.538710'}}, {'_index': 'match', '_type': 'score', '_id': 'LvL122cBjpp4O0gQGkqt', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '1', 'ewp': '8', 'contract': '3NT', 'by': 'N', 'tricks': '10', 'nsscore': '430', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:32.405998'}}, {'_index': 'match', '_type': 'score', '_id': 'M_L122cBjpp4O0gQHUqg', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '6', 'ewp': '13', 'contract': '4S', 'by': 'S', 'tricks': '11', 'nsscore': '480', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:33.411104'}}]}}
python
edited Jan 2 at 12:07
Tagc
5,11553071
asked Jan 2 at 11:55
user1903663user1903663
5241931
add a comment |
I am attempting to create a contract bridge match point scoring system. In the list below the 1st, 3rd, etc. numbers are the pair numbers (players) and the 2nd, 4th etc. numbers are the scores achieved by each pair. So pair 2 scored 430, pair 3 scored 420 and so on.
I want to loop through the list and score as follows:
for each pair score that pair 2 beats they receive 2 points, for each they tie 1 point and where they don't beat they get 0 points. The loop then continues and compares each pair's score in the same way. In the example below, pair 2 gets 7 points (beating 3 other pairs and a tie with 1), pair 7 gets 0 points, pair 6 gets 12 points beating every other pair.
My list (generated from an elasticsearch json object) is:
['2', '430', '3', '420', '4', '460', '5', '400', '7', '0', '1', '430', '6', '480']
The python code I have tried (after multiple variations) is:
nsp_mp = 0
ewp_mp = 0
ns_list =
for row in arr["hits"]["hits"]:
nsp = row["_source"]["nsp"]
nsscore = row["_source"]["nsscore"]
ns_list.append(nsp)
ns_list.append(nsscore)
print(ns_list)
x = ns_list[1]
for i in range(6): #number of competing pairs
if x > ns_list[1::2][i]:
nsp_mp = nsp_mp + 2
elif x == ns_list[1::2][i]:
nsp_mp = nsp_mp
else:
nsp_mp = nsp_mp + 1
print(nsp_mp)
which produces:
['2', '430', '3', '420', '4', '460', '5', '400', '7', '0', '1', '430', '6', '480']
7
which as per calculation above is correct. But when I try to execute a loop it does not return the correct results.
Maybe the approach is wrong. What is the correct way to do this?
The elasticsearch json object is:
arr = {'took': 0, 'timed_out': False, '_shards': {'total': 5, 'successful': 5, 'skipped': 0, 'failed': 0}, 'hits': {'total': 7, 'max_score': 1.0, 'hits': [{'_index': 'match', '_type': 'score', '_id': 'L_L122cBjpp4O0gQG0qd', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '2', 'ewp': '9', 'contract': '3NT', 'by': 'S', 'tricks': '10', 'nsscore': '430', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:32.896151'}}, {'_index': 'match', '_type': 'score', '_id': 'MPL122cBjpp4O0gQHEog', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '3', 'ewp': '10', 'contract': '4S', 'by': 'N', 'tricks': '10', 'nsscore': '420', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:33.027631'}}, {'_index': 'match', '_type': 'score', '_id': 'MfL122cBjpp4O0gQHEqk', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '4', 'ewp': '11', 'contract': '3NT', 'by': 'N', 'tricks': '11', 'nsscore': '460', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:33.158060'}}, {'_index': 'match', '_type': 'score', '_id': 'MvL122cBjpp4O0gQHUoj', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '5', 'ewp': '12', 'contract': '3NT', 'by': 'S', 'tricks': '10', 'nsscore': '400', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:33.285460'}}, {'_index': 'match', '_type': 'score', '_id': 'NPL122cBjpp4O0gQHkof', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '7', 'ewp': '14', 'contract': '3NT', 'by': 'S', 'tricks': '8', 'nsscore': '0', 'ewscore': '50', 'timestamp': '2018-12-23T16:45:33.538710'}}, {'_index': 'match', '_type': 'score', '_id': 'LvL122cBjpp4O0gQGkqt', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '1', 'ewp': '8', 'contract': '3NT', 'by': 'N', 'tricks': '10', 'nsscore': '430', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:32.405998'}}, {'_index': 'match', '_type': 'score', '_id': 'M_L122cBjpp4O0gQHUqg', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '6', 'ewp': '13', 'contract': '4S', 'by': 'S', 'tricks': '11', 'nsscore': '480', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:33.411104'}}]}}
python
edited Jan 2 at 12:07
Tagc
5,11553071
asked Jan 2 at 11:55
user1903663user1903663
5241931
I am attempting to create a contract bridge match point scoring system. In the list below the 1st, 3rd, etc. numbers are the pair numbers (players) and the 2nd, 4th etc. numbers are the scores achieved by each pair. So pair 2 scored 430, pair 3 scored 420 and so on.
I want to loop through the list and score as follows:
for each pair score that pair 2 beats they receive 2 points, for each they tie 1 point and where they don't beat they get 0 points. The loop then continues and compares each pair's score in the same way. In the example below, pair 2 gets 7 points (beating 3 other pairs and a tie with 1), pair 7 gets 0 points, pair 6 gets 12 points beating every other pair.
My list (generated from an elasticsearch json object) is:
['2', '430', '3', '420', '4', '460', '5', '400', '7', '0', '1', '430', '6', '480']
The python code I have tried (after multiple variations) is:
nsp_mp = 0
ewp_mp = 0
ns_list =
for row in arr["hits"]["hits"]:
nsp = row["_source"]["nsp"]
nsscore = row["_source"]["nsscore"]
ns_list.append(nsp)
ns_list.append(nsscore)
print(ns_list)
x = ns_list[1]
for i in range(6): #number of competing pairs
if x > ns_list[1::2][i]:
nsp_mp = nsp_mp + 2
elif x == ns_list[1::2][i]:
nsp_mp = nsp_mp
else:
nsp_mp = nsp_mp + 1
print(nsp_mp)
which produces:
['2', '430', '3', '420', '4', '460', '5', '400', '7', '0', '1', '430', '6', '480']
7
which as per calculation above is correct. But when I try to execute a loop it does not return the correct results.
Maybe the approach is wrong. What is the correct way to do this?
The elasticsearch json object is:
arr = {'took': 0, 'timed_out': False, '_shards': {'total': 5, 'successful': 5, 'skipped': 0, 'failed': 0}, 'hits': {'total': 7, 'max_score': 1.0, 'hits': [{'_index': 'match', '_type': 'score', '_id': 'L_L122cBjpp4O0gQG0qd', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '2', 'ewp': '9', 'contract': '3NT', 'by': 'S', 'tricks': '10', 'nsscore': '430', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:32.896151'}}, {'_index': 'match', '_type': 'score', '_id': 'MPL122cBjpp4O0gQHEog', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '3', 'ewp': '10', 'contract': '4S', 'by': 'N', 'tricks': '10', 'nsscore': '420', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:33.027631'}}, {'_index': 'match', '_type': 'score', '_id': 'MfL122cBjpp4O0gQHEqk', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '4', 'ewp': '11', 'contract': '3NT', 'by': 'N', 'tricks': '11', 'nsscore': '460', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:33.158060'}}, {'_index': 'match', '_type': 'score', '_id': 'MvL122cBjpp4O0gQHUoj', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '5', 'ewp': '12', 'contract': '3NT', 'by': 'S', 'tricks': '10', 'nsscore': '400', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:33.285460'}}, {'_index': 'match', '_type': 'score', '_id': 'NPL122cBjpp4O0gQHkof', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '7', 'ewp': '14', 'contract': '3NT', 'by': 'S', 'tricks': '8', 'nsscore': '0', 'ewscore': '50', 'timestamp': '2018-12-23T16:45:33.538710'}}, {'_index': 'match', '_type': 'score', '_id': 'LvL122cBjpp4O0gQGkqt', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '1', 'ewp': '8', 'contract': '3NT', 'by': 'N', 'tricks': '10', 'nsscore': '430', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:32.405998'}}, {'_index': 'match', '_type': 'score', '_id': 'M_L122cBjpp4O0gQHUqg', '_score': 1.0, '_source': {'tournament_id': 1, 'board_number': '1', 'nsp': '6', 'ewp': '13', 'contract': '4S', 'by': 'S', 'tricks': '11', 'nsscore': '480', 'ewscore': '0', 'timestamp': '2018-12-23T16:45:33.411104'}}]}}
python
python
edited Jan 2 at 12:07
Tagc
5,11553071
asked Jan 2 at 11:55
user1903663user1903663
5241931
edited Jan 2 at 12:07
Tagc
5,11553071
asked Jan 2 at 11:55
user1903663user1903663
5241931
edited Jan 2 at 12:07
Tagc
5,11553071
edited Jan 2 at 12:07
Tagc
5,11553071
edited Jan 2 at 12:07
Tagc
5,11553071
5,11553071
asked Jan 2 at 11:55
user1903663user1903663
5241931
asked Jan 2 at 11:55
user1903663user1903663
5241931
asked Jan 2 at 11:55
user1903663user1903663
5241931
5241931
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
List appears to be a poor data structure for this, I think you are making everything worse by flattening your elasticsearch object.
Note there are a few minor mistakes in listings below - to make sure
I'm not solving someone's homework for free. I also realize this is
not the most efficient way of doing so.
Try with dicts:
1) convert elasticsearch json you have to a dict with a better structure:
scores = {}
for row in arr["hits"]["hits"]:
nsp = row["_source"]["nsp"]
nsscore = row["_source"]["nsscore"]
scores[nsp] = nsscore
This will give you something like this:
{'1': '430',
'2': '430',
'3': '420',
'4': '460',
'5': '400',
'6': '480',
'7': '0'}
2) write a function to calculate pair score:
def calculate_score(pair, scores):
score = 0
for p in scores:
if p == pair:
continue
if scores[p] < scores[pair]:
score += 2 # win
elif scores[p] == scores[pair]:
score += 1
return score
This should give you something like this:
In [13]: calculate_score('1', scores)
Out[13]: 7
In [14]: calculate_score('7', scores)
Out[14]: 0
3) loop over all pairs, calculating scores. I'll leave this as exercise.
answered Jan 2 at 12:17
rvsrvs
952821
awesome, thank you! It works perfectly.
– user1903663
Jan 2 at 13:13
the complete picture is: board_num = arr["hits"]["total"] top = (board_num - 1) * 2 result_score = {} for row in arr["hits"]["hits"]: nsp = row["_source"]["nsp"] ewp = row["_source"]["ewp"] res = calculate_score(nsp, scores) ew_mp_score = top - res result_score.update({'nsp':nsp, 'ns_mp_score': res, 'ewp': ewp, 'ew_mp_score': ew_mp_score}) print(result_score)
– user1903663
Jan 2 at 13:39
add a comment |
The main problem with your code is, that the loop is one short, you have 7 entries. Then you should convert the numbers to int, so that the comparison is correct. In your code, you get for ties 0 points.
Instead of having a list, with flattend pairs, you should use tuple pairs.
ns_list =
for row in arr["hits"]["hits"]:
nsp = int(row["_source"]["nsp"])
nsscore = int(row["_source"]["nsscore"])
ns_list.append((nsp, nsscore))
print(ns_list)
x = ns_list[0][1]
nsp_mp = 0
for nsp, nsscore in ns_list:
if x > nsscore:
nsp_mp += 2
elif x == nsscore:
nsp_mp += 1
print(nsp_mp)
answered Jan 2 at 12:16
DanielDaniel
32.6k42961
thank you very much, this works too.
– user1903663
Jan 2 at 13:41
the reason it is /was 7 is because one of the scores belongs to each pair so the comparison should be with the other 6.
– user1903663
Jan 2 at 13:57
add a comment |
So we can do it like so:
import itertools
d = [(i['_source']['nsp'], i['_source']['nsscore']) for i in arr['hits']['hits']]
d
[('2', '430'),
('3', '420'),
('4', '460'),
('5', '400'),
('7', '0'),
('1', '430'),
('6', '480')]
c = itertools.combinations(d, 2)
counts = {}
for tup in c:
p1, p2 = tup
if not counts.get(p1[0]):
counts[p1[0]] = 0
if int(p1[1]) > int(p2[1]):
counts[p1[0]] += 1
counts
{'2': 3, '3': 2, '4': 3, '5': 1, '7': 0, '1': 0}
answered Jan 2 at 12:20
aws_apprenticeaws_apprentice
3,6681723
I need to study how this can be applied, thank you. I will comment again once I have experimented with it.
– user1903663
Jan 2 at 13:14
If you use combinations (which is a good idea), you also need to testint(p1[1]) < int(p2[1])to get the correct results (e.g. team 1 has two wins, and the most winning team 6 is not even in your result).
– myrmica
Jan 2 at 14:09
I am unsure what the counts dictionary elements are.
– user1903663
Jan 2 at 16:41
add a comment |
I first convert the list of your score to a dictionary object using itertools, then iterating through each key, and for each key, compare the values available in the list
and add accordingly the score you provided and since in this approach you will always add the value 1 because you will always compare it with itself so at end i decrease 1 from the final score there may be a better approach for this
ls = ['2', '430', '3', '420', '4', '460', '5', '400', '7', '0', '1', '430', '6', '480']
d = dict(itertools.zip_longest(*[iter(ls)] * 2, fillvalue=""))
values= d.values()
for item in d.keys():
score=0
for i in values:
if d[item]>i:
score+=2
elif d[item]==i:
score+=1
else:
pass
print(item,":",score-1)
Output:
2 : 7
3 : 4
4 : 10
5 : 2
7 : 0
1 : 7
6 : 12
answered Jan 2 at 12:42
GauravGaurav
180110
1
also wonderful, thank you. So many cool answers to this problem which I have been struggling with for days. Thank you.
– user1903663
Jan 2 at 14:20
1
The zipping part is clever, took me a moment to understand. Now I see it's in the itertools recipes asgrouper. A simple builtinzipwould work as well in this case.
– myrmica
Jan 2 at 14:59
@myrmica Yes, actually i also don't know this before, but while searching for the solution how to convert list into dict i found that response. and learned new thing with it
– Gaurav
Jan 2 at 16:15
add a comment |
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List appears to be a poor data structure for this, I think you are making everything worse by flattening your elasticsearch object.
Note there are a few minor mistakes in listings below - to make sure
I'm not solving someone's homework for free. I also realize this is
not the most efficient way of doing so.
Try with dicts:
1) convert elasticsearch json you have to a dict with a better structure:
scores = {}
for row in arr["hits"]["hits"]:
nsp = row["_source"]["nsp"]
nsscore = row["_source"]["nsscore"]
scores[nsp] = nsscore
This will give you something like this:
{'1': '430',
'2': '430',
'3': '420',
'4': '460',
'5': '400',
'6': '480',
'7': '0'}
2) write a function to calculate pair score:
def calculate_score(pair, scores):
score = 0
for p in scores:
if p == pair:
continue
if scores[p] < scores[pair]:
score += 2 # win
elif scores[p] == scores[pair]:
score += 1
return score
This should give you something like this:
In [13]: calculate_score('1', scores)
Out[13]: 7
In [14]: calculate_score('7', scores)
Out[14]: 0
3) loop over all pairs, calculating scores. I'll leave this as exercise.
answered Jan 2 at 12:17
rvsrvs
952821
awesome, thank you! It works perfectly.
– user1903663
Jan 2 at 13:13
the complete picture is: board_num = arr["hits"]["total"] top = (board_num - 1) * 2 result_score = {} for row in arr["hits"]["hits"]: nsp = row["_source"]["nsp"] ewp = row["_source"]["ewp"] res = calculate_score(nsp, scores) ew_mp_score = top - res result_score.update({'nsp':nsp, 'ns_mp_score': res, 'ewp': ewp, 'ew_mp_score': ew_mp_score}) print(result_score)
– user1903663
Jan 2 at 13:39
add a comment |
List appears to be a poor data structure for this, I think you are making everything worse by flattening your elasticsearch object.
Note there are a few minor mistakes in listings below - to make sure
I'm not solving someone's homework for free. I also realize this is
not the most efficient way of doing so.
Try with dicts:
1) convert elasticsearch json you have to a dict with a better structure:
scores = {}
for row in arr["hits"]["hits"]:
nsp = row["_source"]["nsp"]
nsscore = row["_source"]["nsscore"]
scores[nsp] = nsscore
This will give you something like this:
{'1': '430',
'2': '430',
'3': '420',
'4': '460',
'5': '400',
'6': '480',
'7': '0'}
2) write a function to calculate pair score:
def calculate_score(pair, scores):
score = 0
for p in scores:
if p == pair:
continue
if scores[p] < scores[pair]:
score += 2 # win
elif scores[p] == scores[pair]:
score += 1
return score
This should give you something like this:
In [13]: calculate_score('1', scores)
Out[13]: 7
In [14]: calculate_score('7', scores)
Out[14]: 0
3) loop over all pairs, calculating scores. I'll leave this as exercise.
answered Jan 2 at 12:17
rvsrvs
952821
awesome, thank you! It works perfectly.
– user1903663
Jan 2 at 13:13
the complete picture is: board_num = arr["hits"]["total"] top = (board_num - 1) * 2 result_score = {} for row in arr["hits"]["hits"]: nsp = row["_source"]["nsp"] ewp = row["_source"]["ewp"] res = calculate_score(nsp, scores) ew_mp_score = top - res result_score.update({'nsp':nsp, 'ns_mp_score': res, 'ewp': ewp, 'ew_mp_score': ew_mp_score}) print(result_score)
– user1903663
Jan 2 at 13:39
add a comment |
List appears to be a poor data structure for this, I think you are making everything worse by flattening your elasticsearch object.
Note there are a few minor mistakes in listings below - to make sure
I'm not solving someone's homework for free. I also realize this is
not the most efficient way of doing so.
Try with dicts:
1) convert elasticsearch json you have to a dict with a better structure:
scores = {}
for row in arr["hits"]["hits"]:
nsp = row["_source"]["nsp"]
nsscore = row["_source"]["nsscore"]
scores[nsp] = nsscore
This will give you something like this:
{'1': '430',
'2': '430',
'3': '420',
'4': '460',
'5': '400',
'6': '480',
'7': '0'}
2) write a function to calculate pair score:
def calculate_score(pair, scores):
score = 0
for p in scores:
if p == pair:
continue
if scores[p] < scores[pair]:
score += 2 # win
elif scores[p] == scores[pair]:
score += 1
return score
This should give you something like this:
In [13]: calculate_score('1', scores)
Out[13]: 7
In [14]: calculate_score('7', scores)
Out[14]: 0
3) loop over all pairs, calculating scores. I'll leave this as exercise.
answered Jan 2 at 12:17
rvsrvs
952821
List appears to be a poor data structure for this, I think you are making everything worse by flattening your elasticsearch object.
Note there are a few minor mistakes in listings below - to make sure
I'm not solving someone's homework for free. I also realize this is
not the most efficient way of doing so.
Try with dicts:
1) convert elasticsearch json you have to a dict with a better structure:
scores = {}
for row in arr["hits"]["hits"]:
nsp = row["_source"]["nsp"]
nsscore = row["_source"]["nsscore"]
scores[nsp] = nsscore
This will give you something like this:
{'1': '430',
'2': '430',
'3': '420',
'4': '460',
'5': '400',
'6': '480',
'7': '0'}
2) write a function to calculate pair score:
def calculate_score(pair, scores):
score = 0
for p in scores:
if p == pair:
continue
if scores[p] < scores[pair]:
score += 2 # win
elif scores[p] == scores[pair]:
score += 1
return score
This should give you something like this:
In [13]: calculate_score('1', scores)
Out[13]: 7
In [14]: calculate_score('7', scores)
Out[14]: 0
3) loop over all pairs, calculating scores. I'll leave this as exercise.
answered Jan 2 at 12:17
rvsrvs
952821
answered Jan 2 at 12:17
rvsrvs
952821
answered Jan 2 at 12:17
rvsrvs
952821
answered Jan 2 at 12:17
rvsrvs
952821
952821
awesome, thank you! It works perfectly.
– user1903663
Jan 2 at 13:13
the complete picture is: board_num = arr["hits"]["total"] top = (board_num - 1) * 2 result_score = {} for row in arr["hits"]["hits"]: nsp = row["_source"]["nsp"] ewp = row["_source"]["ewp"] res = calculate_score(nsp, scores) ew_mp_score = top - res result_score.update({'nsp':nsp, 'ns_mp_score': res, 'ewp': ewp, 'ew_mp_score': ew_mp_score}) print(result_score)
– user1903663
Jan 2 at 13:39
add a comment |
awesome, thank you! It works perfectly.
– user1903663
Jan 2 at 13:13
the complete picture is: board_num = arr["hits"]["total"] top = (board_num - 1) * 2 result_score = {} for row in arr["hits"]["hits"]: nsp = row["_source"]["nsp"] ewp = row["_source"]["ewp"] res = calculate_score(nsp, scores) ew_mp_score = top - res result_score.update({'nsp':nsp, 'ns_mp_score': res, 'ewp': ewp, 'ew_mp_score': ew_mp_score}) print(result_score)
– user1903663
Jan 2 at 13:39
awesome, thank you! It works perfectly.
– user1903663
Jan 2 at 13:13
awesome, thank you! It works perfectly.
– user1903663
Jan 2 at 13:13
the complete picture is: board_num = arr["hits"]["total"] top = (board_num - 1) * 2 result_score = {} for row in arr["hits"]["hits"]: nsp = row["_source"]["nsp"] ewp = row["_source"]["ewp"] res = calculate_score(nsp, scores) ew_mp_score = top - res result_score.update({'nsp':nsp, 'ns_mp_score': res, 'ewp': ewp, 'ew_mp_score': ew_mp_score}) print(result_score)
– user1903663
Jan 2 at 13:39
the complete picture is: board_num = arr["hits"]["total"] top = (board_num - 1) * 2 result_score = {} for row in arr["hits"]["hits"]: nsp = row["_source"]["nsp"] ewp = row["_source"]["ewp"] res = calculate_score(nsp, scores) ew_mp_score = top - res result_score.update({'nsp':nsp, 'ns_mp_score': res, 'ewp': ewp, 'ew_mp_score': ew_mp_score}) print(result_score)
– user1903663
Jan 2 at 13:39
add a comment |
The main problem with your code is, that the loop is one short, you have 7 entries. Then you should convert the numbers to int, so that the comparison is correct. In your code, you get for ties 0 points.
Instead of having a list, with flattend pairs, you should use tuple pairs.
ns_list =
for row in arr["hits"]["hits"]:
nsp = int(row["_source"]["nsp"])
nsscore = int(row["_source"]["nsscore"])
ns_list.append((nsp, nsscore))
print(ns_list)
x = ns_list[0][1]
nsp_mp = 0
for nsp, nsscore in ns_list:
if x > nsscore:
nsp_mp += 2
elif x == nsscore:
nsp_mp += 1
print(nsp_mp)
answered Jan 2 at 12:16
DanielDaniel
32.6k42961
thank you very much, this works too.
– user1903663
Jan 2 at 13:41
the reason it is /was 7 is because one of the scores belongs to each pair so the comparison should be with the other 6.
– user1903663
Jan 2 at 13:57
add a comment |
The main problem with your code is, that the loop is one short, you have 7 entries. Then you should convert the numbers to int, so that the comparison is correct. In your code, you get for ties 0 points.
Instead of having a list, with flattend pairs, you should use tuple pairs.
ns_list =
for row in arr["hits"]["hits"]:
nsp = int(row["_source"]["nsp"])
nsscore = int(row["_source"]["nsscore"])
ns_list.append((nsp, nsscore))
print(ns_list)
x = ns_list[0][1]
nsp_mp = 0
for nsp, nsscore in ns_list:
if x > nsscore:
nsp_mp += 2
elif x == nsscore:
nsp_mp += 1
print(nsp_mp)
answered Jan 2 at 12:16
DanielDaniel
32.6k42961
thank you very much, this works too.
– user1903663
Jan 2 at 13:41
the reason it is /was 7 is because one of the scores belongs to each pair so the comparison should be with the other 6.
– user1903663
Jan 2 at 13:57
add a comment |
The main problem with your code is, that the loop is one short, you have 7 entries. Then you should convert the numbers to int, so that the comparison is correct. In your code, you get for ties 0 points.
Instead of having a list, with flattend pairs, you should use tuple pairs.
ns_list =
for row in arr["hits"]["hits"]:
nsp = int(row["_source"]["nsp"])
nsscore = int(row["_source"]["nsscore"])
ns_list.append((nsp, nsscore))
print(ns_list)
x = ns_list[0][1]
nsp_mp = 0
for nsp, nsscore in ns_list:
if x > nsscore:
nsp_mp += 2
elif x == nsscore:
nsp_mp += 1
print(nsp_mp)
answered Jan 2 at 12:16
DanielDaniel
32.6k42961
The main problem with your code is, that the loop is one short, you have 7 entries. Then you should convert the numbers to int, so that the comparison is correct. In your code, you get for ties 0 points.
Instead of having a list, with flattend pairs, you should use tuple pairs.
ns_list =
for row in arr["hits"]["hits"]:
nsp = int(row["_source"]["nsp"])
nsscore = int(row["_source"]["nsscore"])
ns_list.append((nsp, nsscore))
print(ns_list)
x = ns_list[0][1]
nsp_mp = 0
for nsp, nsscore in ns_list:
if x > nsscore:
nsp_mp += 2
elif x == nsscore:
nsp_mp += 1
print(nsp_mp)
answered Jan 2 at 12:16
DanielDaniel
32.6k42961
answered Jan 2 at 12:16
DanielDaniel
32.6k42961
answered Jan 2 at 12:16
DanielDaniel
32.6k42961
answered Jan 2 at 12:16
DanielDaniel
32.6k42961
32.6k42961
thank you very much, this works too.
– user1903663
Jan 2 at 13:41
the reason it is /was 7 is because one of the scores belongs to each pair so the comparison should be with the other 6.
– user1903663
Jan 2 at 13:57
add a comment |
thank you very much, this works too.
– user1903663
Jan 2 at 13:41
the reason it is /was 7 is because one of the scores belongs to each pair so the comparison should be with the other 6.
– user1903663
Jan 2 at 13:57
thank you very much, this works too.
– user1903663
Jan 2 at 13:41
thank you very much, this works too.
– user1903663
Jan 2 at 13:41
the reason it is /was 7 is because one of the scores belongs to each pair so the comparison should be with the other 6.
– user1903663
Jan 2 at 13:57
the reason it is /was 7 is because one of the scores belongs to each pair so the comparison should be with the other 6.
– user1903663
Jan 2 at 13:57
add a comment |
So we can do it like so:
import itertools
d = [(i['_source']['nsp'], i['_source']['nsscore']) for i in arr['hits']['hits']]
d
[('2', '430'),
('3', '420'),
('4', '460'),
('5', '400'),
('7', '0'),
('1', '430'),
('6', '480')]
c = itertools.combinations(d, 2)
counts = {}
for tup in c:
p1, p2 = tup
if not counts.get(p1[0]):
counts[p1[0]] = 0
if int(p1[1]) > int(p2[1]):
counts[p1[0]] += 1
counts
{'2': 3, '3': 2, '4': 3, '5': 1, '7': 0, '1': 0}
answered Jan 2 at 12:20
aws_apprenticeaws_apprentice
3,6681723
I need to study how this can be applied, thank you. I will comment again once I have experimented with it.
– user1903663
Jan 2 at 13:14
If you use combinations (which is a good idea), you also need to testint(p1[1]) < int(p2[1])to get the correct results (e.g. team 1 has two wins, and the most winning team 6 is not even in your result).
– myrmica
Jan 2 at 14:09
I am unsure what the counts dictionary elements are.
– user1903663
Jan 2 at 16:41
add a comment |
So we can do it like so:
import itertools
d = [(i['_source']['nsp'], i['_source']['nsscore']) for i in arr['hits']['hits']]
d
[('2', '430'),
('3', '420'),
('4', '460'),
('5', '400'),
('7', '0'),
('1', '430'),
('6', '480')]
c = itertools.combinations(d, 2)
counts = {}
for tup in c:
p1, p2 = tup
if not counts.get(p1[0]):
counts[p1[0]] = 0
if int(p1[1]) > int(p2[1]):
counts[p1[0]] += 1
counts
{'2': 3, '3': 2, '4': 3, '5': 1, '7': 0, '1': 0}
answered Jan 2 at 12:20
aws_apprenticeaws_apprentice
3,6681723
I need to study how this can be applied, thank you. I will comment again once I have experimented with it.
– user1903663
Jan 2 at 13:14
If you use combinations (which is a good idea), you also need to testint(p1[1]) < int(p2[1])to get the correct results (e.g. team 1 has two wins, and the most winning team 6 is not even in your result).
– myrmica
Jan 2 at 14:09
I am unsure what the counts dictionary elements are.
– user1903663
Jan 2 at 16:41
add a comment |
So we can do it like so:
import itertools
d = [(i['_source']['nsp'], i['_source']['nsscore']) for i in arr['hits']['hits']]
d
[('2', '430'),
('3', '420'),
('4', '460'),
('5', '400'),
('7', '0'),
('1', '430'),
('6', '480')]
c = itertools.combinations(d, 2)
counts = {}
for tup in c:
p1, p2 = tup
if not counts.get(p1[0]):
counts[p1[0]] = 0
if int(p1[1]) > int(p2[1]):
counts[p1[0]] += 1
counts
{'2': 3, '3': 2, '4': 3, '5': 1, '7': 0, '1': 0}
answered Jan 2 at 12:20
aws_apprenticeaws_apprentice
3,6681723
So we can do it like so:
import itertools
d = [(i['_source']['nsp'], i['_source']['nsscore']) for i in arr['hits']['hits']]
d
[('2', '430'),
('3', '420'),
('4', '460'),
('5', '400'),
('7', '0'),
('1', '430'),
('6', '480')]
c = itertools.combinations(d, 2)
counts = {}
for tup in c:
p1, p2 = tup
if not counts.get(p1[0]):
counts[p1[0]] = 0
if int(p1[1]) > int(p2[1]):
counts[p1[0]] += 1
counts
{'2': 3, '3': 2, '4': 3, '5': 1, '7': 0, '1': 0}
answered Jan 2 at 12:20
aws_apprenticeaws_apprentice
3,6681723
answered Jan 2 at 12:20
aws_apprenticeaws_apprentice
3,6681723
answered Jan 2 at 12:20
aws_apprenticeaws_apprentice
3,6681723
answered Jan 2 at 12:20
aws_apprenticeaws_apprentice
3,6681723
3,6681723
I need to study how this can be applied, thank you. I will comment again once I have experimented with it.
– user1903663
Jan 2 at 13:14
If you use combinations (which is a good idea), you also need to testint(p1[1]) < int(p2[1])to get the correct results (e.g. team 1 has two wins, and the most winning team 6 is not even in your result).
– myrmica
Jan 2 at 14:09
I am unsure what the counts dictionary elements are.
– user1903663
Jan 2 at 16:41
add a comment |
I need to study how this can be applied, thank you. I will comment again once I have experimented with it.
– user1903663
Jan 2 at 13:14
If you use combinations (which is a good idea), you also need to testint(p1[1]) < int(p2[1])to get the correct results (e.g. team 1 has two wins, and the most winning team 6 is not even in your result).
– myrmica
Jan 2 at 14:09
I am unsure what the counts dictionary elements are.
– user1903663
Jan 2 at 16:41
I need to study how this can be applied, thank you. I will comment again once I have experimented with it.
– user1903663
Jan 2 at 13:14
I need to study how this can be applied, thank you. I will comment again once I have experimented with it.
– user1903663
Jan 2 at 13:14
If you use combinations (which is a good idea), you also need to test
int(p1[1]) < int(p2[1]) to get the correct results (e.g. team 1 has two wins, and the most winning team 6 is not even in your result).– myrmica
Jan 2 at 14:09
If you use combinations (which is a good idea), you also need to test
int(p1[1]) < int(p2[1]) to get the correct results (e.g. team 1 has two wins, and the most winning team 6 is not even in your result).– myrmica
Jan 2 at 14:09
I am unsure what the counts dictionary elements are.
– user1903663
Jan 2 at 16:41
I am unsure what the counts dictionary elements are.
– user1903663
Jan 2 at 16:41
add a comment |
I first convert the list of your score to a dictionary object using itertools, then iterating through each key, and for each key, compare the values available in the list
and add accordingly the score you provided and since in this approach you will always add the value 1 because you will always compare it with itself so at end i decrease 1 from the final score there may be a better approach for this
ls = ['2', '430', '3', '420', '4', '460', '5', '400', '7', '0', '1', '430', '6', '480']
d = dict(itertools.zip_longest(*[iter(ls)] * 2, fillvalue=""))
values= d.values()
for item in d.keys():
score=0
for i in values:
if d[item]>i:
score+=2
elif d[item]==i:
score+=1
else:
pass
print(item,":",score-1)
Output:
2 : 7
3 : 4
4 : 10
5 : 2
7 : 0
1 : 7
6 : 12
answered Jan 2 at 12:42
GauravGaurav
180110
1
also wonderful, thank you. So many cool answers to this problem which I have been struggling with for days. Thank you.
– user1903663
Jan 2 at 14:20
1
The zipping part is clever, took me a moment to understand. Now I see it's in the itertools recipes asgrouper. A simple builtinzipwould work as well in this case.
– myrmica
Jan 2 at 14:59
@myrmica Yes, actually i also don't know this before, but while searching for the solution how to convert list into dict i found that response. and learned new thing with it
– Gaurav
Jan 2 at 16:15
add a comment |
I first convert the list of your score to a dictionary object using itertools, then iterating through each key, and for each key, compare the values available in the list
and add accordingly the score you provided and since in this approach you will always add the value 1 because you will always compare it with itself so at end i decrease 1 from the final score there may be a better approach for this
ls = ['2', '430', '3', '420', '4', '460', '5', '400', '7', '0', '1', '430', '6', '480']
d = dict(itertools.zip_longest(*[iter(ls)] * 2, fillvalue=""))
values= d.values()
for item in d.keys():
score=0
for i in values:
if d[item]>i:
score+=2
elif d[item]==i:
score+=1
else:
pass
print(item,":",score-1)
Output:
2 : 7
3 : 4
4 : 10
5 : 2
7 : 0
1 : 7
6 : 12
answered Jan 2 at 12:42
GauravGaurav
180110
1
also wonderful, thank you. So many cool answers to this problem which I have been struggling with for days. Thank you.
– user1903663
Jan 2 at 14:20
1
The zipping part is clever, took me a moment to understand. Now I see it's in the itertools recipes asgrouper. A simple builtinzipwould work as well in this case.
– myrmica
Jan 2 at 14:59
@myrmica Yes, actually i also don't know this before, but while searching for the solution how to convert list into dict i found that response. and learned new thing with it
– Gaurav
Jan 2 at 16:15
add a comment |
I first convert the list of your score to a dictionary object using itertools, then iterating through each key, and for each key, compare the values available in the list
and add accordingly the score you provided and since in this approach you will always add the value 1 because you will always compare it with itself so at end i decrease 1 from the final score there may be a better approach for this
ls = ['2', '430', '3', '420', '4', '460', '5', '400', '7', '0', '1', '430', '6', '480']
d = dict(itertools.zip_longest(*[iter(ls)] * 2, fillvalue=""))
values= d.values()
for item in d.keys():
score=0
for i in values:
if d[item]>i:
score+=2
elif d[item]==i:
score+=1
else:
pass
print(item,":",score-1)
Output:
2 : 7
3 : 4
4 : 10
5 : 2
7 : 0
1 : 7
6 : 12
answered Jan 2 at 12:42
GauravGaurav
180110
I first convert the list of your score to a dictionary object using itertools, then iterating through each key, and for each key, compare the values available in the list
and add accordingly the score you provided and since in this approach you will always add the value 1 because you will always compare it with itself so at end i decrease 1 from the final score there may be a better approach for this
ls = ['2', '430', '3', '420', '4', '460', '5', '400', '7', '0', '1', '430', '6', '480']
d = dict(itertools.zip_longest(*[iter(ls)] * 2, fillvalue=""))
values= d.values()
for item in d.keys():
score=0
for i in values:
if d[item]>i:
score+=2
elif d[item]==i:
score+=1
else:
pass
print(item,":",score-1)
Output:
2 : 7
3 : 4
4 : 10
5 : 2
7 : 0
1 : 7
6 : 12
answered Jan 2 at 12:42
GauravGaurav
180110
answered Jan 2 at 12:42
GauravGaurav
180110
answered Jan 2 at 12:42
GauravGaurav
180110
answered Jan 2 at 12:42
GauravGaurav
180110
180110
1
also wonderful, thank you. So many cool answers to this problem which I have been struggling with for days. Thank you.
– user1903663
Jan 2 at 14:20
1
The zipping part is clever, took me a moment to understand. Now I see it's in the itertools recipes asgrouper. A simple builtinzipwould work as well in this case.
– myrmica
Jan 2 at 14:59
@myrmica Yes, actually i also don't know this before, but while searching for the solution how to convert list into dict i found that response. and learned new thing with it
– Gaurav
Jan 2 at 16:15
add a comment |
1
also wonderful, thank you. So many cool answers to this problem which I have been struggling with for days. Thank you.
– user1903663
Jan 2 at 14:20
1
The zipping part is clever, took me a moment to understand. Now I see it's in the itertools recipes asgrouper. A simple builtinzipwould work as well in this case.
– myrmica
Jan 2 at 14:59
@myrmica Yes, actually i also don't know this before, but while searching for the solution how to convert list into dict i found that response. and learned new thing with it
– Gaurav
Jan 2 at 16:15
1
1
also wonderful, thank you. So many cool answers to this problem which I have been struggling with for days. Thank you.
– user1903663
Jan 2 at 14:20
also wonderful, thank you. So many cool answers to this problem which I have been struggling with for days. Thank you.
– user1903663
Jan 2 at 14:20
1
1
The zipping part is clever, took me a moment to understand. Now I see it's in the itertools recipes as
grouper. A simple builtin zip would work as well in this case.– myrmica
Jan 2 at 14:59
The zipping part is clever, took me a moment to understand. Now I see it's in the itertools recipes as
grouper. A simple builtin zip would work as well in this case.– myrmica
Jan 2 at 14:59
@myrmica Yes, actually i also don't know this before, but while searching for the solution how to convert list into dict i found that response. and learned new thing with it
– Gaurav
Jan 2 at 16:15
@myrmica Yes, actually i also don't know this before, but while searching for the solution how to convert list into dict i found that response. and learned new thing with it
– Gaurav
Jan 2 at 16:15
add a comment |
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