Question

我有一个列表，其中我要提取某些索引的所有元素或有问题的元素在另一个列表中：

list = ['foo', 'bar', 'spam', 'baz', 'qux']
indices = [0, -1]
other_list = ['spam']
processed_list = magic(list, indices, other_list)
processed_list == ['foo', 'spam', 'qux']

我知道我可以通过列表理解（类似processed_list = [list[x] for x in indices]）来实现其中任何一个，但我找不到合并它们的方法。

Answer 1

这是一种方法。注意在Python中索引从0开始，所以我相应地更改了输入。

lst = ['foo', 'bar', 'spam', 'baz', 'qux']
indices = [0, -1]
other_list = ['spam']

def magic(lst, indices, other):

    n = len(lst)
    idx = {k if k >= 0 else n+k for k in indices}
    other = set(other)

    return [j for i, j in enumerate(lst) if (i in idx) or (j in other)]

processed_list = magic(lst, indices, other_list)

# ['foo', 'spam', 'qux']

Answer 2

一个简单的过程：

>>> processed_list = [l[i] for i in indices]
>>> processed_list.extend([ele for ele in other_list if ele in l])

或单个衬垫，虽然它感觉不对。

>>> processed_list = [l[i] for i in indices] + [ele for ele in other_list if ele in l]

由于元素可能会重复，因此如果需要，请稍后使用set。

#driver values：

IN : indices = [0, -1]
IN : other_list = ['spam']
OUT : ['foo', 'qux', 'spam']

Python：使用与另一个列表匹配的indices元素筛选列表

2 个答案: