从 Pandas Column 解压字典 [英] Unpack dictionary from Pandas Column

问题描述

我有一个数据框，其中一列作为字典.我想将它解包成多列(即代码，数量是下面原始列格式中的单独列).以下代码用于使用 pandas v0.22，现在 (0.23) 给出索引错误:

I have a dataframe that has one of the columns as a dictionary. I want to unpack it into multiple columns (i.e. code, amount are separate columns in the below Raw column format). The following code used to work with pandas v0.22, now (0.23) giving an index error:

pd.DataFrame.from_records(df.col_name.fillna(pd.Series([{'code':'not applicable'}], index=df.index)).values.tolist())

ValueError: Length of passed values is 1, index implies x

我在 google/stack overflow 上搜索了几个小时，之前提供的其他解决方案都不再有效.

I searched google/stack overflow for hours and none of the other solutions previously presented work anymore.

原始列格式:

     dict_codes
0   {'code': 'xx', 'amount': '10.00',...
1   {'code': 'yy', 'amount': '20.00'...
2   {'code': 'bb', 'amount': '30.00'...
3   {'code': 'aa', 'amount': '40.00'...
10  {'code': 'zz', 'amount': '50.00'...
11                            NaN
12                            NaN
13                            NaN

大家有什么建议吗?

谢谢

推荐答案

设置

df = pd.DataFrame(dict(
    codes=[
        {'amount': 12, 'code': 'a'},
        {'amount': 19, 'code': 'x'},
        {'amount': 37, 'code': 'm'},
        np.nan,
        np.nan,
        np.nan,
    ]
))

df

                         codes
0  {'amount': 12, 'code': 'a'}
1  {'amount': 19, 'code': 'x'}
2  {'amount': 37, 'code': 'm'}
3                          NaN
4                          NaN
5                          NaN

<小时>

apply 与 pd.Series

确保先dropna

df.codes.dropna().apply(pd.Series)

   amount code
0      12    a
1      19    x
2      37    m

<小时>

df.drop('codes', 1).assign(**df.codes.dropna().apply(pd.Series))

   amount code
0    12.0    a
1    19.0    x
2    37.0    m
3     NaN  NaN
4     NaN  NaN
5     NaN  NaN

<小时>

tolist 和 from_records

同样的想法，但跳过apply

pd.DataFrame.from_records(df.codes.dropna().tolist())

   amount code
0      12    a
1      19    x
2      37    m

<小时>

df.drop('codes', 1).assign(**pd.DataFrame.from_records(df.codes.dropna().tolist()))

   amount code
0    12.0    a
1    19.0    x
2    37.0    m
3     NaN  NaN
4     NaN  NaN
5     NaN  NaN

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