从 Pandas Column 解压字典
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Unpack dictionary from Pandas Column
提问by DBa
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:
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，金额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：
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
I searched google/stack overflow for hours and none of the other solutions previously presented work anymore.
我在 google/stack overflow 上搜索了几个小时，以前的其他解决方案都没有工作了。
Raw column format:
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
Does anyone have any suggestions?
有没有人有什么建议？
Thanks
谢谢
采纳答案by piRSquared
Setup
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
applywith pd.Series
apply和 pd.Series
Make sure to dropnafirst
确保dropna先
df.codes.dropna().apply(pd.Series)
amount code
0 12 a
1 19 x
2 37 m
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
tolistand from_records
tolist和 from_records
Same idea but skip the apply
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
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
回答by user3483203
Setup
设置
codes
0 {'amount': 12, 'code': 10}
1 {'amount': 3, 'code': 3}
applywith pd.Series
apply和 pd.Series
df.codes.apply(pd.Series)
amount code
0 12 10
1 3 3
