Pandas groupby - 计算与相对点的距离 [英] Pandas groupby - calculating distance from relative point

问题描述

假设我有一个看起来像这样的东西

Lets say I have something that looks like this

df = pd.DataFrame({'Event':['A','A','A','A', 'A' ,'B','B','B','B','B'],  'Number':[1,2,3,4,5,6,7,8,9,10],'Ref':[False,False,False,False,True,False,False,False,True,False]})

我想要做的是创建一个新列，它是 Number 与 ref 中 True 的差异.因此，对于 A 组，True 是最后一个，因此该列将显示为 -4,-3,-2,-1,0.我一直在考虑做以下事情:

What I want to do is create a new column which is the difference in Number from the True in ref. So for the A group, the True is the last one, so the column would read -4,-3,-2,-1,0. I have been thinking to do the following:

for col in df.groupby('Event'):
    temp = col[1]
    reference = temp[temp.Ref==True]
    dist1 = temp.apply(lambda x:x.Number-reference.Number,axis=1)

这似乎为每个组正确计算，但我不确定如何将结果加入 df.

This seems to correctly calculate for each group, but I am not sure how to join the result into the df.

推荐答案

您可以执行以下操作:

df["new"] = df.Number - df.Number[df.groupby('Event')['Ref'].transform('idxmax')].reset_index(drop=True)
print(df)

输出

  Event  Number    Ref  new
0     A       1  False   -4
1     A       2  False   -3
2     A       3  False   -2
3     A       4  False   -1
4     A       5   True    0
5     B       6  False   -3
6     B       7  False   -2
7     B       8  False   -1
8     B       9   True    0
9     B      10  False    1

这个:df.groupby('Event')['Ref'].transform('idxmax') 填充按组查找索引，其中 Ref 为 True.基本上它会找到最大值的索引，因此假设 True = 1 和 False = 0，它会找到 True 值的索引.

This: df.groupby('Event')['Ref'].transform('idxmax') fill find the indices by group where Ref is True. Basically it finds the indices of the max values, so given that True = 1, and False = 0, it find the indices of the True values.

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