Mixing
How to iterate through two lists in parallel?
I have two iterables in Python, and I want to go over them in pairs:
foo = (1, 2, 3)
bar = (4, 5, 6)
for (f, b) in some_iterator(foo, bar):
print "f: ", f, "; b: ", b
It should result in:
f: 1; b: 4
f: 2; b: 5
f: 3; b: 6
One way to do it is to iterate over the indices:
for i in xrange(len(foo)):
print "f: ", foo[i], "; b: ", b[i]
But that seems somewhat unpythonic to me. Is there a better way to do it?
python iterator
edited Sep 30 '17 at 20:24
martineau
68.9k1091186
asked Nov 2 '09 at 21:26
Nathan FellmanNathan Fellman
61.9k80225292
add a comment |
I have two iterables in Python, and I want to go over them in pairs:
foo = (1, 2, 3)
bar = (4, 5, 6)
for (f, b) in some_iterator(foo, bar):
print "f: ", f, "; b: ", b
It should result in:
f: 1; b: 4
f: 2; b: 5
f: 3; b: 6
One way to do it is to iterate over the indices:
for i in xrange(len(foo)):
print "f: ", foo[i], "; b: ", b[i]
But that seems somewhat unpythonic to me. Is there a better way to do it?
python iterator
edited Sep 30 '17 at 20:24
martineau
68.9k1091186
asked Nov 2 '09 at 21:26
Nathan FellmanNathan Fellman
61.9k80225292
add a comment |
I have two iterables in Python, and I want to go over them in pairs:
foo = (1, 2, 3)
bar = (4, 5, 6)
for (f, b) in some_iterator(foo, bar):
print "f: ", f, "; b: ", b
It should result in:
f: 1; b: 4
f: 2; b: 5
f: 3; b: 6
One way to do it is to iterate over the indices:
for i in xrange(len(foo)):
print "f: ", foo[i], "; b: ", b[i]
But that seems somewhat unpythonic to me. Is there a better way to do it?
python iterator
edited Sep 30 '17 at 20:24
martineau
68.9k1091186
asked Nov 2 '09 at 21:26
Nathan FellmanNathan Fellman
61.9k80225292
I have two iterables in Python, and I want to go over them in pairs:
foo = (1, 2, 3)
bar = (4, 5, 6)
for (f, b) in some_iterator(foo, bar):
print "f: ", f, "; b: ", b
It should result in:
f: 1; b: 4
f: 2; b: 5
f: 3; b: 6
One way to do it is to iterate over the indices:
for i in xrange(len(foo)):
print "f: ", foo[i], "; b: ", b[i]
But that seems somewhat unpythonic to me. Is there a better way to do it?
python iterator
python iterator
edited Sep 30 '17 at 20:24
martineau
68.9k1091186
asked Nov 2 '09 at 21:26
Nathan FellmanNathan Fellman
61.9k80225292
edited Sep 30 '17 at 20:24
martineau
68.9k1091186
asked Nov 2 '09 at 21:26
Nathan FellmanNathan Fellman
61.9k80225292
edited Sep 30 '17 at 20:24
martineau
68.9k1091186
edited Sep 30 '17 at 20:24
martineau
68.9k1091186
edited Sep 30 '17 at 20:24
martineau
68.9k1091186
68.9k1091186
asked Nov 2 '09 at 21:26
Nathan FellmanNathan Fellman
61.9k80225292
asked Nov 2 '09 at 21:26
Nathan FellmanNathan Fellman
61.9k80225292
asked Nov 2 '09 at 21:26
Nathan FellmanNathan Fellman
61.9k80225292
61.9k80225292
add a comment |
add a comment |
8 Answers
8
active
oldest
votes
for f, b in zip(foo, bar):
print(f, b)
zip stops when the shorter of foo or bar stops.
In Python 2, zip
returns a list of tuples. This is fine when foo and bar are not massive. If
they are both massive then forming zip(foo,bar) is an unnecessarily massive
temporary variable, and should be replaced by itertools.izip or
itertools.izip_longest, which returns an iterator instead of a list.
import itertools
for f,b in itertools.izip(foo,bar):
print(f,b)
for f,b in itertools.izip_longest(foo,bar):
print(f,b)
izip stops when either foo or bar is exhausted.
izip_longest stops when both foo and bar are exhausted.
When the shorter iterator(s) are exhausted, izip_longest yields a tuple with None in the position corresponding to that iterator. You can also set a different fillvalue besides None if you wish. See here for the full story.
In Python 3, zip
returns an iterator of tuples, like itertools.izip in Python2. To get a list
of tuples, use list(zip(foo, bar)). And to zip until both iterators are
exhausted, you would use
itertools.zip_longest.
Note also that zip and its zip-like brethen can accept an arbitrary number of iterables as arguments. For example,
for num, cheese, color in zip([1,2,3], ['manchego', 'stilton', 'brie'],
['red', 'blue', 'green']):
print('{} {} {}'.format(num, color, cheese))
prints
1 red manchego
2 blue stilton
3 green brie
answered Nov 2 '09 at 21:28
unutbuunutbu
557k10511961253
3
@unutbu In Python 3, the function name isitertools.zip_longest, instead ofitertools.izip_longest(basicallyzip...instead ofizip...in theitertoolsmodule). It's a one character edit, otherwise I'd edit the super minor correction into your answer myself.
– Michael A
Oct 15 '14 at 20:04
@unutbu Why would I prefer OP's method over theizipone (even though theizip/ziplooks much cleaner)?
– armundle
Mar 14 '16 at 19:23
2
You might want to mention Python 3 first, as it's probably more future-proof. Moreover, it*s worth pointing out that in Python 3, zip() has exactly that advantage that only itertools.izip() had in Python 2 and thus it is usually the way to go.
– Daniel S.
Jun 14 '16 at 17:40
2
May I ask you to update your answer to explicitly state thatzipandzip-like functions fromitertoolsaccept any number of iterables and not just 2? This question is canonical now and your answer is the only one worth updating.
– vaultah
Jul 11 '16 at 15:01
1
@CharlieParker: Yes you can, but then you would usefor i, (f, b) in enumerate(zip(foo, bar)).
– unutbu
Mar 6 '18 at 19:20
|
show 6 more comments
You want the zip function.
for (f,b) in zip(foo, bar):
print "f: ", f ,"; b: ", b
answered Nov 2 '09 at 21:27
Karl GuertinKarl Guertin
3,07721719
10
Before Python 3.0 you'd want to useitertools.izipif you have large numbers of elements.
– Georg Schölly
Nov 2 '09 at 21:35
add a comment |
The builtin zip does exactly what you want. If you want the same over iterables instead of lists you could look at itertools.izip which does the same thing but gives results one at a time.
answered Nov 2 '09 at 21:28
robincerobince
9,32522542
add a comment |
What you're looking for is called zip.
answered Nov 2 '09 at 21:28
Alex MartelliAlex Martelli
630k12810421284
add a comment |
You should use 'zip' function. Here is an example how your own zip function can look like
def custom_zip(seq1, seq2):
it1 = iter(seq1)
it2 = iter(seq2)
while True:
yield next(it1), next(it2)
answered Apr 23 '17 at 11:09
Vlad BezdenVlad Bezden
30.4k10134115
Doesn't this have exactly the same result aszip(seq1, seq2)?
– Niklas Mertsch
Jun 6 '18 at 9:35
@NiklasMertsch yes it has exactly the same result. I just provided example how zip function looks like
– Vlad Bezden
Jun 6 '18 at 15:40
add a comment |
zip function solves the issue
Docs: ZIP Library function
AIM: To put the output side by side
Problem:
#value1 is a list
value1 = driver.find_elements_by_class_name("review-text")
#value2 is a list
value2 = driver.find_elements_by_class_name("review-date")
for val1 in value1:
print(val1.text)
print "n"
for val2 in value2:
print(val2.text)
print "n"
Output:
review1
review2
review3
date1
date2
date3
Solution:
for val1, val2 in zip(value1,value2):
print (val1.text+':'+val2.text)
print "n"
Output:
review1:date1
review2:date2
review3:date3
answered Mar 28 '17 at 6:39
Manojit BallavManojit Ballav
5315
add a comment |
you can use 3 type in one dictionary :
def construct_dictionary_from_lists(names, ages, scores):
end_str_dic = {}
for item_name, item_age, score_item in zip(names, ages, scores):
end_str_dic[item_name] = item_age, score_item
return end_str_dic
print(
construct_dictionary_from_lists(
["paul", "saul", "steve", "chimpy"],
[28, 59, 22, 5],
[59, 85, 55, 60]
)
)
edited Jan 14 at 21:12
Ahmed Nour Jamal El-Din
7891614
answered Dec 23 '18 at 9:46
shervin haririshervin hariri
16
add a comment |
def ncustom_zip(seq1,seq2,max_length):
length= len(seq1) if len(seq1)>len(seq2) else len(seq2) if max_length else len(seq1) if len(seq1)<len(seq2) else len(seq2)
for i in range(length):
x= seq1[i] if len(seq1)>i else None
y= seq2[i] if len(seq2)>i else None
yield x,y
l=[12,2,3,9]
p=[89,8,92,5,7]
for i,j in ncustom_zip(l,p,True):
print i,j
for i,j in ncustom_zip(l,p,False):
print i,j
answered Dec 24 '17 at 14:01
Sagar TSagar T
271110
is this my answer is right?
– Sagar T
Dec 24 '17 at 14:09
what is wrong in my answer.?
– Sagar T
Dec 25 '17 at 5:12
6
As explained in the first comment, it is lacking explanation what your code is doing, why it is better/different from the other options, what its performance and memory implication are etc. Just a code dump is not a good answer on Stack Overflow
– rene
Dec 25 '17 at 10:41
add a comment |
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8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
for f, b in zip(foo, bar):
print(f, b)
zip stops when the shorter of foo or bar stops.
In Python 2, zip
returns a list of tuples. This is fine when foo and bar are not massive. If
they are both massive then forming zip(foo,bar) is an unnecessarily massive
temporary variable, and should be replaced by itertools.izip or
itertools.izip_longest, which returns an iterator instead of a list.
import itertools
for f,b in itertools.izip(foo,bar):
print(f,b)
for f,b in itertools.izip_longest(foo,bar):
print(f,b)
izip stops when either foo or bar is exhausted.
izip_longest stops when both foo and bar are exhausted.
When the shorter iterator(s) are exhausted, izip_longest yields a tuple with None in the position corresponding to that iterator. You can also set a different fillvalue besides None if you wish. See here for the full story.
In Python 3, zip
returns an iterator of tuples, like itertools.izip in Python2. To get a list
of tuples, use list(zip(foo, bar)). And to zip until both iterators are
exhausted, you would use
itertools.zip_longest.
Note also that zip and its zip-like brethen can accept an arbitrary number of iterables as arguments. For example,
for num, cheese, color in zip([1,2,3], ['manchego', 'stilton', 'brie'],
['red', 'blue', 'green']):
print('{} {} {}'.format(num, color, cheese))
prints
1 red manchego
2 blue stilton
3 green brie
answered Nov 2 '09 at 21:28
unutbuunutbu
557k10511961253
3
@unutbu In Python 3, the function name isitertools.zip_longest, instead ofitertools.izip_longest(basicallyzip...instead ofizip...in theitertoolsmodule). It's a one character edit, otherwise I'd edit the super minor correction into your answer myself.
– Michael A
Oct 15 '14 at 20:04
@unutbu Why would I prefer OP's method over theizipone (even though theizip/ziplooks much cleaner)?
– armundle
Mar 14 '16 at 19:23
2
You might want to mention Python 3 first, as it's probably more future-proof. Moreover, it*s worth pointing out that in Python 3, zip() has exactly that advantage that only itertools.izip() had in Python 2 and thus it is usually the way to go.
– Daniel S.
Jun 14 '16 at 17:40
2
May I ask you to update your answer to explicitly state thatzipandzip-like functions fromitertoolsaccept any number of iterables and not just 2? This question is canonical now and your answer is the only one worth updating.
– vaultah
Jul 11 '16 at 15:01
1
@CharlieParker: Yes you can, but then you would usefor i, (f, b) in enumerate(zip(foo, bar)).
– unutbu
Mar 6 '18 at 19:20
|
show 6 more comments
for f, b in zip(foo, bar):
print(f, b)
zip stops when the shorter of foo or bar stops.
In Python 2, zip
returns a list of tuples. This is fine when foo and bar are not massive. If
they are both massive then forming zip(foo,bar) is an unnecessarily massive
temporary variable, and should be replaced by itertools.izip or
itertools.izip_longest, which returns an iterator instead of a list.
import itertools
for f,b in itertools.izip(foo,bar):
print(f,b)
for f,b in itertools.izip_longest(foo,bar):
print(f,b)
izip stops when either foo or bar is exhausted.
izip_longest stops when both foo and bar are exhausted.
When the shorter iterator(s) are exhausted, izip_longest yields a tuple with None in the position corresponding to that iterator. You can also set a different fillvalue besides None if you wish. See here for the full story.
In Python 3, zip
returns an iterator of tuples, like itertools.izip in Python2. To get a list
of tuples, use list(zip(foo, bar)). And to zip until both iterators are
exhausted, you would use
itertools.zip_longest.
Note also that zip and its zip-like brethen can accept an arbitrary number of iterables as arguments. For example,
for num, cheese, color in zip([1,2,3], ['manchego', 'stilton', 'brie'],
['red', 'blue', 'green']):
print('{} {} {}'.format(num, color, cheese))
prints
1 red manchego
2 blue stilton
3 green brie
answered Nov 2 '09 at 21:28
unutbuunutbu
557k10511961253
3
@unutbu In Python 3, the function name isitertools.zip_longest, instead ofitertools.izip_longest(basicallyzip...instead ofizip...in theitertoolsmodule). It's a one character edit, otherwise I'd edit the super minor correction into your answer myself.
– Michael A
Oct 15 '14 at 20:04
@unutbu Why would I prefer OP's method over theizipone (even though theizip/ziplooks much cleaner)?
– armundle
Mar 14 '16 at 19:23
2
You might want to mention Python 3 first, as it's probably more future-proof. Moreover, it*s worth pointing out that in Python 3, zip() has exactly that advantage that only itertools.izip() had in Python 2 and thus it is usually the way to go.
– Daniel S.
Jun 14 '16 at 17:40
2
May I ask you to update your answer to explicitly state thatzipandzip-like functions fromitertoolsaccept any number of iterables and not just 2? This question is canonical now and your answer is the only one worth updating.
– vaultah
Jul 11 '16 at 15:01
1
@CharlieParker: Yes you can, but then you would usefor i, (f, b) in enumerate(zip(foo, bar)).
– unutbu
Mar 6 '18 at 19:20
|
show 6 more comments
for f, b in zip(foo, bar):
print(f, b)
zip stops when the shorter of foo or bar stops.
In Python 2, zip
returns a list of tuples. This is fine when foo and bar are not massive. If
they are both massive then forming zip(foo,bar) is an unnecessarily massive
temporary variable, and should be replaced by itertools.izip or
itertools.izip_longest, which returns an iterator instead of a list.
import itertools
for f,b in itertools.izip(foo,bar):
print(f,b)
for f,b in itertools.izip_longest(foo,bar):
print(f,b)
izip stops when either foo or bar is exhausted.
izip_longest stops when both foo and bar are exhausted.
When the shorter iterator(s) are exhausted, izip_longest yields a tuple with None in the position corresponding to that iterator. You can also set a different fillvalue besides None if you wish. See here for the full story.
In Python 3, zip
returns an iterator of tuples, like itertools.izip in Python2. To get a list
of tuples, use list(zip(foo, bar)). And to zip until both iterators are
exhausted, you would use
itertools.zip_longest.
Note also that zip and its zip-like brethen can accept an arbitrary number of iterables as arguments. For example,
for num, cheese, color in zip([1,2,3], ['manchego', 'stilton', 'brie'],
['red', 'blue', 'green']):
print('{} {} {}'.format(num, color, cheese))
prints
1 red manchego
2 blue stilton
3 green brie
answered Nov 2 '09 at 21:28
unutbuunutbu
557k10511961253
for f, b in zip(foo, bar):
print(f, b)
zip stops when the shorter of foo or bar stops.
In Python 2, zip
returns a list of tuples. This is fine when foo and bar are not massive. If
they are both massive then forming zip(foo,bar) is an unnecessarily massive
temporary variable, and should be replaced by itertools.izip or
itertools.izip_longest, which returns an iterator instead of a list.
import itertools
for f,b in itertools.izip(foo,bar):
print(f,b)
for f,b in itertools.izip_longest(foo,bar):
print(f,b)
izip stops when either foo or bar is exhausted.
izip_longest stops when both foo and bar are exhausted.
When the shorter iterator(s) are exhausted, izip_longest yields a tuple with None in the position corresponding to that iterator. You can also set a different fillvalue besides None if you wish. See here for the full story.
In Python 3, zip
returns an iterator of tuples, like itertools.izip in Python2. To get a list
of tuples, use list(zip(foo, bar)). And to zip until both iterators are
exhausted, you would use
itertools.zip_longest.
Note also that zip and its zip-like brethen can accept an arbitrary number of iterables as arguments. For example,
for num, cheese, color in zip([1,2,3], ['manchego', 'stilton', 'brie'],
['red', 'blue', 'green']):
print('{} {} {}'.format(num, color, cheese))
prints
1 red manchego
2 blue stilton
3 green brie
answered Nov 2 '09 at 21:28
unutbuunutbu
557k10511961253
edited Jul 11 '16 at 17:13
answered Nov 2 '09 at 21:28
unutbuunutbu
557k10511961253
answered Nov 2 '09 at 21:28
unutbuunutbu
557k10511961253
answered Nov 2 '09 at 21:28
unutbuunutbu
557k10511961253
557k10511961253
3
@unutbu In Python 3, the function name isitertools.zip_longest, instead ofitertools.izip_longest(basicallyzip...instead ofizip...in theitertoolsmodule). It's a one character edit, otherwise I'd edit the super minor correction into your answer myself.
– Michael A
Oct 15 '14 at 20:04
@unutbu Why would I prefer OP's method over theizipone (even though theizip/ziplooks much cleaner)?
– armundle
Mar 14 '16 at 19:23
2
You might want to mention Python 3 first, as it's probably more future-proof. Moreover, it*s worth pointing out that in Python 3, zip() has exactly that advantage that only itertools.izip() had in Python 2 and thus it is usually the way to go.
– Daniel S.
Jun 14 '16 at 17:40
2
May I ask you to update your answer to explicitly state thatzipandzip-like functions fromitertoolsaccept any number of iterables and not just 2? This question is canonical now and your answer is the only one worth updating.
– vaultah
Jul 11 '16 at 15:01
1
@CharlieParker: Yes you can, but then you would usefor i, (f, b) in enumerate(zip(foo, bar)).
– unutbu
Mar 6 '18 at 19:20
|
show 6 more comments
3
@unutbu In Python 3, the function name isitertools.zip_longest, instead ofitertools.izip_longest(basicallyzip...instead ofizip...in theitertoolsmodule). It's a one character edit, otherwise I'd edit the super minor correction into your answer myself.
– Michael A
Oct 15 '14 at 20:04
@unutbu Why would I prefer OP's method over theizipone (even though theizip/ziplooks much cleaner)?
– armundle
Mar 14 '16 at 19:23
2
You might want to mention Python 3 first, as it's probably more future-proof. Moreover, it*s worth pointing out that in Python 3, zip() has exactly that advantage that only itertools.izip() had in Python 2 and thus it is usually the way to go.
– Daniel S.
Jun 14 '16 at 17:40
2
May I ask you to update your answer to explicitly state thatzipandzip-like functions fromitertoolsaccept any number of iterables and not just 2? This question is canonical now and your answer is the only one worth updating.
– vaultah
Jul 11 '16 at 15:01
1
@CharlieParker: Yes you can, but then you would usefor i, (f, b) in enumerate(zip(foo, bar)).
– unutbu
Mar 6 '18 at 19:20
3
3
@unutbu In Python 3, the function name is
itertools.zip_longest, instead of itertools.izip_longest (basically zip... instead of izip... in the itertools module). It's a one character edit, otherwise I'd edit the super minor correction into your answer myself.– Michael A
Oct 15 '14 at 20:04
@unutbu In Python 3, the function name is
itertools.zip_longest, instead of itertools.izip_longest (basically zip... instead of izip... in the itertools module). It's a one character edit, otherwise I'd edit the super minor correction into your answer myself.– Michael A
Oct 15 '14 at 20:04
@unutbu Why would I prefer OP's method over the
izip one (even though the izip/ zip looks much cleaner)?– armundle
Mar 14 '16 at 19:23
@unutbu Why would I prefer OP's method over the
izip one (even though the izip/ zip looks much cleaner)?– armundle
Mar 14 '16 at 19:23
2
2
You might want to mention Python 3 first, as it's probably more future-proof. Moreover, it*s worth pointing out that in Python 3, zip() has exactly that advantage that only itertools.izip() had in Python 2 and thus it is usually the way to go.
– Daniel S.
Jun 14 '16 at 17:40
You might want to mention Python 3 first, as it's probably more future-proof. Moreover, it*s worth pointing out that in Python 3, zip() has exactly that advantage that only itertools.izip() had in Python 2 and thus it is usually the way to go.
– Daniel S.
Jun 14 '16 at 17:40
2
2
May I ask you to update your answer to explicitly state that
zip and zip-like functions from itertools accept any number of iterables and not just 2? This question is canonical now and your answer is the only one worth updating.– vaultah
Jul 11 '16 at 15:01
May I ask you to update your answer to explicitly state that
zip and zip-like functions from itertools accept any number of iterables and not just 2? This question is canonical now and your answer is the only one worth updating.– vaultah
Jul 11 '16 at 15:01
1
1
@CharlieParker: Yes you can, but then you would use
for i, (f, b) in enumerate(zip(foo, bar)).– unutbu
Mar 6 '18 at 19:20
@CharlieParker: Yes you can, but then you would use
for i, (f, b) in enumerate(zip(foo, bar)).– unutbu
Mar 6 '18 at 19:20
|
show 6 more comments
You want the zip function.
for (f,b) in zip(foo, bar):
print "f: ", f ,"; b: ", b
answered Nov 2 '09 at 21:27
Karl GuertinKarl Guertin
3,07721719
10
Before Python 3.0 you'd want to useitertools.izipif you have large numbers of elements.
– Georg Schölly
Nov 2 '09 at 21:35
add a comment |
You want the zip function.
for (f,b) in zip(foo, bar):
print "f: ", f ,"; b: ", b
answered Nov 2 '09 at 21:27
Karl GuertinKarl Guertin
3,07721719
10
Before Python 3.0 you'd want to useitertools.izipif you have large numbers of elements.
– Georg Schölly
Nov 2 '09 at 21:35
add a comment |
You want the zip function.
for (f,b) in zip(foo, bar):
print "f: ", f ,"; b: ", b
answered Nov 2 '09 at 21:27
Karl GuertinKarl Guertin
3,07721719
You want the zip function.
for (f,b) in zip(foo, bar):
print "f: ", f ,"; b: ", b
answered Nov 2 '09 at 21:27
Karl GuertinKarl Guertin
3,07721719
answered Nov 2 '09 at 21:27
Karl GuertinKarl Guertin
3,07721719
answered Nov 2 '09 at 21:27
Karl GuertinKarl Guertin
3,07721719
answered Nov 2 '09 at 21:27
Karl GuertinKarl Guertin
3,07721719
3,07721719
10
Before Python 3.0 you'd want to useitertools.izipif you have large numbers of elements.
– Georg Schölly
Nov 2 '09 at 21:35
add a comment |
10
Before Python 3.0 you'd want to useitertools.izipif you have large numbers of elements.
– Georg Schölly
Nov 2 '09 at 21:35
10
10
Before Python 3.0 you'd want to use
itertools.izip if you have large numbers of elements.– Georg Schölly
Nov 2 '09 at 21:35
Before Python 3.0 you'd want to use
itertools.izip if you have large numbers of elements.– Georg Schölly
Nov 2 '09 at 21:35
add a comment |
The builtin zip does exactly what you want. If you want the same over iterables instead of lists you could look at itertools.izip which does the same thing but gives results one at a time.
answered Nov 2 '09 at 21:28
robincerobince
9,32522542
add a comment |
The builtin zip does exactly what you want. If you want the same over iterables instead of lists you could look at itertools.izip which does the same thing but gives results one at a time.
answered Nov 2 '09 at 21:28
robincerobince
9,32522542
add a comment |
The builtin zip does exactly what you want. If you want the same over iterables instead of lists you could look at itertools.izip which does the same thing but gives results one at a time.
answered Nov 2 '09 at 21:28
robincerobince
9,32522542
The builtin zip does exactly what you want. If you want the same over iterables instead of lists you could look at itertools.izip which does the same thing but gives results one at a time.
answered Nov 2 '09 at 21:28
robincerobince
9,32522542
answered Nov 2 '09 at 21:28
robincerobince
9,32522542
answered Nov 2 '09 at 21:28
robincerobince
9,32522542
answered Nov 2 '09 at 21:28
robincerobince
9,32522542
9,32522542
add a comment |
add a comment |
What you're looking for is called zip.
answered Nov 2 '09 at 21:28
Alex MartelliAlex Martelli
630k12810421284
add a comment |
What you're looking for is called zip.
answered Nov 2 '09 at 21:28
Alex MartelliAlex Martelli
630k12810421284
add a comment |
What you're looking for is called zip.
answered Nov 2 '09 at 21:28
Alex MartelliAlex Martelli
630k12810421284
What you're looking for is called zip.
answered Nov 2 '09 at 21:28
Alex MartelliAlex Martelli
630k12810421284
answered Nov 2 '09 at 21:28
Alex MartelliAlex Martelli
630k12810421284
answered Nov 2 '09 at 21:28
Alex MartelliAlex Martelli
630k12810421284
answered Nov 2 '09 at 21:28
Alex MartelliAlex Martelli
630k12810421284
630k12810421284
add a comment |
add a comment |
You should use 'zip' function. Here is an example how your own zip function can look like
def custom_zip(seq1, seq2):
it1 = iter(seq1)
it2 = iter(seq2)
while True:
yield next(it1), next(it2)
answered Apr 23 '17 at 11:09
Vlad BezdenVlad Bezden
30.4k10134115
Doesn't this have exactly the same result aszip(seq1, seq2)?
– Niklas Mertsch
Jun 6 '18 at 9:35
@NiklasMertsch yes it has exactly the same result. I just provided example how zip function looks like
– Vlad Bezden
Jun 6 '18 at 15:40
add a comment |
You should use 'zip' function. Here is an example how your own zip function can look like
def custom_zip(seq1, seq2):
it1 = iter(seq1)
it2 = iter(seq2)
while True:
yield next(it1), next(it2)
answered Apr 23 '17 at 11:09
Vlad BezdenVlad Bezden
30.4k10134115
Doesn't this have exactly the same result aszip(seq1, seq2)?
– Niklas Mertsch
Jun 6 '18 at 9:35
@NiklasMertsch yes it has exactly the same result. I just provided example how zip function looks like
– Vlad Bezden
Jun 6 '18 at 15:40
add a comment |
You should use 'zip' function. Here is an example how your own zip function can look like
def custom_zip(seq1, seq2):
it1 = iter(seq1)
it2 = iter(seq2)
while True:
yield next(it1), next(it2)
answered Apr 23 '17 at 11:09
Vlad BezdenVlad Bezden
30.4k10134115
You should use 'zip' function. Here is an example how your own zip function can look like
def custom_zip(seq1, seq2):
it1 = iter(seq1)
it2 = iter(seq2)
while True:
yield next(it1), next(it2)
answered Apr 23 '17 at 11:09
Vlad BezdenVlad Bezden
30.4k10134115
answered Apr 23 '17 at 11:09
Vlad BezdenVlad Bezden
30.4k10134115
answered Apr 23 '17 at 11:09
Vlad BezdenVlad Bezden
30.4k10134115
answered Apr 23 '17 at 11:09
Vlad BezdenVlad Bezden
30.4k10134115
30.4k10134115
Doesn't this have exactly the same result aszip(seq1, seq2)?
– Niklas Mertsch
Jun 6 '18 at 9:35
@NiklasMertsch yes it has exactly the same result. I just provided example how zip function looks like
– Vlad Bezden
Jun 6 '18 at 15:40
add a comment |
Doesn't this have exactly the same result aszip(seq1, seq2)?
– Niklas Mertsch
Jun 6 '18 at 9:35
@NiklasMertsch yes it has exactly the same result. I just provided example how zip function looks like
– Vlad Bezden
Jun 6 '18 at 15:40
Doesn't this have exactly the same result as
zip(seq1, seq2)?– Niklas Mertsch
Jun 6 '18 at 9:35
Doesn't this have exactly the same result as
zip(seq1, seq2)?– Niklas Mertsch
Jun 6 '18 at 9:35
@NiklasMertsch yes it has exactly the same result. I just provided example how zip function looks like
– Vlad Bezden
Jun 6 '18 at 15:40
@NiklasMertsch yes it has exactly the same result. I just provided example how zip function looks like
– Vlad Bezden
Jun 6 '18 at 15:40
add a comment |
zip function solves the issue
Docs: ZIP Library function
AIM: To put the output side by side
Problem:
#value1 is a list
value1 = driver.find_elements_by_class_name("review-text")
#value2 is a list
value2 = driver.find_elements_by_class_name("review-date")
for val1 in value1:
print(val1.text)
print "n"
for val2 in value2:
print(val2.text)
print "n"
Output:
review1
review2
review3
date1
date2
date3
Solution:
for val1, val2 in zip(value1,value2):
print (val1.text+':'+val2.text)
print "n"
Output:
review1:date1
review2:date2
review3:date3
answered Mar 28 '17 at 6:39
Manojit BallavManojit Ballav
5315
add a comment |
zip function solves the issue
Docs: ZIP Library function
AIM: To put the output side by side
Problem:
#value1 is a list
value1 = driver.find_elements_by_class_name("review-text")
#value2 is a list
value2 = driver.find_elements_by_class_name("review-date")
for val1 in value1:
print(val1.text)
print "n"
for val2 in value2:
print(val2.text)
print "n"
Output:
review1
review2
review3
date1
date2
date3
Solution:
for val1, val2 in zip(value1,value2):
print (val1.text+':'+val2.text)
print "n"
Output:
review1:date1
review2:date2
review3:date3
answered Mar 28 '17 at 6:39
Manojit BallavManojit Ballav
5315
add a comment |
zip function solves the issue
Docs: ZIP Library function
AIM: To put the output side by side
Problem:
#value1 is a list
value1 = driver.find_elements_by_class_name("review-text")
#value2 is a list
value2 = driver.find_elements_by_class_name("review-date")
for val1 in value1:
print(val1.text)
print "n"
for val2 in value2:
print(val2.text)
print "n"
Output:
review1
review2
review3
date1
date2
date3
Solution:
for val1, val2 in zip(value1,value2):
print (val1.text+':'+val2.text)
print "n"
Output:
review1:date1
review2:date2
review3:date3
answered Mar 28 '17 at 6:39
Manojit BallavManojit Ballav
5315
zip function solves the issue
Docs: ZIP Library function
AIM: To put the output side by side
Problem:
#value1 is a list
value1 = driver.find_elements_by_class_name("review-text")
#value2 is a list
value2 = driver.find_elements_by_class_name("review-date")
for val1 in value1:
print(val1.text)
print "n"
for val2 in value2:
print(val2.text)
print "n"
Output:
review1
review2
review3
date1
date2
date3
Solution:
for val1, val2 in zip(value1,value2):
print (val1.text+':'+val2.text)
print "n"
Output:
review1:date1
review2:date2
review3:date3
answered Mar 28 '17 at 6:39
Manojit BallavManojit Ballav
5315
answered Mar 28 '17 at 6:39
Manojit BallavManojit Ballav
5315
answered Mar 28 '17 at 6:39
Manojit BallavManojit Ballav
5315
answered Mar 28 '17 at 6:39
Manojit BallavManojit Ballav
5315
5315
add a comment |
add a comment |
you can use 3 type in one dictionary :
def construct_dictionary_from_lists(names, ages, scores):
end_str_dic = {}
for item_name, item_age, score_item in zip(names, ages, scores):
end_str_dic[item_name] = item_age, score_item
return end_str_dic
print(
construct_dictionary_from_lists(
["paul", "saul", "steve", "chimpy"],
[28, 59, 22, 5],
[59, 85, 55, 60]
)
)
edited Jan 14 at 21:12
Ahmed Nour Jamal El-Din
7891614
answered Dec 23 '18 at 9:46
shervin haririshervin hariri
16
add a comment |
you can use 3 type in one dictionary :
def construct_dictionary_from_lists(names, ages, scores):
end_str_dic = {}
for item_name, item_age, score_item in zip(names, ages, scores):
end_str_dic[item_name] = item_age, score_item
return end_str_dic
print(
construct_dictionary_from_lists(
["paul", "saul", "steve", "chimpy"],
[28, 59, 22, 5],
[59, 85, 55, 60]
)
)
edited Jan 14 at 21:12
Ahmed Nour Jamal El-Din
7891614
answered Dec 23 '18 at 9:46
shervin haririshervin hariri
16
add a comment |
you can use 3 type in one dictionary :
def construct_dictionary_from_lists(names, ages, scores):
end_str_dic = {}
for item_name, item_age, score_item in zip(names, ages, scores):
end_str_dic[item_name] = item_age, score_item
return end_str_dic
print(
construct_dictionary_from_lists(
["paul", "saul", "steve", "chimpy"],
[28, 59, 22, 5],
[59, 85, 55, 60]
)
)
edited Jan 14 at 21:12
Ahmed Nour Jamal El-Din
7891614
answered Dec 23 '18 at 9:46
shervin haririshervin hariri
16
you can use 3 type in one dictionary :
def construct_dictionary_from_lists(names, ages, scores):
end_str_dic = {}
for item_name, item_age, score_item in zip(names, ages, scores):
end_str_dic[item_name] = item_age, score_item
return end_str_dic
print(
construct_dictionary_from_lists(
["paul", "saul", "steve", "chimpy"],
[28, 59, 22, 5],
[59, 85, 55, 60]
)
)
edited Jan 14 at 21:12
Ahmed Nour Jamal El-Din
7891614
answered Dec 23 '18 at 9:46
shervin haririshervin hariri
16
edited Jan 14 at 21:12
Ahmed Nour Jamal El-Din
7891614
edited Jan 14 at 21:12
Ahmed Nour Jamal El-Din
7891614
edited Jan 14 at 21:12
Ahmed Nour Jamal El-Din
7891614
7891614
answered Dec 23 '18 at 9:46
shervin haririshervin hariri
16
answered Dec 23 '18 at 9:46
shervin haririshervin hariri
16
answered Dec 23 '18 at 9:46
shervin haririshervin hariri
16
16
add a comment |
add a comment |
def ncustom_zip(seq1,seq2,max_length):
length= len(seq1) if len(seq1)>len(seq2) else len(seq2) if max_length else len(seq1) if len(seq1)<len(seq2) else len(seq2)
for i in range(length):
x= seq1[i] if len(seq1)>i else None
y= seq2[i] if len(seq2)>i else None
yield x,y
l=[12,2,3,9]
p=[89,8,92,5,7]
for i,j in ncustom_zip(l,p,True):
print i,j
for i,j in ncustom_zip(l,p,False):
print i,j
answered Dec 24 '17 at 14:01
Sagar TSagar T
271110
is this my answer is right?
– Sagar T
Dec 24 '17 at 14:09
what is wrong in my answer.?
– Sagar T
Dec 25 '17 at 5:12
6
As explained in the first comment, it is lacking explanation what your code is doing, why it is better/different from the other options, what its performance and memory implication are etc. Just a code dump is not a good answer on Stack Overflow
– rene
Dec 25 '17 at 10:41
add a comment |
def ncustom_zip(seq1,seq2,max_length):
length= len(seq1) if len(seq1)>len(seq2) else len(seq2) if max_length else len(seq1) if len(seq1)<len(seq2) else len(seq2)
for i in range(length):
x= seq1[i] if len(seq1)>i else None
y= seq2[i] if len(seq2)>i else None
yield x,y
l=[12,2,3,9]
p=[89,8,92,5,7]
for i,j in ncustom_zip(l,p,True):
print i,j
for i,j in ncustom_zip(l,p,False):
print i,j
answered Dec 24 '17 at 14:01
Sagar TSagar T
271110
is this my answer is right?
– Sagar T
Dec 24 '17 at 14:09
what is wrong in my answer.?
– Sagar T
Dec 25 '17 at 5:12
6
As explained in the first comment, it is lacking explanation what your code is doing, why it is better/different from the other options, what its performance and memory implication are etc. Just a code dump is not a good answer on Stack Overflow
– rene
Dec 25 '17 at 10:41
add a comment |
def ncustom_zip(seq1,seq2,max_length):
length= len(seq1) if len(seq1)>len(seq2) else len(seq2) if max_length else len(seq1) if len(seq1)<len(seq2) else len(seq2)
for i in range(length):
x= seq1[i] if len(seq1)>i else None
y= seq2[i] if len(seq2)>i else None
yield x,y
l=[12,2,3,9]
p=[89,8,92,5,7]
for i,j in ncustom_zip(l,p,True):
print i,j
for i,j in ncustom_zip(l,p,False):
print i,j
answered Dec 24 '17 at 14:01
Sagar TSagar T
271110
def ncustom_zip(seq1,seq2,max_length):
length= len(seq1) if len(seq1)>len(seq2) else len(seq2) if max_length else len(seq1) if len(seq1)<len(seq2) else len(seq2)
for i in range(length):
x= seq1[i] if len(seq1)>i else None
y= seq2[i] if len(seq2)>i else None
yield x,y
l=[12,2,3,9]
p=[89,8,92,5,7]
for i,j in ncustom_zip(l,p,True):
print i,j
for i,j in ncustom_zip(l,p,False):
print i,j
answered Dec 24 '17 at 14:01
Sagar TSagar T
271110
edited Dec 25 '17 at 8:16
answered Dec 24 '17 at 14:01
Sagar TSagar T
271110
answered Dec 24 '17 at 14:01
Sagar TSagar T
271110
answered Dec 24 '17 at 14:01
Sagar TSagar T
271110
271110
is this my answer is right?
– Sagar T
Dec 24 '17 at 14:09
what is wrong in my answer.?
– Sagar T
Dec 25 '17 at 5:12
6
As explained in the first comment, it is lacking explanation what your code is doing, why it is better/different from the other options, what its performance and memory implication are etc. Just a code dump is not a good answer on Stack Overflow
– rene
Dec 25 '17 at 10:41
add a comment |
is this my answer is right?
– Sagar T
Dec 24 '17 at 14:09
what is wrong in my answer.?
– Sagar T
Dec 25 '17 at 5:12
6
As explained in the first comment, it is lacking explanation what your code is doing, why it is better/different from the other options, what its performance and memory implication are etc. Just a code dump is not a good answer on Stack Overflow
– rene
Dec 25 '17 at 10:41
is this my answer is right?
– Sagar T
Dec 24 '17 at 14:09
is this my answer is right?
– Sagar T
Dec 24 '17 at 14:09
what is wrong in my answer.?
– Sagar T
Dec 25 '17 at 5:12
what is wrong in my answer.?
– Sagar T
Dec 25 '17 at 5:12
6
6
As explained in the first comment, it is lacking explanation what your code is doing, why it is better/different from the other options, what its performance and memory implication are etc. Just a code dump is not a good answer on Stack Overflow
– rene
Dec 25 '17 at 10:41
As explained in the first comment, it is lacking explanation what your code is doing, why it is better/different from the other options, what its performance and memory implication are etc. Just a code dump is not a good answer on Stack Overflow
– rene
Dec 25 '17 at 10:41
add a comment |
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