Pandas GroupBy 并选择特定列中具有最小值的行 [英] Pandas GroupBy and select rows with the minimum value in a specific column

问题描述

我正在按 A 列对我的数据集进行分组，然后想取 B 列中的最小值和 C 列中的相应值.

I am grouping my dataset by column A and then would like to take the minimum value in column B and the corresponding value in column C.

data = pd.DataFrame({'A': [1, 2], 'B':[ 2, 4], 'C':[10, 4]})
data  
    A   B   C
0   1   4   3
1   1   5   4
2   1   2   10
3   2   7   2
4   2   4   4
5   2   6   6  

我想得到:

    A   B   C
0   1   2   10
1   2   4   4

目前我按 A 分组，并创建一个值来指示我将保留在数据集中的行:

For the moment I am grouping by A, and creating a value that indicates me the rows I will keep in my dataset:

a = data.groupby('A').min()
a['A'] = a.index
to_keep = [str(x[0]) + str(x[1]) for x in a[['A', 'B']].values]
data['id'] = data['A'].astype(str) + data['B'].astype('str')
data[data['id'].isin(to_keep)]

我相信有一种更直接的方法可以做到这一点.我在这里看到了许多使用多索引的答案，但我想这样做而不向我的数据帧添加多索引.感谢您的帮助.

I am sure that there is a much more straight forward way to do this. I have seen many answers here that use multi-indexing but I would like to do this without adding multi-index to my dataframe. Thank you for your help.

推荐答案

我觉得你想多了.只需使用 groupby 和 idxmin:

I feel like you're overthinking this. Just use groupby and idxmin:

df.loc[df.groupby('A').B.idxmin()]

   A  B   C
2  1  2  10
4  2  4   4

<小时>

df.loc[df.groupby('A').B.idxmin()].reset_index(drop=True)

   A  B   C
0  1  2  10
1  2  4   4

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