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在这里，我有这个数据框，我试图从第 2 列中的每个数组中删除重复元素，如下所示，并在第 3 列中删除结果数组。

Column1    Column 2                                                      Column3
0        [ABC|QWER|12345, ABC|QWER|12345]                         [ABC|QWER|12345] 
1        [TBC|WERT|567890,TBC|WERT|567890]                        [TBC|WERT|567890]
2        [ERT|TYIO|9845366, ERT|TYIO|9845366,ERT|TYIO|5]   [ERT|TYIO|9845366, ERT|TYIO|5]
3        NaN                                               NaN
4        [SAR|QWPO|34564557,SAR|QWPO|3456455]             [SAR|QWPO|34564557,SAR|QWPO|3456455]
5        NaN                                              NaN
6        [SE|WERT|12233412]                                [SE|WERT|12233412]
7        NaN                                               NaN

我正在使用以下代码，但它显示了格式错误的节点或字符串的错误。请帮助解决这个问题。

import ast
    def ddpe(a):
    return list(dict.fromkeys(ast.literal_eval(a)))

  df['column3'] = df['column2'].apply(ddpe)
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1 回答 1

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我假设 'column2' 的值是字符串，因为您尝试使用ast.literal_eval. 在这种情况下，试试这个

import pandas as pd
import numpy as np

def ddpe(str_val):
    if pd.isna(str_val):  # return NaN if value is NaN
        return np.nan  
    # Remove the square brackets, split on ',' and strip possible
    # whitespaces between elements   
    vals = [v.strip() for v in str_val.strip('[]').split(',')]
    # remove duplicates keeping the original order
    return list(dict.fromkeys(vals))

df['column3'] = df['column2'].apply(ddpe)

如果列值已经是列表，您只需要

def ddpe(lst_val):
    # return NaN is value is not a list. 
    # Assuming those are only the two options.
    if not isinstance(lst_val, list):   
        return np.nan  
    return list(dict.fromkeys(lst_val))

df['column3'] = df['column2'].apply(ddpe)
于 2021-11-17T19:05:39.190 回答