为元素列表中的每个单词返回“元素列表”的索引(单词)

Question

我有这样一个列表，其中每个元素的字符串中的第一个数字正好是每个元素的索引：

list = [" ","1- make your choice", "2- put something and make", "3- make something happens", "4- giulio took his choice so make","5- make your choice", "6- put something and make", "7- make something happens", "8- giulio took his choice so make","9- make your choice", "10- put something and make", "11- make something happens", "12- giulio took his choice so make"]

我希望为元素列表中的每个单词返回“元素列表”的索引，其中(单词)位于：

for x in list:
    ....

我的意思是这样的：

position_of_word_in_all_elements_list = set("make": 1,2,3,4,5,6,7,8,9,10,11,12)    

position_of_word_in_all_elements_list = set("your": 1,5,9)

position_of_word_in_all_elements_list = set("giulio":4,8,12)

有什么建议吗？

Answer 1

这将发现输入中的所有字符串都会出现，甚至是"1-“等。但是，从结果中过滤不喜欢的记录并不是什么大事：

# find the set of all words (sequences separated by a space) in input
s = set(" ".join(list).split(" "))

# for each word go through input and add index to the 
# list if word is in the element. output list into a dict with
# the word as a key
res = dict((key, [ i for i, value in enumerate(list) if key in value.split(" ")]) for key in s)

{“：”、“和”：2、6、10、‘8 -’、' 11 -'：11、‘6 -’、‘6 -’、‘某事’：2、3、6、7、10、11、‘你’：1、5、9、‘发生’：3、7、11、'giulio'：4、8、12、'make'：1、2、3、4、5、6、7、8、9、10、11、12、‘4 -’“2-”：2，“他的”：4，8，12，“9-”：9，'10-'：10，'7-'：7，‘12-’12，‘拿走’：4，8，12，'put'：2，6，10，‘选择’：1，4，5，8，9，12，‘5-’‘5，’5‘，'so'：4，8，12，'3-'：3，'1-'：1}

Answer 2

首先，将列表重命名为不干扰Python内置的内容，所以

>>> from collections import defaultdict
>>> li = [" ","1- make your choice", "2- put something and make", "3- make something happens", "4- giulio took his choice so make","5- make your choice", "6- put something and make", "7- make something happens", "8- giulio took his choice so make","9- make your choice", "10- put something and make", "11- make something happens", "12- giulio took his choice so make"]`
>>> dd = defaultdict(list)
>>> for l in li:
        try: # this is ugly hack to skip the " " value
            index,words = l.split('-')
        except ValueError:
            continue
        word_list = words.strip().split()
        for word in word_list:
            dd[word].append(index)
>>> dd['make']
['1', '2', '3', '4', '5', '6', '7', '8', '9', '10', '11', '12']

defaultdict所做的工作:只要关键字(在我们的例子中)存在于字典中，它就像普通的字典一样工作。如果键不存在，它将创建它，其值对应于(在本例中为空列表)，正如声明为dd = defaultdict(list)时所指定的那样。我不是最好的解释者，所以如果不清楚的话，我建议在其他地方阅读默认的内容:)

Answer 3

@Oleg写了一个很棒的书呆子解决方案。我想出了以下解决这个问题的简单方法。

def findIndex(st, lis):
    positions = []
    j = 0
    for x in lis:
        if st in x: 
            positions.append(j)
            j += 1
    return positions

$>>> findIndex(“您”，列表) 1、5、9