Python - 将列表中的项与另一个列表中的项连接起来

Question

我需要将列表中的项与另一个列表中的项连接起来.在我的例子中,该项是一个字符串(更准确的路径).在连接之后,我想获得一个列表,其中包含由连接产生的所有可能的项目.

例:

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

list2 = ['FileA', 'FileB']

我想获得这样的列表:

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

谢谢!

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']

或者,稍微更便携(也许更强大):

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']