如何根据pandas数据帧中的条件移动列中的值

Question

嗨我有这样的df，

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

我正在尝试更改分数列中的值。值应移至df[sl no]==1 or df[Name] is the first occurence of a name

的单元格

我的预期输出应该是，

    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

请帮忙。

Answer 1

列表理解中的

next

有条件地在列表解析中的迭代器上调用next。

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

如果您的assert声明失败，那么您的数据就会出现问题，而您提出的问题也不可行。

loc - 基于作业

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

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

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

Answer 2

您可以尝试：

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

首先使用相同的值填充score列：

    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

然后删除列sl no不是1

的位置

    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