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Accepted
Trulimon
Asked:2022-07-20 16:46:19 +0000 UTC2022-07-20 16:46:19 +0000 UTC

如何从逗号分隔子字符串的列中删除重复值？

  • 772

在一个表中，列值用行表示：

df = pd.DataFrame({ 'a':['female, female, female, female, male, female', 'female, male, female, female', 'female, female, female', 'male, male, male']})

       a
0   female, female, female, female, male, female
1   female, male, female, female
2   female, female, female
3   male, male, male

我使用该方法的解决方案set()：

f = lambda x: [set(y) for y in x.split('; ')]
df['b'] = df['a'].apply(f)

给出以下结果：

      a                                              b
0   female, female, female, female, male, female    [{m, , e, ,, a, f, l}]
1   female, male, female, female                    [{m, , e, ,, a, f, l}]
2   female, female, female                          [{m, , e, ,, a, f, l}]
3   male, male, male                                [{m, , e, ,, a, l}]

你需要：

     a                                               b
0   female, female, female, female, male, female    female, male
1   female, male, female, female                    female, male
2   female, female, female                          female
3   male, male, male                                male
python

1 个回答

  1. Best Answer

    选项1：

    df["res"] = [", ".join(set(x.split(", "))) for x in df["a"]]

    选项 2：

    df["res"] = df["a"].str.split(", ").apply(lambda x: ", ".join(set(x)))

    选项 3：

    df["res"] = df["a"].str.split(", ").explode().groupby(level=0).apply(lambda x: ", ".join(set(x)))

    结果：

    In [93]: df
Out[93]:
                                              a           res
0  female, female, female, female, male, female  male, female
1                  female, male, female, female  male, female
2                        female, female, female        female
3                              male, male, male          male
    • 4

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