Mixing
Python: Symmetrical Difference Between List of Sets of Strings
I have a list that contains multiple sets of strings, and I would like to find the symmetric difference between each string and the other strings in the set.
For example, I have the following list:
targets = [{'B', 'C', 'A'}, {'E', 'C', 'D'}, {'F', 'E', 'D'}]
For the above, desired output is:
[2, 0, 1]
because in the first set, A and B are not found in any of the other sets, for the second set, there are no unique elements to the set, and for the third set, F is not found in any of the other sets.
I thought about approaching this backwards; finding the intersection of each set and subtracting the length of the intersection from the length of the list, but set.intersection(*) does not appear to work on strings, so I'm stuck:
set1 = {'A', 'B', 'C'}
set2 = {'C', 'D', 'E'}
set3 = {'D', 'E', 'F'}
targets = [set1, set2, set3]
>>> set.intersection(*targets)
set()
python-3.x set set-intersection symmetric-difference
asked Nov 20 '18 at 0:49
SummerElaSummerEla
782922
add a comment |
I have a list that contains multiple sets of strings, and I would like to find the symmetric difference between each string and the other strings in the set.
For example, I have the following list:
targets = [{'B', 'C', 'A'}, {'E', 'C', 'D'}, {'F', 'E', 'D'}]
For the above, desired output is:
[2, 0, 1]
because in the first set, A and B are not found in any of the other sets, for the second set, there are no unique elements to the set, and for the third set, F is not found in any of the other sets.
I thought about approaching this backwards; finding the intersection of each set and subtracting the length of the intersection from the length of the list, but set.intersection(*) does not appear to work on strings, so I'm stuck:
set1 = {'A', 'B', 'C'}
set2 = {'C', 'D', 'E'}
set3 = {'D', 'E', 'F'}
targets = [set1, set2, set3]
>>> set.intersection(*targets)
set()
python-3.x set set-intersection symmetric-difference
asked Nov 20 '18 at 0:49
SummerElaSummerEla
782922
Is each string a single character? If not, are you going by whether the entire string occurs elsewhere, or just the character by character differences?
– Unsolved Cypher
Nov 20 '18 at 1:00
1
Each string contains at least six characters, and I would like to test the entire occurrence of the string. Perhaps I oversimplified my example, but I wasn't sure that would make a difference.
– SummerEla
Nov 20 '18 at 1:08
add a comment |
I have a list that contains multiple sets of strings, and I would like to find the symmetric difference between each string and the other strings in the set.
For example, I have the following list:
targets = [{'B', 'C', 'A'}, {'E', 'C', 'D'}, {'F', 'E', 'D'}]
For the above, desired output is:
[2, 0, 1]
because in the first set, A and B are not found in any of the other sets, for the second set, there are no unique elements to the set, and for the third set, F is not found in any of the other sets.
I thought about approaching this backwards; finding the intersection of each set and subtracting the length of the intersection from the length of the list, but set.intersection(*) does not appear to work on strings, so I'm stuck:
set1 = {'A', 'B', 'C'}
set2 = {'C', 'D', 'E'}
set3 = {'D', 'E', 'F'}
targets = [set1, set2, set3]
>>> set.intersection(*targets)
set()
python-3.x set set-intersection symmetric-difference
asked Nov 20 '18 at 0:49
SummerElaSummerEla
782922
I have a list that contains multiple sets of strings, and I would like to find the symmetric difference between each string and the other strings in the set.
For example, I have the following list:
targets = [{'B', 'C', 'A'}, {'E', 'C', 'D'}, {'F', 'E', 'D'}]
For the above, desired output is:
[2, 0, 1]
because in the first set, A and B are not found in any of the other sets, for the second set, there are no unique elements to the set, and for the third set, F is not found in any of the other sets.
I thought about approaching this backwards; finding the intersection of each set and subtracting the length of the intersection from the length of the list, but set.intersection(*) does not appear to work on strings, so I'm stuck:
set1 = {'A', 'B', 'C'}
set2 = {'C', 'D', 'E'}
set3 = {'D', 'E', 'F'}
targets = [set1, set2, set3]
>>> set.intersection(*targets)
set()
python-3.x set set-intersection symmetric-difference
python-3.x set set-intersection symmetric-difference
asked Nov 20 '18 at 0:49
SummerElaSummerEla
782922
asked Nov 20 '18 at 0:49
SummerElaSummerEla
782922
asked Nov 20 '18 at 0:49
SummerElaSummerEla
782922
asked Nov 20 '18 at 0:49
SummerElaSummerEla
782922
asked Nov 20 '18 at 0:49
SummerElaSummerEla
782922
782922
Is each string a single character? If not, are you going by whether the entire string occurs elsewhere, or just the character by character differences?
– Unsolved Cypher
Nov 20 '18 at 1:00
1
Each string contains at least six characters, and I would like to test the entire occurrence of the string. Perhaps I oversimplified my example, but I wasn't sure that would make a difference.
– SummerEla
Nov 20 '18 at 1:08
add a comment |
Is each string a single character? If not, are you going by whether the entire string occurs elsewhere, or just the character by character differences?
– Unsolved Cypher
Nov 20 '18 at 1:00
1
Each string contains at least six characters, and I would like to test the entire occurrence of the string. Perhaps I oversimplified my example, but I wasn't sure that would make a difference.
– SummerEla
Nov 20 '18 at 1:08
Is each string a single character? If not, are you going by whether the entire string occurs elsewhere, or just the character by character differences?
– Unsolved Cypher
Nov 20 '18 at 1:00
Is each string a single character? If not, are you going by whether the entire string occurs elsewhere, or just the character by character differences?
– Unsolved Cypher
Nov 20 '18 at 1:00
1
1
Each string contains at least six characters, and I would like to test the entire occurrence of the string. Perhaps I oversimplified my example, but I wasn't sure that would make a difference.
– SummerEla
Nov 20 '18 at 1:08
Each string contains at least six characters, and I would like to test the entire occurrence of the string. Perhaps I oversimplified my example, but I wasn't sure that would make a difference.
– SummerEla
Nov 20 '18 at 1:08
add a comment |
3 Answers
3
active
oldest
votes
The issue you're having is that there are no strings shared by all three sets, so your intersection comes up empty. That's not a string issue, it would work the same with numbers or anything else you can put in a set.
The only way I see to do a global calculation over all the sets, then use that to find the number of unique values in each one is to first count all the values (using collections.Counter), then for each set, count the number of values that showed up only once in the global count.
from collections import Counter
def unique_count(sets):
count = Counter()
for s in sets:
count.update(s)
return [sum(count[x] == 1 for x in s) for s in sets]
answered Nov 20 '18 at 1:34
BlckknghtBlckknght
62.4k556100
I've found that this answer still works even when there is no difference between sets. @Unsolved Cypher's did not :(
– SummerEla
Dec 18 '18 at 20:56
add a comment |
Try something like below:
Get symmetric difference with every set. Then intersect with the given input set.
def symVal(index,targets):
bseSet = targets[index]
symSet = bseSet
for j in range(len(targets)):
if index != j:
symSet = symSet ^ targets[j]
print(len(symSet & bseSet))
for i in range(len(targets)):
symVal(i,targets)
answered Nov 20 '18 at 1:29
NaiduNaidu
4,9593817
I'm pretty sure this calculation will get the wrong answer if any values are in all three sets (or appear any odd number of times if there are more sets than that).
– Blckknght
Dec 18 '18 at 21:15
add a comment |
Your code example doesn't work because it's finding the intersection between all of the sets, which is 0 (since no element occurs everywhere). You want to find the difference between each set and the union of all other sets. For example:
set1 = {'A', 'B', 'C'}
set2 = {'C', 'D', 'E'}
set3 = {'D', 'E', 'F'}
targets = [set1, set2, set3]
result =
for set_element in targets:
result.append(len(set_element.difference(set.union(*[x for x in targets if x is not set_element]))))
print(result)
(note that the [x for x in targets if x != set_element] is just the set of all other sets)
answered Nov 20 '18 at 1:30
Unsolved CypherUnsolved Cypher
495314
1
This works great, and quickly. Thank you!
– SummerEla
Nov 20 '18 at 2:46
Unfortunately, I found an error in your solution. It returns an error when there is no difference between sets.
– SummerEla
Dec 18 '18 at 20:57
1
@SummerEla Good catch! I've updated the code, and now it works in that situation as well. I changed the!=to anis not, so that it's every list is checked against other lists that are different objects rather than just those that have different values (since lists can be different objects but look identical)
– Unsolved Cypher
Dec 18 '18 at 23:43
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
The issue you're having is that there are no strings shared by all three sets, so your intersection comes up empty. That's not a string issue, it would work the same with numbers or anything else you can put in a set.
The only way I see to do a global calculation over all the sets, then use that to find the number of unique values in each one is to first count all the values (using collections.Counter), then for each set, count the number of values that showed up only once in the global count.
from collections import Counter
def unique_count(sets):
count = Counter()
for s in sets:
count.update(s)
return [sum(count[x] == 1 for x in s) for s in sets]
answered Nov 20 '18 at 1:34
BlckknghtBlckknght
62.4k556100
I've found that this answer still works even when there is no difference between sets. @Unsolved Cypher's did not :(
– SummerEla
Dec 18 '18 at 20:56
add a comment |
The issue you're having is that there are no strings shared by all three sets, so your intersection comes up empty. That's not a string issue, it would work the same with numbers or anything else you can put in a set.
The only way I see to do a global calculation over all the sets, then use that to find the number of unique values in each one is to first count all the values (using collections.Counter), then for each set, count the number of values that showed up only once in the global count.
from collections import Counter
def unique_count(sets):
count = Counter()
for s in sets:
count.update(s)
return [sum(count[x] == 1 for x in s) for s in sets]
answered Nov 20 '18 at 1:34
BlckknghtBlckknght
62.4k556100
I've found that this answer still works even when there is no difference between sets. @Unsolved Cypher's did not :(
– SummerEla
Dec 18 '18 at 20:56
add a comment |
The issue you're having is that there are no strings shared by all three sets, so your intersection comes up empty. That's not a string issue, it would work the same with numbers or anything else you can put in a set.
The only way I see to do a global calculation over all the sets, then use that to find the number of unique values in each one is to first count all the values (using collections.Counter), then for each set, count the number of values that showed up only once in the global count.
from collections import Counter
def unique_count(sets):
count = Counter()
for s in sets:
count.update(s)
return [sum(count[x] == 1 for x in s) for s in sets]
answered Nov 20 '18 at 1:34
BlckknghtBlckknght
62.4k556100
The issue you're having is that there are no strings shared by all three sets, so your intersection comes up empty. That's not a string issue, it would work the same with numbers or anything else you can put in a set.
The only way I see to do a global calculation over all the sets, then use that to find the number of unique values in each one is to first count all the values (using collections.Counter), then for each set, count the number of values that showed up only once in the global count.
from collections import Counter
def unique_count(sets):
count = Counter()
for s in sets:
count.update(s)
return [sum(count[x] == 1 for x in s) for s in sets]
answered Nov 20 '18 at 1:34
BlckknghtBlckknght
62.4k556100
answered Nov 20 '18 at 1:34
BlckknghtBlckknght
62.4k556100
answered Nov 20 '18 at 1:34
BlckknghtBlckknght
62.4k556100
answered Nov 20 '18 at 1:34
BlckknghtBlckknght
62.4k556100
62.4k556100
I've found that this answer still works even when there is no difference between sets. @Unsolved Cypher's did not :(
– SummerEla
Dec 18 '18 at 20:56
add a comment |
I've found that this answer still works even when there is no difference between sets. @Unsolved Cypher's did not :(
– SummerEla
Dec 18 '18 at 20:56
I've found that this answer still works even when there is no difference between sets. @Unsolved Cypher's did not :(
– SummerEla
Dec 18 '18 at 20:56
I've found that this answer still works even when there is no difference between sets. @Unsolved Cypher's did not :(
– SummerEla
Dec 18 '18 at 20:56
add a comment |
Try something like below:
Get symmetric difference with every set. Then intersect with the given input set.
def symVal(index,targets):
bseSet = targets[index]
symSet = bseSet
for j in range(len(targets)):
if index != j:
symSet = symSet ^ targets[j]
print(len(symSet & bseSet))
for i in range(len(targets)):
symVal(i,targets)
answered Nov 20 '18 at 1:29
NaiduNaidu
4,9593817
I'm pretty sure this calculation will get the wrong answer if any values are in all three sets (or appear any odd number of times if there are more sets than that).
– Blckknght
Dec 18 '18 at 21:15
add a comment |
Try something like below:
Get symmetric difference with every set. Then intersect with the given input set.
def symVal(index,targets):
bseSet = targets[index]
symSet = bseSet
for j in range(len(targets)):
if index != j:
symSet = symSet ^ targets[j]
print(len(symSet & bseSet))
for i in range(len(targets)):
symVal(i,targets)
answered Nov 20 '18 at 1:29
NaiduNaidu
4,9593817
I'm pretty sure this calculation will get the wrong answer if any values are in all three sets (or appear any odd number of times if there are more sets than that).
– Blckknght
Dec 18 '18 at 21:15
add a comment |
Try something like below:
Get symmetric difference with every set. Then intersect with the given input set.
def symVal(index,targets):
bseSet = targets[index]
symSet = bseSet
for j in range(len(targets)):
if index != j:
symSet = symSet ^ targets[j]
print(len(symSet & bseSet))
for i in range(len(targets)):
symVal(i,targets)
answered Nov 20 '18 at 1:29
NaiduNaidu
4,9593817
Try something like below:
Get symmetric difference with every set. Then intersect with the given input set.
def symVal(index,targets):
bseSet = targets[index]
symSet = bseSet
for j in range(len(targets)):
if index != j:
symSet = symSet ^ targets[j]
print(len(symSet & bseSet))
for i in range(len(targets)):
symVal(i,targets)
answered Nov 20 '18 at 1:29
NaiduNaidu
4,9593817
answered Nov 20 '18 at 1:29
NaiduNaidu
4,9593817
answered Nov 20 '18 at 1:29
NaiduNaidu
4,9593817
answered Nov 20 '18 at 1:29
NaiduNaidu
4,9593817
4,9593817
I'm pretty sure this calculation will get the wrong answer if any values are in all three sets (or appear any odd number of times if there are more sets than that).
– Blckknght
Dec 18 '18 at 21:15
add a comment |
I'm pretty sure this calculation will get the wrong answer if any values are in all three sets (or appear any odd number of times if there are more sets than that).
– Blckknght
Dec 18 '18 at 21:15
I'm pretty sure this calculation will get the wrong answer if any values are in all three sets (or appear any odd number of times if there are more sets than that).
– Blckknght
Dec 18 '18 at 21:15
I'm pretty sure this calculation will get the wrong answer if any values are in all three sets (or appear any odd number of times if there are more sets than that).
– Blckknght
Dec 18 '18 at 21:15
add a comment |
Your code example doesn't work because it's finding the intersection between all of the sets, which is 0 (since no element occurs everywhere). You want to find the difference between each set and the union of all other sets. For example:
set1 = {'A', 'B', 'C'}
set2 = {'C', 'D', 'E'}
set3 = {'D', 'E', 'F'}
targets = [set1, set2, set3]
result =
for set_element in targets:
result.append(len(set_element.difference(set.union(*[x for x in targets if x is not set_element]))))
print(result)
(note that the [x for x in targets if x != set_element] is just the set of all other sets)
answered Nov 20 '18 at 1:30
Unsolved CypherUnsolved Cypher
495314
1
This works great, and quickly. Thank you!
– SummerEla
Nov 20 '18 at 2:46
Unfortunately, I found an error in your solution. It returns an error when there is no difference between sets.
– SummerEla
Dec 18 '18 at 20:57
1
@SummerEla Good catch! I've updated the code, and now it works in that situation as well. I changed the!=to anis not, so that it's every list is checked against other lists that are different objects rather than just those that have different values (since lists can be different objects but look identical)
– Unsolved Cypher
Dec 18 '18 at 23:43
add a comment |
Your code example doesn't work because it's finding the intersection between all of the sets, which is 0 (since no element occurs everywhere). You want to find the difference between each set and the union of all other sets. For example:
set1 = {'A', 'B', 'C'}
set2 = {'C', 'D', 'E'}
set3 = {'D', 'E', 'F'}
targets = [set1, set2, set3]
result =
for set_element in targets:
result.append(len(set_element.difference(set.union(*[x for x in targets if x is not set_element]))))
print(result)
(note that the [x for x in targets if x != set_element] is just the set of all other sets)
answered Nov 20 '18 at 1:30
Unsolved CypherUnsolved Cypher
495314
1
This works great, and quickly. Thank you!
– SummerEla
Nov 20 '18 at 2:46
Unfortunately, I found an error in your solution. It returns an error when there is no difference between sets.
– SummerEla
Dec 18 '18 at 20:57
1
@SummerEla Good catch! I've updated the code, and now it works in that situation as well. I changed the!=to anis not, so that it's every list is checked against other lists that are different objects rather than just those that have different values (since lists can be different objects but look identical)
– Unsolved Cypher
Dec 18 '18 at 23:43
add a comment |
Your code example doesn't work because it's finding the intersection between all of the sets, which is 0 (since no element occurs everywhere). You want to find the difference between each set and the union of all other sets. For example:
set1 = {'A', 'B', 'C'}
set2 = {'C', 'D', 'E'}
set3 = {'D', 'E', 'F'}
targets = [set1, set2, set3]
result =
for set_element in targets:
result.append(len(set_element.difference(set.union(*[x for x in targets if x is not set_element]))))
print(result)
(note that the [x for x in targets if x != set_element] is just the set of all other sets)
answered Nov 20 '18 at 1:30
Unsolved CypherUnsolved Cypher
495314
Your code example doesn't work because it's finding the intersection between all of the sets, which is 0 (since no element occurs everywhere). You want to find the difference between each set and the union of all other sets. For example:
set1 = {'A', 'B', 'C'}
set2 = {'C', 'D', 'E'}
set3 = {'D', 'E', 'F'}
targets = [set1, set2, set3]
result =
for set_element in targets:
result.append(len(set_element.difference(set.union(*[x for x in targets if x is not set_element]))))
print(result)
(note that the [x for x in targets if x != set_element] is just the set of all other sets)
answered Nov 20 '18 at 1:30
Unsolved CypherUnsolved Cypher
495314
edited Dec 18 '18 at 23:42
answered Nov 20 '18 at 1:30
Unsolved CypherUnsolved Cypher
495314
answered Nov 20 '18 at 1:30
Unsolved CypherUnsolved Cypher
495314
answered Nov 20 '18 at 1:30
Unsolved CypherUnsolved Cypher
495314
495314
1
This works great, and quickly. Thank you!
– SummerEla
Nov 20 '18 at 2:46
Unfortunately, I found an error in your solution. It returns an error when there is no difference between sets.
– SummerEla
Dec 18 '18 at 20:57
1
@SummerEla Good catch! I've updated the code, and now it works in that situation as well. I changed the!=to anis not, so that it's every list is checked against other lists that are different objects rather than just those that have different values (since lists can be different objects but look identical)
– Unsolved Cypher
Dec 18 '18 at 23:43
add a comment |
1
This works great, and quickly. Thank you!
– SummerEla
Nov 20 '18 at 2:46
Unfortunately, I found an error in your solution. It returns an error when there is no difference between sets.
– SummerEla
Dec 18 '18 at 20:57
1
@SummerEla Good catch! I've updated the code, and now it works in that situation as well. I changed the!=to anis not, so that it's every list is checked against other lists that are different objects rather than just those that have different values (since lists can be different objects but look identical)
– Unsolved Cypher
Dec 18 '18 at 23:43
1
1
This works great, and quickly. Thank you!
– SummerEla
Nov 20 '18 at 2:46
This works great, and quickly. Thank you!
– SummerEla
Nov 20 '18 at 2:46
Unfortunately, I found an error in your solution. It returns an error when there is no difference between sets.
– SummerEla
Dec 18 '18 at 20:57
Unfortunately, I found an error in your solution. It returns an error when there is no difference between sets.
– SummerEla
Dec 18 '18 at 20:57
1
1
@SummerEla Good catch! I've updated the code, and now it works in that situation as well. I changed the
!= to an is not, so that it's every list is checked against other lists that are different objects rather than just those that have different values (since lists can be different objects but look identical)– Unsolved Cypher
Dec 18 '18 at 23:43
@SummerEla Good catch! I've updated the code, and now it works in that situation as well. I changed the
!= to an is not, so that it's every list is checked against other lists that are different objects rather than just those that have different values (since lists can be different objects but look identical)– Unsolved Cypher
Dec 18 '18 at 23:43
add a comment |
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Is each string a single character? If not, are you going by whether the entire string occurs elsewhere, or just the character by character differences?
– Unsolved Cypher
Nov 20 '18 at 1:00
1
Each string contains at least six characters, and I would like to test the entire occurrence of the string. Perhaps I oversimplified my example, but I wasn't sure that would make a difference.
– SummerEla
Nov 20 '18 at 1:08