Python逗号分隔列表

Question

I need help, I'm trying to code a comma separated list but the results to be printed in a different way.

Here is my example code:

a = ('name1', 'name2', 'name3')
b = ('url1', 'url2', 'url3')
results = (a, b)
print(results)

This will show:

(('name1', 'name2', 'name3'), ('url1', 'url2', 'url3'))

However, I would like it to show:

name1, url1, name2, url2, name3, url3

Any help appreciated

Thanks

Answer 1

You can first use zip(..) to generate 2-tuples for each name and url, then we can use a generator or list comprehension to join these together, like:

print(', '.join(y for x in zip(a,b) for y in x))

This generates:

>>> print(', '.join(y for x in zip(a,b) for y in x))
name1, url1, name2, url2, name3, url3

Or a more declarative style:

from itertools import chain

print(', '.join(chain(*zip(a,b))))

EDIT :

If you want to print the url on separate lines, you can use:

for n,u in zip(a,b):
    print(n,u,sep=',')

Or if you want to produce a string first:

print('\n'.join('{}, {}'.format(*t) for t in zip(a,b))

Answer 2

This should do the trick.

a = ('name1', 'name2', 'name3')
b = ('url1', 'url2', 'url3')

from itertools import chain

ab = tuple(chain.from_iterable(zip(a, b)))

lst = ', '.join(ab)

Answer 3

There are a plethora of ways to do this, here is what first came to my mind.

import operator
results = reduce(operator.add, zip(a, b))

Here is another more 'broken down' maybe ugly way of doing it, however it's easier to see what's going on.

results = []
while True:
    try:
         results.append(a.pop(0))
         results.append(b.pop(0))
    except IndexError:
        break

Answer 4

Willem's answer is the way to go for your original question.

If you want to do more than just display the elements, you might want to use a dict for your data:

>>> dict(zip(a,b))
{'name2': 'url2', 'name3': 'url3', 'name1': 'url1'}

If the order is important, you can use an OrderedDict :

>>> from collections import OrderedDict
>>> OrderedDict(zip(a,b))
OrderedDict([('name1', 'url1'), ('name2', 'url2'), ('name3', 'url3')])