Question

我有一个pandas数据帧，如下所示。

DF1 =

sid                 path
 1    '["rome","is","in","province","lazio"]'   
 1    "['rome', 'is', 'in', 'province', 'naples']"
 1     ['N']
 1    "['rome', 'is', 'in', 'province', 'in', 'campania']"
 ....

我想删除列path中所有不必要的字符，因此结果应如下所示：

DF2 =

    sid                  path
     1         rome is in province lazio
     1         rome is in province naples
     1                    N
     1         rome is in province in campania
 ....

我尝试替换所有不必要的字符：

 DF1["path"].replace("[","").replace("]","").replace('"',"").replace(","," ").replace("'","")

但它没有用。我想这是由于条目["N"]

我该怎么做？任何帮助表示赞赏！

Answer 1

您可以使用ast.literal_eval安全地将列表输出作为字符串读取。记录真实列表的一种方法是捕获ValueError。

请注意，如果可能的话，您应该尝试在问题到达数据框之前对这些问题进行上游排序。

from ast import literal_eval

df = pd.DataFrame({'sid': [1, 1, 1, 1],
                   'path': ['["rome","is","in","province","lazio"]',
                            "['rome', 'is', 'in', 'province', 'naples']",
                            ['N'],
                            "['rome', 'is', 'in', 'province', 'in', 'campania']"]})

def converter(x):
    try:
        return ' '.join(literal_eval(x))
    except ValueError:
        return ' '.join(x)

df['path'] = df['path'].apply(converter)

print(df)

                              path  sid
0        rome is in province lazio    1
1       rome is in province naples    1
2                                N    1
3  rome is in province in campania    1

Answer 2

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<强>演示：

str.join

<强>输出：

import pandas as pd
import ast
df = pd.DataFrame({"path": ['["rome","is","in","province","lazio"]', "['rome', 'is', 'in', 'province', 'naples']", ['N']]})
df['path'] = df['path'].astype(str).apply(ast.literal_eval).apply(lambda x: " ".join(x))
print(df)

从Pandas DataFrame列中删除字符串

2 个答案: