Art*_*hur 2 python indexing replace list
我有一个类似的列表:
\nmylist = ['La', 'domestication', "d'un", 'animal', 'ou', "d'un", 'v\xc3\xa9g\xc3\xa9tal,', 'necessite', "l'acquisition", "d'une", 'ferme']\n
Run Code Online (Sandbox Code Playgroud)\n我想拆分具有“ ”的元素 ”的元素拆分为 2 个元素,并将它们的索引保留在原始列表中。
\nOUTPUT REQUESTED : my_new_list = ['La', 'domestication', "d'" ,'un', 'animal', 'ou', "d'", 'un', 'v\xc3\xa9g\xc3\xa9tal,', 'necessite', "l'", 'acquisition', "d'", 'une', 'ferme']\n
Run Code Online (Sandbox Code Playgroud)\n我尝试了一些东西,但我承认我不得不在正确的索引中替换两个新的分割元素,这是我尝试过的代码:
\nfor word in mylist:\n if "'" in word:\n new_words = word.split("'")\n mylist[mylist.index(word)] = (new_words[0]+"'")\n mylist.insert(mylist.index((new_words[0]+"'")+1), new_words[1]) \n\nprint(mylist)\n
Run Code Online (Sandbox Code Playgroud)\n感谢您的时间和帮助:)
\n假设您对创建新列表感到满意,实现此目的的一种方法是将现有列表与空格连接在一起,然后按空格或 进行拆分\'
:
import re\n\nmylist = [\n \'La\', \'domestication\', "d\'un",\n \'animal\', \'ou\', "d\'un", \'v\xc3\xa9g\xc3\xa9tal,\',\n \'necessite\', "l\'acquisition",\n "d\'une", \'ferme\'\n]\n\nmy_new_list = re.split(r" |(?<=\')", \' \'.join(mylist))\n
Run Code Online (Sandbox Code Playgroud)\n输出:
\n[\n \'La\', \'domestication\', "d\'", \'un\',\n \'animal\', \'ou\', "d\'", \'un\', \'v\xc3\xa9g\xc3\xa9tal,\',\n \'necessite\', "l\'", \'acquisition\',\n "d\'", \'une\', \'ferme\'\n]\n
Run Code Online (Sandbox Code Playgroud)\n请注意,这假设列表中的单词中没有空格;如果可能的话,您可以将代码中的空格替换为单词中未出现的字符(或字符序列),例如\\0
:
my_new_list = re.split(r"\\0|(?<=\')", \'\\0\'.join(mylist))\n
Run Code Online (Sandbox Code Playgroud)\n
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