如何在嵌套列表中就地拆分单词

Question

这是我的代码：

abc是一个多维列表

输入列表：abc = [['word1_word1.5','word2'],['word1_word1.5','word2']]

for i in range(len(abc)):
    for j in range(len(abc[i])):
        if "word1_word1.5" in abc[i][j]:
                newword = abc[i][j].split("_")
                // Replace the abc[i][j] with "word1" and "word1.5"

预期输出：

abc = [['word1','word1.5','word2'],['word1','word1.5','word2']]

Answer 1

for item in abc:
    for ele in item:
        if ele in xyz:
            split_word = ele.split("_")

此逻辑 return 您以存储在 split_word 中的列表形式拆分单词。现在将 abc 中的原始单词替换为 split_word.

希望对您有所帮助:)

Answer 2

for lst in abc:
    for n in range(len(lst)):
        if lst[n] in xyz:
            wrd = lst.pop(n)
            lst += wrd.split("_")

这会将存在于 xyz 中的单词替换为其拆分版本，在 abc

中就地

Answer 3

鉴于此：

>>> abc = [['word1_word1.5', 'word2', 'word3', 'word4_word4.5'], ['word5_word5.5', 'word2', 'word3', 'word1_word1.5']]

拆分每个单词：

>>> [[w.split('_') for w in l] for l in abc]
[[['word1', 'word1.5'], ['word2'], ['word3'], ['word4', 'word4.5']], [['word5', 'word5.5'], ['word2'], ['word3'], ['word1', 'word1.5']]]

连接子列表：

>>> from itertools import chain
>>> [list(chain.from_iterable(w.split('_') for w in l)) for l in abc]
[['word1', 'word1.5', 'word2', 'word3', 'word4', 'word4.5'], ['word5', 'word5.5', 'word2', 'word3', 'word1', 'word1.5']]

仅当 w 出现在 xyz 时拆分：

>>> xyz = {'word1_word1.5', 'word5_word5.5', 'word4_word4.5'}
>>> [list(chain.from_iterable(w.split('_') if w in xyz else [w] for w in l)) for l in abc]
[['word1', 'word1.5', 'word2', 'word3', 'word4', 'word4.5'], ['word5', 'word5.5', 'word2', 'word3', 'word1', 'word1.5']]

如何在嵌套列表中就地拆分单词

How to split words in-place, within a nested list

python

replace

nested-lists