如何复制 pandas DataFrame 行并定期更改一列

Question

我有 df

pd.DataFrame([["A1"     "B1",      "C1", "P"],
              ["A2"     "B2",      "C2", "P"],
              ["A3"     "B3",      "C3", "P"]], columns=["col_a"  "col_b",   "col_c", "col_d"])


col_a  col_b   col_c col_d
A1     B1      C1    P
A2     B2      C2    P
A3     B3      C3    P
...

我需要的结果基本上是重复并确保列在 col_d 中对于每个唯一行出现

都有 P Q R 扩展

col_a  col_b   col_c col_d
A1     B1      C1    P
A1     B1      C1    Q
A1     B1      C1    R

A2     B2      C2    P
A2     B2      C2    Q
A2     B2      C2    R

A3     B3      C3    P
A3     B3      C3    Q
A3     B3      C3    R
...

目前我只有：

new_df = pd.DataFrame(np.repeat(df.values, 3, axis=0), columns=df.columns)

这会导致这些值重复，但 col_d 没有变化

编辑：

现在我偶然发现了另一个需求，对于每个唯一的 col_a 和 col_b 我需要将“S”添加到 col_d

结果例如：

col_a  col_b   col_c col_d
A1     B1      C1    P
A1     B1      C1    Q
A1     B1      C1    R
A1     B1       T    S

A2     B2      C2    P
A2     B2      C2    Q
A2     B2      C2    R
A2     B2       T    S

非常感谢您的帮助！

Answer 1

向 col_d 列添加值 DataFrame.assign with numpy.tile:

L = ['P','Q','R']
new_df = (pd.DataFrame(np.repeat(df.values, 3, axis=0), columns=df.columns)
           .assign(col_d = np.tile(L, len(df))))

print (new_df)
  col_acol_b col_c col_d
0       A1B1    C1     P
1       A1B1    C1     Q
2       A1B1    C1     R
3       A2B2    C2     P
4       A2B2    C2     Q
5       A2B2    C2     R
6       A3B3    C3     P
7       A3B3    C3     Q
8       A3B3    C3     R

另一个类似的想法是通过 DataFrame.loc:

重复索引和重复的行

L = ['P','Q','R']
new_df = (df.loc[df.index.repeat(3)]
            .assign(col_d = np.tile(L, len(df)))
            .reset_index(drop=True))

print (new_df)
  col_acol_b col_c col_d
0       A1B1    C1     P
1       A1B1    C1     Q
2       A1B1    C1     R
3       A2B2    C2     P
4       A2B2    C2     Q
5       A2B2    C2     R
6       A3B3    C3     P
7       A3B3    C3     Q
8       A3B3    C3     R

编辑：

L = ['P','Q','R','S']
new_df = (pd.DataFrame(np.repeat(df.values, len(L), axis=0), columns=df.columns)
           .assign(col_d = np.tile(L, len(df)),
                   col_c = lambda x: x['col_c'].mask(x['col_d'].eq('S'), 'T')))

print (new_df)
   col_acol_b col_c col_d
0        A1B1    C1     P
1        A1B1    C1     Q
2        A1B1    C1     R
3        A1B1     T     S
4        A2B2    C2     P
5        A2B2    C2     Q
6        A2B2    C2     R
7        A2B2     T     S
8        A3B3    C3     P
9        A3B3    C3     Q
10       A3B3    C3     R
11       A3B3     T     S

Answer 2

如果您已经有了第一个数据框，您可以 assign and explode:

l= ['P','Q','R']
new_df = df.assign(col_d=[l]*len(df)).explode('col_d')

或merge:

new_df = df.drop(columns='col_d').merge(pd.Series(l, name='col_d'), how='cross')

输出：

  col_acol_b col_c col_d
0       A1B1    C1     P
1       A1B1    C1     Q
2       A1B1    C1     R
3       A2B2    C2     P
4       A2B2    C2     Q
5       A2B2    C2     R
6       A3B3    C3     P
7       A3B3    C3     Q
8       A3B3    C3     R

Answer 3

您可以使用 complete from pyjanitor 轻松生成组合：

# pip install pyjanitor
import pandas as pd
import janitor 

df.complete(['col_a', 'col_b', 'col_c'], {'col_d': ['P','Q','R']})

  col_a col_b col_c col_d
0    A1    B1    C1     P
1    A1    B1    C1     Q
2    A1    B1    C1     R
3    A2    B2    C2     P
4    A2    B2    C2     Q
5    A2    B2    C2     R
6    A3    B3    C3     P
7    A3    B3    C3     Q
8    A3    B3    C3     R

基本上，您将 ['col_a', 'col_b', 'col_c'] 与 {'col_d': ['P','Q','R']} 组合在一起；使用字典允许您将新值引入数据。

对于需要引入S的场景，可以分解步骤：

(df
.complete(['col_a', 'col_b'], {'col_d': ['P','Q','R', 'S']})
.assign(col_c = lambda df: np.where(df.col_d.eq('S'), 'T', df.col_c))
.ffill()
)

   col_a col_b col_c col_d
0     A1    B1    C1     P
1     A1    B1    C1     Q
2     A1    B1    C1     R
3     A1    B1     T     S
4     A2    B2    C2     P
5     A2    B2    C2     Q
6     A2    B2    C2     R
7     A2    B2     T     S
8     A3    B3    C3     P
9     A3    B3    C3     Q
10    A3    B3    C3     R
11    A3    B3     T     S

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