Mixing
How to get the longest string length from an array of lists from each lists matching index?
I have an array of lists in the form
list = [['hello','hi','hey'],['where','when','why'],['him','herself','themselves']]
I want to compare the length of list[0][0] to list[1][0] and list[2][0], basically all the first indexes, and obtain the length of the longest string size.
it must iterate through the list because the number of items and number of lists in the list can be any size.
for example, the answer of this should be
length1 = 5
length2 = 6 #('herself' is longer than 'hi' and 'when')
length3 = 10
TIA!
python string list
edited Nov 20 '18 at 14:50
jpp
97.7k2159109
asked Nov 20 '18 at 14:35
Python newbiePython newbie
144
add a comment |
I have an array of lists in the form
list = [['hello','hi','hey'],['where','when','why'],['him','herself','themselves']]
I want to compare the length of list[0][0] to list[1][0] and list[2][0], basically all the first indexes, and obtain the length of the longest string size.
it must iterate through the list because the number of items and number of lists in the list can be any size.
for example, the answer of this should be
length1 = 5
length2 = 6 #('herself' is longer than 'hi' and 'when')
length3 = 10
TIA!
python string list
edited Nov 20 '18 at 14:50
jpp
97.7k2159109
asked Nov 20 '18 at 14:35
Python newbiePython newbie
144
Please include your code that is not producing the desired output
– dfundako
Nov 20 '18 at 14:37
Please show the code you've tried so far and the results you got.
– Owen
Nov 20 '18 at 14:37
I think you wantlength2 = 7as'herself'has 7 characters.
– jpp
Nov 20 '18 at 14:50
add a comment |
I have an array of lists in the form
list = [['hello','hi','hey'],['where','when','why'],['him','herself','themselves']]
I want to compare the length of list[0][0] to list[1][0] and list[2][0], basically all the first indexes, and obtain the length of the longest string size.
it must iterate through the list because the number of items and number of lists in the list can be any size.
for example, the answer of this should be
length1 = 5
length2 = 6 #('herself' is longer than 'hi' and 'when')
length3 = 10
TIA!
python string list
edited Nov 20 '18 at 14:50
jpp
97.7k2159109
asked Nov 20 '18 at 14:35
Python newbiePython newbie
144
I have an array of lists in the form
list = [['hello','hi','hey'],['where','when','why'],['him','herself','themselves']]
I want to compare the length of list[0][0] to list[1][0] and list[2][0], basically all the first indexes, and obtain the length of the longest string size.
it must iterate through the list because the number of items and number of lists in the list can be any size.
for example, the answer of this should be
length1 = 5
length2 = 6 #('herself' is longer than 'hi' and 'when')
length3 = 10
TIA!
python string list
python string list
edited Nov 20 '18 at 14:50
jpp
97.7k2159109
asked Nov 20 '18 at 14:35
Python newbiePython newbie
144
edited Nov 20 '18 at 14:50
jpp
97.7k2159109
asked Nov 20 '18 at 14:35
Python newbiePython newbie
144
edited Nov 20 '18 at 14:50
jpp
97.7k2159109
edited Nov 20 '18 at 14:50
jpp
97.7k2159109
edited Nov 20 '18 at 14:50
jpp
97.7k2159109
97.7k2159109
asked Nov 20 '18 at 14:35
Python newbiePython newbie
144
asked Nov 20 '18 at 14:35
Python newbiePython newbie
144
asked Nov 20 '18 at 14:35
Python newbiePython newbie
144
144
Please include your code that is not producing the desired output
– dfundako
Nov 20 '18 at 14:37
Please show the code you've tried so far and the results you got.
– Owen
Nov 20 '18 at 14:37
I think you wantlength2 = 7as'herself'has 7 characters.
– jpp
Nov 20 '18 at 14:50
add a comment |
Please include your code that is not producing the desired output
– dfundako
Nov 20 '18 at 14:37
Please show the code you've tried so far and the results you got.
– Owen
Nov 20 '18 at 14:37
I think you wantlength2 = 7as'herself'has 7 characters.
– jpp
Nov 20 '18 at 14:50
Please include your code that is not producing the desired output
– dfundako
Nov 20 '18 at 14:37
Please include your code that is not producing the desired output
– dfundako
Nov 20 '18 at 14:37
Please show the code you've tried so far and the results you got.
– Owen
Nov 20 '18 at 14:37
Please show the code you've tried so far and the results you got.
– Owen
Nov 20 '18 at 14:37
I think you want
length2 = 7 as 'herself' has 7 characters.– jpp
Nov 20 '18 at 14:50
I think you want
length2 = 7 as 'herself' has 7 characters.– jpp
Nov 20 '18 at 14:50
add a comment |
3 Answers
3
active
oldest
votes
You don't need to create a variable number of variables. You can use either a list comprehension or a dictionary:
L = [['hello','hi','hey'],['where','when','why'],['him','herself','themselves']]
# list comprehension
res_list = [max(map(len, i)) for i in zip(*L)]
[5, 7, 10]
# dictionary from enumerated generator expression
res_dict = dict(enumerate((max(map(len, i)) for i in zip(*L)), 1))
{1: 5, 2: 7, 3: 10}
answered Nov 20 '18 at 14:45
jppjpp
97.7k2159109
add a comment |
Lots of ways to do it in Python.
array = [['hello','hi','hey'],
['where','when','why'],
['him','herself','themselves']]
length1 = 0
for elem in array:
if length1 < len(elem[0]):
length1 = len(elem[0])
length2 = max(array, key=lambda elem: len(elem[1]))
from itertools import accumulate
length3 = accumulate(array,
lambda e1, e2: max(len(e1[2]), len(e2[2]))
As a side note it is generally not recommended to assign something to standard identifiers, like list.
answered Nov 20 '18 at 14:43
bipllbipll
8,0061925
add a comment |
Just go through the triples from zip() and print out the length of the longest word:
lst = [['hello','hi','hey'],['where','when','why'],['him','herself','themselves']]
for i, triple in enumerate(zip(*lst), start=1):
print('length%d = %d' % (i, len(max(triple, key=len))))
# length1 = 5
# length2 = 7
# length3 = 10
Or as a dictionary:
{'length%d' % i: len(max(e, key=len)) for i, e in enumerate(zip(*lst), start=1)}
# {'length1': 5, 'length2': 7, 'length3': 10}
Which is nicer than storing variables for each length.
answered Nov 20 '18 at 14:40
RoadRunnerRoadRunner
11.2k31340
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
You don't need to create a variable number of variables. You can use either a list comprehension or a dictionary:
L = [['hello','hi','hey'],['where','when','why'],['him','herself','themselves']]
# list comprehension
res_list = [max(map(len, i)) for i in zip(*L)]
[5, 7, 10]
# dictionary from enumerated generator expression
res_dict = dict(enumerate((max(map(len, i)) for i in zip(*L)), 1))
{1: 5, 2: 7, 3: 10}
answered Nov 20 '18 at 14:45
jppjpp
97.7k2159109
add a comment |
You don't need to create a variable number of variables. You can use either a list comprehension or a dictionary:
L = [['hello','hi','hey'],['where','when','why'],['him','herself','themselves']]
# list comprehension
res_list = [max(map(len, i)) for i in zip(*L)]
[5, 7, 10]
# dictionary from enumerated generator expression
res_dict = dict(enumerate((max(map(len, i)) for i in zip(*L)), 1))
{1: 5, 2: 7, 3: 10}
answered Nov 20 '18 at 14:45
jppjpp
97.7k2159109
add a comment |
You don't need to create a variable number of variables. You can use either a list comprehension or a dictionary:
L = [['hello','hi','hey'],['where','when','why'],['him','herself','themselves']]
# list comprehension
res_list = [max(map(len, i)) for i in zip(*L)]
[5, 7, 10]
# dictionary from enumerated generator expression
res_dict = dict(enumerate((max(map(len, i)) for i in zip(*L)), 1))
{1: 5, 2: 7, 3: 10}
answered Nov 20 '18 at 14:45
jppjpp
97.7k2159109
You don't need to create a variable number of variables. You can use either a list comprehension or a dictionary:
L = [['hello','hi','hey'],['where','when','why'],['him','herself','themselves']]
# list comprehension
res_list = [max(map(len, i)) for i in zip(*L)]
[5, 7, 10]
# dictionary from enumerated generator expression
res_dict = dict(enumerate((max(map(len, i)) for i in zip(*L)), 1))
{1: 5, 2: 7, 3: 10}
answered Nov 20 '18 at 14:45
jppjpp
97.7k2159109
edited Nov 20 '18 at 15:25
answered Nov 20 '18 at 14:45
jppjpp
97.7k2159109
answered Nov 20 '18 at 14:45
jppjpp
97.7k2159109
answered Nov 20 '18 at 14:45
jppjpp
97.7k2159109
97.7k2159109
add a comment |
add a comment |
Lots of ways to do it in Python.
array = [['hello','hi','hey'],
['where','when','why'],
['him','herself','themselves']]
length1 = 0
for elem in array:
if length1 < len(elem[0]):
length1 = len(elem[0])
length2 = max(array, key=lambda elem: len(elem[1]))
from itertools import accumulate
length3 = accumulate(array,
lambda e1, e2: max(len(e1[2]), len(e2[2]))
As a side note it is generally not recommended to assign something to standard identifiers, like list.
answered Nov 20 '18 at 14:43
bipllbipll
8,0061925
add a comment |
Lots of ways to do it in Python.
array = [['hello','hi','hey'],
['where','when','why'],
['him','herself','themselves']]
length1 = 0
for elem in array:
if length1 < len(elem[0]):
length1 = len(elem[0])
length2 = max(array, key=lambda elem: len(elem[1]))
from itertools import accumulate
length3 = accumulate(array,
lambda e1, e2: max(len(e1[2]), len(e2[2]))
As a side note it is generally not recommended to assign something to standard identifiers, like list.
answered Nov 20 '18 at 14:43
bipllbipll
8,0061925
add a comment |
Lots of ways to do it in Python.
array = [['hello','hi','hey'],
['where','when','why'],
['him','herself','themselves']]
length1 = 0
for elem in array:
if length1 < len(elem[0]):
length1 = len(elem[0])
length2 = max(array, key=lambda elem: len(elem[1]))
from itertools import accumulate
length3 = accumulate(array,
lambda e1, e2: max(len(e1[2]), len(e2[2]))
As a side note it is generally not recommended to assign something to standard identifiers, like list.
answered Nov 20 '18 at 14:43
bipllbipll
8,0061925
Lots of ways to do it in Python.
array = [['hello','hi','hey'],
['where','when','why'],
['him','herself','themselves']]
length1 = 0
for elem in array:
if length1 < len(elem[0]):
length1 = len(elem[0])
length2 = max(array, key=lambda elem: len(elem[1]))
from itertools import accumulate
length3 = accumulate(array,
lambda e1, e2: max(len(e1[2]), len(e2[2]))
As a side note it is generally not recommended to assign something to standard identifiers, like list.
answered Nov 20 '18 at 14:43
bipllbipll
8,0061925
answered Nov 20 '18 at 14:43
bipllbipll
8,0061925
answered Nov 20 '18 at 14:43
bipllbipll
8,0061925
answered Nov 20 '18 at 14:43
bipllbipll
8,0061925
8,0061925
add a comment |
add a comment |
Just go through the triples from zip() and print out the length of the longest word:
lst = [['hello','hi','hey'],['where','when','why'],['him','herself','themselves']]
for i, triple in enumerate(zip(*lst), start=1):
print('length%d = %d' % (i, len(max(triple, key=len))))
# length1 = 5
# length2 = 7
# length3 = 10
Or as a dictionary:
{'length%d' % i: len(max(e, key=len)) for i, e in enumerate(zip(*lst), start=1)}
# {'length1': 5, 'length2': 7, 'length3': 10}
Which is nicer than storing variables for each length.
answered Nov 20 '18 at 14:40
RoadRunnerRoadRunner
11.2k31340
add a comment |
Just go through the triples from zip() and print out the length of the longest word:
lst = [['hello','hi','hey'],['where','when','why'],['him','herself','themselves']]
for i, triple in enumerate(zip(*lst), start=1):
print('length%d = %d' % (i, len(max(triple, key=len))))
# length1 = 5
# length2 = 7
# length3 = 10
Or as a dictionary:
{'length%d' % i: len(max(e, key=len)) for i, e in enumerate(zip(*lst), start=1)}
# {'length1': 5, 'length2': 7, 'length3': 10}
Which is nicer than storing variables for each length.
answered Nov 20 '18 at 14:40
RoadRunnerRoadRunner
11.2k31340
add a comment |
Just go through the triples from zip() and print out the length of the longest word:
lst = [['hello','hi','hey'],['where','when','why'],['him','herself','themselves']]
for i, triple in enumerate(zip(*lst), start=1):
print('length%d = %d' % (i, len(max(triple, key=len))))
# length1 = 5
# length2 = 7
# length3 = 10
Or as a dictionary:
{'length%d' % i: len(max(e, key=len)) for i, e in enumerate(zip(*lst), start=1)}
# {'length1': 5, 'length2': 7, 'length3': 10}
Which is nicer than storing variables for each length.
answered Nov 20 '18 at 14:40
RoadRunnerRoadRunner
11.2k31340
Just go through the triples from zip() and print out the length of the longest word:
lst = [['hello','hi','hey'],['where','when','why'],['him','herself','themselves']]
for i, triple in enumerate(zip(*lst), start=1):
print('length%d = %d' % (i, len(max(triple, key=len))))
# length1 = 5
# length2 = 7
# length3 = 10
Or as a dictionary:
{'length%d' % i: len(max(e, key=len)) for i, e in enumerate(zip(*lst), start=1)}
# {'length1': 5, 'length2': 7, 'length3': 10}
Which is nicer than storing variables for each length.
answered Nov 20 '18 at 14:40
RoadRunnerRoadRunner
11.2k31340
edited Nov 20 '18 at 14:57
answered Nov 20 '18 at 14:40
RoadRunnerRoadRunner
11.2k31340
answered Nov 20 '18 at 14:40
RoadRunnerRoadRunner
11.2k31340
answered Nov 20 '18 at 14:40
RoadRunnerRoadRunner
11.2k31340
11.2k31340
add a comment |
add a comment |
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Please include your code that is not producing the desired output
– dfundako
Nov 20 '18 at 14:37
Please show the code you've tried so far and the results you got.
– Owen
Nov 20 '18 at 14:37
I think you want
length2 = 7as'herself'has 7 characters.– jpp
Nov 20 '18 at 14:50