Question

您好问题是，通过搜索单词列表来遍历一列，然后如果找到任何搜索到的列表单词，则创建一个布尔列。这是我的代码

# Code naf related to sport.
code = ["3230Z","4764Z","7721Z","8551Z","9311Z", "9312Z", "9313Z", "9319Z",
        "9329Z", "364Z" "524W", "714B", "804C", "926A", "926C", "930L", "927C",
        "923K"]

# check keywords of code into "Code_Naf" column
for branch in code:
    df_codeNaf["topNAF"] = df_codeNaf["Code_NAF"].str.contains("3230Z" or "4764Z" or "7721Z" or "8551Z"
                                                                      or "9311Z" or "9312Z" or "9313Z" or "9319Z"
                                                                      or "9329Z" or "364Z" "524W" or "714B" or
                                                                      "804C" or "926A" or "926C" or "930L" or
                                                                      "927C" or "923K")

当我在topNaf栏中查看时，我发现只有2个True，但实际上有两个以上。我的代码有什么问题？谢谢

Answer 1

您的问题是您在$DIR_LIST中的每个df_codeNaf['topNAF']中都更改了banch。您的代码可以通过以下方式固定：

code

但是更好的是，您可以在一行中将df_codeNaf['topNAF'] = False for branch in code: df_codeNaf['topNAF'] = df_codeNaf['topNAF'] | df_codeNaf['Code_NAF'].str.contains(branch).与regex一起尝试：

contains

Answer 2

这是使用lambda的方法

code = ["3230Z","4764Z","7721Z","8551Z","9311Z", "9312Z", "9313Z", "9319Z",
        "9329Z", "364Z" "524W", "714B", "804C", "926A", "926C", "930L", "927C",
        "923K"]

df_codeNaf["topNAF"] = df_codeNaf["Code_NAF"].apply(lambda x: True if x in code else False)

在列中搜索单词列表，如果找到单词，则创建一个布尔列

2 个答案: