How to iterate through two lists in parallel?

655

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?

edited Sep 30 '17 at 20:24

martineau

68.9k1091186

asked Nov 2 '09 at 21:26

Nathan Fellman

61.9k80225292

add a comment |

655

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?

edited Sep 30 '17 at 20:24

martineau

68.9k1091186

asked Nov 2 '09 at 21:26

Nathan Fellman

61.9k80225292

add a comment |

655

202

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?

edited Sep 30 '17 at 20:24

martineau

68.9k1091186

asked Nov 2 '09 at 21:26

Nathan 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

edited Sep 30 '17 at 20:24

martineau

68.9k1091186

asked Nov 2 '09 at 21:26

Nathan Fellman

61.9k80225292

edited Sep 30 '17 at 20:24

martineau

68.9k1091186

asked Nov 2 '09 at 21:26

Nathan 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

asked Nov 2 '09 at 21:26

Nathan Fellman

61.9k80225292

asked Nov 2 '09 at 21:26

Nathan Fellman

61.9k80225292

asked Nov 2 '09 at 21:26

Nathan Fellman

61.9k80225292

add a comment |

8 Answers
8

active

oldest

votes

1034

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

edited Jul 11 '16 at 17:13

answered Nov 2 '09 at 21:28

unutbu

557k10511961253

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 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

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 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

@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 Guertin

3,07721719

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

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

robince

9,32522542

add a comment |

What you're looking for is called zip.

answered Nov 2 '09 at 21:28

Alex 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 Bezden

30.4k10134115

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

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 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 hariri

add a comment |

-4

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

edited Dec 25 '17 at 8:16

answered Dec 24 '17 at 14:01

Sagar 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

1034

for f, b in zip(foo, bar):

    print(f, b)

zip stops when the shorter of foo or bar stops.

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)

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

edited Jul 11 '16 at 17:13

answered Nov 2 '09 at 21:28

unutbu

557k10511961253

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 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

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 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

@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

1034

for f, b in zip(foo, bar):

    print(f, b)

zip stops when the shorter of foo or bar stops.

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)

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

edited Jul 11 '16 at 17:13

answered Nov 2 '09 at 21:28

unutbu

557k10511961253

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 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

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 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

@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

1034

for f, b in zip(foo, bar):

    print(f, b)

zip stops when the shorter of foo or bar stops.

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)

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

edited Jul 11 '16 at 17:13

answered Nov 2 '09 at 21:28

unutbu

557k10511961253

for f, b in zip(foo, bar):

    print(f, b)

zip stops when the shorter of foo or bar stops.

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)

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

edited Jul 11 '16 at 17:13

answered Nov 2 '09 at 21:28

unutbu

557k10511961253

edited Jul 11 '16 at 17:13

answered Nov 2 '09 at 21:28

unutbu

557k10511961253

answered Nov 2 '09 at 21:28

unutbu

557k10511961253

answered Nov 2 '09 at 21:28

unutbu

557k10511961253

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 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

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 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

@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

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 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

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 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

@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

@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

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

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

@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 Guertin

3,07721719

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

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 Guertin

3,07721719

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

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 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 Guertin

3,07721719

answered Nov 2 '09 at 21:27

Karl Guertin

3,07721719

answered Nov 2 '09 at 21:27

Karl Guertin

3,07721719

answered Nov 2 '09 at 21:27

Karl Guertin

3,07721719

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

add a comment |

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

robince

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

robince

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

robince

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

robince

9,32522542

answered Nov 2 '09 at 21:28

robince

9,32522542

answered Nov 2 '09 at 21:28

robince

9,32522542

answered Nov 2 '09 at 21:28

robince

9,32522542

add a comment |

What you're looking for is called zip.

answered Nov 2 '09 at 21:28

Alex Martelli

630k12810421284

add a comment |

What you're looking for is called zip.

answered Nov 2 '09 at 21:28

Alex Martelli

630k12810421284

add a comment |

What you're looking for is called zip.

answered Nov 2 '09 at 21:28

Alex Martelli

630k12810421284

What you're looking for is called zip.

answered Nov 2 '09 at 21:28

Alex Martelli

630k12810421284

answered Nov 2 '09 at 21:28

Alex Martelli

630k12810421284

answered Nov 2 '09 at 21:28

Alex Martelli

630k12810421284

answered Nov 2 '09 at 21:28

Alex 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 Bezden

30.4k10134115

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

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 Bezden

30.4k10134115

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

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 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 Bezden

30.4k10134115

answered Apr 23 '17 at 11:09

Vlad Bezden

30.4k10134115

answered Apr 23 '17 at 11:09

Vlad Bezden

30.4k10134115

answered Apr 23 '17 at 11:09

Vlad Bezden

30.4k10134115

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

add a comment |

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

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

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 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 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 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 Ballav

5315

answered Mar 28 '17 at 6:39

Manojit Ballav

5315

answered Mar 28 '17 at 6:39

Manojit Ballav

5315

answered Mar 28 '17 at 6:39

Manojit 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 hariri

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 hariri

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 hariri

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 hariri

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

answered Dec 23 '18 at 9:46

shervin hariri

answered Dec 23 '18 at 9:46

shervin hariri

answered Dec 23 '18 at 9:46

shervin hariri

add a comment |

-4

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

edited Dec 25 '17 at 8:16

answered Dec 24 '17 at 14:01

Sagar 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 |

-4

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

edited Dec 25 '17 at 8:16

answered Dec 24 '17 at 14:01

Sagar 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 |

-4

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

edited Dec 25 '17 at 8:16

answered Dec 24 '17 at 14:01

Sagar 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

edited Dec 25 '17 at 8:16

answered Dec 24 '17 at 14:01

Sagar T

271110

edited Dec 25 '17 at 8:16

answered Dec 24 '17 at 14:01

Sagar T

271110

answered Dec 24 '17 at 14:01

Sagar T

271110

answered Dec 24 '17 at 14:01

Sagar 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 |

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

what is wrong in my answer.?

– Sagar T
Dec 25 '17 at 5:12

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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