Python 根据其他两列的值，以n次重复一行

我有一个熊猫数据框，有4列，a、B、C、D：

A   B   C   D
1   0   2   ["apple", "pear", "peach"]
2   2   3   ["cherry"]
3   3   3   ["banana", "cherry"]
4   4   7   []
5   1   3   ["apple", "grapes"]

我希望通过将每行重复n次得到以下结果，其中n等于的值（C列-B列），并使用更新后的值（每次重复+1）更新B列中的值

不确定我怎么能做到这一点？谢谢

编辑 列D对于每个B值可能不是唯一的，因此分组方式在此处可能不起作用。

与以下选项一起使用：

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您可以执行index.repeat和groupby.cumcount

import pandas as pd
c = ['A','B','C','D']
d = [[1,   0,   2,   ["apple", "pear", "peach"]],
     [2,   2,   3,   ["cherry"]],
     [3,   3,   3,   ["banana", "cherry"]],
     [4,   4,   7,   []],
     [5,   1,   3,   ["apple", "grapes"]]]
df = pd.DataFrame(d,columns=c)
df = df.loc[df.index.repeat(df['C']-df['B']+1)]
df['B'] = df['B']+df.groupby(['A']).cumcount()
print (df)

输出为：

   A  B  C                     D
0  1  0  2  [apple, pear, peach]
0  1  1  2  [apple, pear, peach]
0  1  2  2  [apple, pear, peach]
1  2  2  3              [cherry]
1  2  3  3              [cherry]
2  3  3  3      [banana, cherry]
3  4  4  7                    []
3  4  5  7                    []
3  4  6  7                    []
3  4  7  7                    []
4  5  1  3       [apple, grapes]
4  5  2  3       [apple, grapes]
4  5  3  3       [apple, grapes]

有关更多详细信息，请参阅stackoverflow帖子：

---

谢谢@Mayank Porwal的回答，但我应该用一个更好的例子来澄清，D列对于每个B值可能不是唯一的。所以groupby可能不起作用。

import pandas as pd
c = ['A','B','C','D']
d = [[1,   0,   2,   ["apple", "pear", "peach"]],
     [2,   2,   3,   ["cherry"]],
     [3,   3,   3,   ["banana", "cherry"]],
     [4,   4,   7,   []],
     [5,   1,   3,   ["apple", "grapes"]]]
df = pd.DataFrame(d,columns=c)
df = df.loc[df.index.repeat(df['C']-df['B']+1)]
df['B'] = df['B']+df.groupby(['A']).cumcount()
print (df)

   A  B  C                     D
0  1  0  2  [apple, pear, peach]
0  1  1  2  [apple, pear, peach]
0  1  2  2  [apple, pear, peach]
1  2  2  3              [cherry]
1  2  3  3              [cherry]
2  3  3  3      [banana, cherry]
3  4  4  7                    []
3  4  5  7                    []
3  4  6  7                    []
3  4  7  7                    []
4  5  1  3       [apple, grapes]
4  5  2  3       [apple, grapes]
4  5  3  3       [apple, grapes]