用空格分割列表列表中的字符串

Question

假設我有以下結構：

t = [['I will','take','care'],['I know','what','to','do']]

正如您在第一個列表中看到的那樣，我有'I will'並且我希望它們分成兩個元素'I'和'will' ，結果是：

[['I', 'will', 'take', 'care'], ['I', 'know', 'what', 'to', 'do']]

快速而骯臟的算法如下：

train_text_new = []


for sent in t:
  new = []
  for word in sent:
    temp = word.split(' ')
    for item2 in temp:
      new.append(item2)


  train_text_new.append(new)

但我想知道是否有更易讀、可能更有效的算法來解決這個問題。

Answer 1

您可以制作一個簡單的生成器來產生拆分，然后在列表理解中使用它：

t = [['I will','take','care'],['I know','what','to','do']]

def splitWords(l):
    for words in l:
        yield from words.split()

[list(splitWords(sublist)) for sublist in t]
# [['I', 'will', 'take', 'care'], ['I', 'know', 'what', 'you', 'to', 'do']]

Answer 2

你可以試試這個。 假設拆分總是發生在子列表的第一個元素上

t = [['I will','take','care'],['I know','what','to','do']]
[start.split()+rest for start,*rest in t]
# [['I', 'will', 'take', 'care'], ['I', 'know', 'what', 'to', 'do']]

如果拆分應該發生在子列表中的任何單詞上，試試這個。

[[j for i in lst for j in i.split()]for lst in t]
# [['I', 'will', 'take', 'care'], ['I', 'know', 'what', 'to', 'do']]

Answer 3

使用join將每個內部列表連接到一個字符串並使用split to list 拆分該字符串就可以了

t = [['I will','take','care'],['I know','what','to','do']]
res = [' '.join(i).split() for i in t]
print(res)
# output [['I', 'will', 'take', 'care'], ['I', 'know', 'what', 'to', 'do']]

Answer 4

您可以使用itertools.chain.from_iterable在拆分后進行展平：

from itertools import chain

t = [['I will','take','care'],['I know','what','to','do']]

print([list(chain.from_iterable(x.split() for x in y)) for y in t])

Output：

[['I', 'will', 'take', 'care'], ['I', 'know', 'what', 'to', 'do']]