带有格式的熊猫组合

Posted 2023-02-25

【中文标题】带有格式的熊猫组合

【英文标题】：Pandas combinations with format

【发布时间】：2018-08-06 04:10:58

【问题描述】：

我的数据集是这样的，

Col1    Col2    Col3
A       10      x1
B       100     x2
C       1000    x3

这就是我希望输出的样子，

Col1    Col2    Col3    Col4    Col5    Col6    Col7    Col8    Col9
A       10      x1      Empty   Empty   Empty   Empty   Empty   Empty
B       100     x2      Empty   Empty   Empty   Empty   Empty   Empty
C       1000    x3      Empty   Empty   Empty   Empty   Empty   Empty
A       10      x1      B       100     x2      Empty   Empty   Empty
B       100     x2      C       1000    x3      Empty   Empty   Empty
A       10      x1      B       100     x2      C       1000    x3

我可以通过 itertools.combinations 获得 A、B、C 的各种组合，但我如何获得这张表？

【参考方案1】：

使用 itertools.combinations 和 itertools.chain.from_iterable：

arr = list(itertools.chain.from_iterable(
    [[j for i in el for j in i] for el in itertools.combinations(df.values.tolist(), i)]
    for i in range(1, len(df)+1)
    )
)

pd.DataFrame(arr)

   0     1   2     3       4     5     6       7     8
0  A    10  x1  None     NaN  None  None     NaN  None
1  B   100  x2  None     NaN  None  None     NaN  None
2  C  1000  x3  None     NaN  None  None     NaN  None
3  A    10  x1     B   100.0    x2  None     NaN  None
4  A    10  x1     C  1000.0    x3  None     NaN  None
5  B   100  x2     C  1000.0    x3  None     NaN  None
6  A    10  x1     B   100.0    x2     C  1000.0    x3

另一个使用concat的选项：

out = pd.concat(
          [pd.DataFrame(list(itertools.combinations(df.values.tolist(), i)))
          for i in range(1, len(df)+1)]
)

out.applymap(lambda x: [] if type(x) == float else x).sum(1).apply(pd.Series)

   0     1   2    3       4    5    6       7    8
0  A    10  x1  NaN     NaN  NaN  NaN     NaN  NaN
1  B   100  x2  NaN     NaN  NaN  NaN     NaN  NaN
2  C  1000  x3  NaN     NaN  NaN  NaN     NaN  NaN
0  A    10  x1    B   100.0   x2  NaN     NaN  NaN
1  A    10  x1    C  1000.0   x3  NaN     NaN  NaN
2  B   100  x2    C  1000.0   x3  NaN     NaN  NaN
0  A    10  x1    B   100.0   x2    C  1000.0   x3