如何根据dataframe中的条件移动列中的值

next 在列表中

有条件调用 下一个 理解列表中的迭代器。

assert df['sl no'].eq(1).sum() == df['score'].notna().sum()

it = iter(df.score.dropna().tolist())
df['score'] = [
    next(it) if i else np.nan for i in df['sl no'].eq(1)
]

df
    Name  sl no                details  score
0    Ram      1   ram is going to ooty    1.5
1    Ram      2         ram sings well    NaN
2   Ravi      1      ravi play cricket    1.0
3   Ravi      2     ravi is in chennai    NaN
4  Kumar      1  kumar passed the exam    3.0
5  Kumar      2       kumar is in town    NaN
6  Kumar      3                he left    NaN

如果你的 assert 声明失败,你的数据有问题,你的要求不可行。

loc -基于任务的分配

v = df.score.dropna().tolist()

df['score'] = np.nan
df.loc[df['sl no'].eq(1), 'score'] = v

df
姓名sl无详细信息评分
0内存1内存将达到1.5
1公羊2公羊唱得很好
2拉维1拉维打板球1.0
3拉维2拉维在金奈南
4库马尔1库马尔通过考试3.0
5库马尔2库马尔在南城
6库马尔3他离开了南

你可以这样做:

df['score'] = (df['score'].replace('',np.nan).groupby(df['Name']).transform(lambda x: x.bfill().ffill()))
df.loc[df['sl no'] != 1, 'score'] = np.NaN

先填一栏 score 使用相同的值:

    Name  sl no               details  score
0    Ram   1     ram is going to ooty    1.5
1    Ram   2           ram sings well    1.5
2   Ravi   1        ravi play cricket    1.0
3   Ravi   2       ravi is in chennai    1.0
4  Kumar   1    kumar passed the exam    3.0
5  Kumar   2         kumar is in town    3.0
6  Kumar   3                  he left    3.0

然后移除柱的位置 sl no 不是1

    Name  sl no              details  score
0    Ram   1    ram is going to ooty    1.5
1    Ram   2          ram sings well    NaN
2   Ravi   1       ravi play cricket    1.0
3   Ravi   2      ravi is in chennai    NaN
4  Kumar   1   kumar passed the exam    3.0
5  Kumar   2        kumar is in town    NaN
6  Kumar   3                 he left    NaN