Pandas-基于与另一列的交叉引用计算新值 [英] Pandas - Calculate New Value Based on Cross Reference with Another Column

问题描述

我正在尝试在其值交叉引用到另一列的列中计算新值.

I'm trying to calculate new values in a column whose values are cross-referenced to another column.

>>> import pandas as pd
>>> df = pd.DataFrame( {"A":[0., 100., 80., 40., 0., 60.], 
                        "B":[12,  12,   3,  19,  3,  19]} )
>>> df
       A   B
0    0.0  12
1  100.0  12
2   80.0   3
3   40.0  19
4    0.0   3
5   60.0  19

我想根据某个功能在A列中找到所有值为0，在B列中找到对应的值，然后更改具有相同B列值的所有A列值.例如，在上面的示例中，我想将列A的前两个值分别为0.和100更改为0.5和99.5，因为df.A[0]为0. B列中的df.B[0] = 12值与df.B[1] = 12相同.

I want to find all values in column A that are 0, find out the corresponding value in column B, then change all column A values that have the same column B value, according to some function. For instance in the example above I would like to change the first two values of column A, df.A[0] and df.A[1], respectively 0. and 100., into 0.5 and 99.5, because df.A[0] is 0. and it has the same value df.B[0] = 12 in column B as df.B[1] = 12.

df
      A   B
0   0.5  12
1  99.5  12
2  79.5   3
3  40.0  19
4   0.5   3
5  60.0  19

我尝试链接loc，aggregate，groupby和mask功能，但没有成功.是通过for循环的唯一方法吗?

I tried chaining loc, aggregate, groupby and mask functionalities, but I'm not succeeding. Is the only way through a for loop?

扩大示例以更好地说明意图.

Broadened example to better illustrate intent.

推荐答案

我找到了可行的解决方案，尽管可能不是最优的.我对分组依据进行链接，过滤和变换以获得所需的序列，然后将结果替换为原始数据帧.

I found a working solution, although probably sub-optimal. I chain groupby, filter and transform to obtain a desired series, and then replace the result in the original dataframe.

import pandas as pd
df = pd.DataFrame( {"A":[0., 100., 80., 40., 0., 60.], 
                    "B":[12,  12,   3,  19,  3,  19]} )
u = ( df.groupby(by="B",  sort=False)
         .filter(lambda x: x.A.min() == 0, dropna=False)
         .A.transform( lambda x: (x+0.5).where(x == 0, x - 0.5) ) 
    )
df.loc[pd.notnull(u), "A"] = u

给出以下结果

print("\ninitial df\n",df,"\n\nintermediate series\n",u,"\n\nfinal result",df)

initial df
        A   B
0    0.0  12
1  100.0  12
2   80.0   3
3   40.0  19
4    0.0   3
5   60.0  19

intermediate series
 0     0.5
1    99.5
2    79.5
3     NaN
4     0.5
5     NaN
Name: A, dtype: float64

final result       A   B
0   0.5  12
1  99.5  12
2  79.5   3
3  40.0  19
4   0.5   3
5  60.0  19

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