Python-将列表中的项目与另一个列表中的项目连接

Question

I need to concatenate an item from a list with an item from another list. In my case the item is a string (a path more exactly). After the concatenation I want to obtain a list with all the possible items resulted from concatenation.

Example:

list1 = ['Library/FolderA/', 'Library/FolderB/', 'Library/FolderC/']

list2 = ['FileA', 'FileB']

I want to obtain a list like this:

[
    'Library/FolderA/FileA', 
    'Library/FolderA/FileB', 
    'Library/FolderB/FileA', 
    'Library/FolderB/FileB', 
    'Library/FolderC/FileA', 
    'Library/FolderC/FileB'
]

Thank you!

Answer 1

In [11]: [d+f for (d,f) in itertools.product(list1, list2)]
Out[11]: 
['Library/FolderA/FileA',
 'Library/FolderA/FileB',
 'Library/FolderB/FileA',
 'Library/FolderB/FileB',
 'Library/FolderC/FileA',
 'Library/FolderC/FileB']

or, slightly more portably (and perhaps robustly):

In [16]: [os.path.join(*p) for p in itertools.product(list1, list2)]
Out[16]: 
['Library/FolderA/FileA',
 'Library/FolderA/FileB',
 'Library/FolderB/FileA',
 'Library/FolderB/FileB',
 'Library/FolderC/FileA',
 'Library/FolderC/FileB']

Answer 2

You can use a list comprehension:

>>> [d + f for d in list1 for f in list2]
['Library/FolderA/FileA', 'Library/FolderA/FileB', 'Library/FolderB/FileA', 'Library/FolderB/FileB', 'Library/FolderC/FileA', 'Library/FolderC/FileB']

You may want to use os.path.join() instead of simple concatenation though.

Answer 3

The built-in itertools module defines a product() function for this:

import itertools
result = itertools.product(list1, list2)

Answer 4

The for loop can do this easily:

my_list, combo = [], ''
list1 = ['Library/FolderA/', 'Library/FolderB/', 'Library/FolderC/']
list2 = ['FileA', 'FileB']
for x in list1:
   for y in list2:
      combo = x + y
      my_list.append(combo)
return my_list

You can also just print them:

list1 = ['Library/FolderA/', 'Library/FolderB/', 'Library/FolderC/']
list2 = ['FileA', 'FileB']
for x in list1:
   for y in list2:
      print str(x + y)