我有兩個列表列表，但是，當實現映射/連接函數時，它們返回了不同的結果。第一個是我想要的結果。list1 = ['I woke up at 6am today.', 'I live in vancouver.', 'I go to gym by 6pm.',]list2 =`[['7am run ??\u200d♂? done ? @kristianevofit @ottowallin @trboxing @btsport @ringtv @frank_warren_official @mtkglobal @marbella.co.uk'],['我已經屈服于#bottlecapchallenge ????? 你怎么看？#bluesteel #scrubs']]`功能：[''.join(x) for x in list1][''.join(x) for x in list2]list1 的結果：['I woke up at 6am today.', 'I live in vancouver.', 'I go to gym by 6pm.',]結果list2：['[', '[', "'", '7', 'a', 'm', ' ', 'r', 'u',]']期望的結果是在 on 上產生與 on 相同的list2結果list1。
2 回答
喵喵時光機
TA貢獻1846條經驗 獲得超7個贊
你list2是一個列表列表，string而你list1是一個字符串列表。因此，您需要展平您list2的以獲得list1如下結果。
import ast
# Convert to list
list2 = ast.literal_eval(list2)
# Flatten nested list/list of list into list
flat_list2 = [y for x in list2 for y in x]
# Then you can use this
result = [''.join(x) for x in List2]
或者，您可以將它們組合起來：
import ast
list2 = ast.literal_eval(list2)
result [''.join(y) for x in list2 for y in x]
當然，您需要確保您的嵌套列表字符串必須遵循正確的 Python 語法。下面是運行的代碼IPython
In [1]: list2 = """[[\'7am run ??\\u200d♂? done ? @kristianevofit @ottowallin @trboxing @btsport @ringtv @frank_warren_official @mtkglobal @marbella.co.uk\'],
...: [\'I have succummed to the #bottlecapchallenge ????? What do you think? #bluesteel #scrubs\']]"""
In [2]: import ast
In [3]: list2 = ast.literal_eval(list2)
In [4]: result = [''.join(y) for x in list2 for y in x]
In [5]: result
Out[5]:
['7am run ??\u200d♂? done ? @kristianevofit @ottowallin @trboxing @btsport @ringtv @frank_warren_official @mtkglobal @marbella.co.uk',
'I have succummed to the #bottlecapchallenge ????? What do you think? #bluesteel #scrubs']
慕田峪9158850
TA貢獻1794條經驗 獲得超8個贊
您可以使用ast.literal_eval這對您很有用：
import ast
list2 = ast.literal_eval(list2)
result2 = [''.join(x) for x in List2]
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