Question

我有两个项目列表，每个项目都是一个字符串。我想绕过两个项目，如果不是一组单词中的一个单词，则基本上删除一个单词。但是，以下代码将所有单词放在一起，而不是创建两个单独的项目。我希望自己的updatedlist有两个项目，每个要更新的原始项目一个：

#stopwords is a variable for a set of words that I dont want in my final updated list
updated_list = []
articles = list_of_articles

for article in articles:
    for word in article:
         if word not in stopwords:
              updated_list.append(word)


articles = [['this, 'is', 'a', 'test'], ['what', 'is', 'your', 'name']]
stopwords = {'is', 'a'}

expected output:
updated_list = [['this, 'test'],['what', 'your', 'name']]

current output:
updated_list = ['this, 'test','what', 'your', 'name']

Answer 1

如果您喜欢列表推导，可以使用以下示例：

articles = [['this', 'is', 'a', 'test'], ['what', 'is', 'your', 'name']]
stopwords = {'is', 'a'}


articles = [[word for word in article if word not in stopwords] for article in articles]
print(articles)

打印：

[['this', 'test'], ['what', 'your', 'name']]

Answer 2

因此，如果我正确理解了您的问题，那么您想将列表追加到列表中。

这应该可以完成工作：

updated_list = []
articles = list_of_articles

for article in articles:
    temp_list = list()
    for word in article:
         if word not in stopwords:
             temp_list.append(word)
    updated_list.append(temp_list)

Answer 3

除了将所有文章的单词添加到一个列表中之外，您需要为每个文章维护单独的列表，最后将它们添加到updated_list中。

Answer 4

您可以执行以下操作：

updated_list = []
stopwords = {'is', 'a'}

articles = [['this', 'is', 'a', 'test'], ['what', 'is', 'your', 'name']]

for article in articles:
    lst = []
    for word in article:
        if word not in stopwords:
            lst.append(word)
    updated_list.append(lst)

print(updated_list)

输出

[['this', 'test'], ['what', 'your', 'name']]

但是我建议您使用以下data type precedence list，因为它被认为是更多 pythonic ：

stopwords = {'is', 'a'}
articles = [['this', 'is', 'a', 'test'], ['what', 'is', 'your', 'name']]

updated_list = [[word for word in article if word not in stopwords] for article in articles]
print(updated_list)

输出

[['this', 'test'], ['what', 'your', 'name']]

for循环内的追加列表

4 个答案: