pandas :将行取消堆叠到新列中

2022-01-22 00:00:00 python pandas pivot

问题描述

我有一个看起来像这样的 df:

i have a df that looks like this:

     a        date    c
0  ABC  2020-06-01  0.1
1  ABC  2020-05-01  0.2
2  DEF  2020-07-01  0.3
3  DEF  2020-01-01  0.4
4  DEF  2020-02-01  0.5
5  DEF  2020-07-01  0.6

我想取消堆叠"；列'a'所以我的新df看起来像这样

i would like to "unstack" column 'a' so my new df looks like this

     a       date1   c1        date2   c2        date3   c3        date4   c4
0  ABC  2020-06-01  0.1   2020-05-01  0.2          nan  nan          nan  nan
1  DEF  2020-07-01  0.3   2020-01-01  0.4   2020-02-01  0.5   2020-07-01  0.6

我该怎么做?

解决方案

使用 GroupBy.cumcount 用于 MultiIndex 的辅助计数器并通过 DataFrame.unstack，然后为了正确的顺序使用 DataFrame.sort_index 和 map 用于展平 MultiIndex:

Use GroupBy.cumcount for helper counter for MultiIndex and reshape by DataFrame.unstack, then for correct order is used DataFrame.sort_index with map for flatten MultiIndex:

df = (df.set_index(['a',df.groupby('a').cumcount().add(1)])
        .unstack()
        .sort_index(axis=1, level=[1, 0], ascending=[True, False]))
df.columns = df.columns.map(lambda x: f'{x[0]}{x[1]}')
df = df.reset_index()
print (df)
     a       date1   c1       date2   c2       date3   c3       date4   c4
0  ABC  2020-06-01  0.1  2020-05-01  0.2         NaN  NaN         NaN  NaN
1  DEF  2020-07-01  0.3  2020-01-01  0.4  2020-02-01  0.5  2020-07-01  0.6

或者如果由于不同的列名称而无法进行排序，一种想法是使用 DataFrame.reindex:

Or if sorting is not possible because different columns names one idea is use DataFrame.reindex:

df1 = df.set_index(['a',df.groupby('a').cumcount().add(1)])
mux = pd.MultiIndex.from_product([df1.index.levels[1], ['date','c']])
df = df1.unstack().swaplevel(1,0, axis=1).reindex(mux, axis=1)
df.columns = df.columns.map(lambda x: f'{x[1]}{x[0]}')
df = df.reset_index()
print (df)
     a       date1   c1       date2   c2       date3   c3       date4   c4
0  ABC  2020-06-01  0.1  2020-05-01  0.2         NaN  NaN         NaN  NaN
1  DEF  2020-07-01  0.3  2020-01-01  0.4  2020-02-01  0.5  2020-07-01  0.6

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