用另一个列表的内容替换列表项 [英] Replacing list item with contents of another list

问题描述

类似于这个问题，但不能代替一个项目与另一个项目，我想用列表的内容替换任何一个项目的出现.

Similar to this question, but instead of replacing one item with another, I'd like to replace any occurrences of one item with the contents of a list.

orig = [ 'a', 'b', 'c', 'd', 'c' ]
repl = [ 'x', 'y', 'z' ]
desired = [ 'a', 'b', 'x', 'y', 'z', 'd', 'x', 'y', 'z' ]

# these are all incorrect, or fail to compile
[ repl if x == 'c' else x for x in orig ]
[ [a for a in orig] if x == 'c' else x for x in orig ]
[ (a for a in orig) if x == 'c' else x for x in orig ]
[ a for a in orig if x == 'c' else x for x in orig ]

很明显，我的意思是替换项目的 all 项，而不仅仅是第一项. (对未在回答中涉及此情况的任何人表示歉意.)

made it clear I meant to replace all occurrences of the item, rather than just the first. (Apologies to anyone who didn't cover that case in their answer.)

推荐答案

不同的方法:当我进行替换时，我更喜欢用字典来思考.所以我会做类似的事情

Different approach: when I'm doing replacements, I prefer to think in terms of dictionaries. So I'd do something like

>>> orig = [ 'a', 'b', 'c', 'd' ]
>>> rep = {'c': ['x', 'y', 'z']}
>>> [i for c in orig for i in rep.get(c, [c])]
['a', 'b', 'x', 'y', 'z', 'd']

最后一行是标准拼合习惯用法.

where the last line is the standard flattening idiom.

这种方法的一个优点(劣势?)是它将处理'c'的多次出现.

One advantage (disadvantage?) of this approach is that it'll handle multiple occurrences of 'c'.

[更新:]

或者，如果您愿意:

>>> from itertools import chain
>>> list(chain.from_iterable(rep.get(c, [c]) for c in orig))
['a', 'b', 'x', 'y', 'z', 'd']

关于修改后的测试用例:

On the revised test case:

>>> orig = [ 'a', 'b', 'c', 'd', 'c' ]
>>> rep = {'c': ['x', 'y', 'z']}
>>> list(chain.from_iterable(rep.get(c, [c]) for c in orig))
['a', 'b', 'x', 'y', 'z', 'd', 'x', 'y', 'z']

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