A列窗口中B列值的平均值

A列窗口中B列值的平均值

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问题描述

如果我在Python中有一个pandas DataFrame,如下所示:

If I have a pandas DataFrame in Python such as follows:

import numpy as np
import pandas as pd

a = np.random.uniform(0,10,20)
b = np.random.uniform(0,1,20)
data = np.vstack([a,b]).T

df = pd.DataFrame(data)
df.columns = ['A','B']
df.sort_values(by=['A'])

           A         B
5   0.057519  0.465408
14  1.610972  0.398077
3   1.725556  0.397708
17  1.734124  0.600723
11  1.944105  0.694152
19  3.265799  0.878538
13  3.352460  0.770505
10  3.865299  0.064723
16  4.137863  0.659662
12  5.597172  0.122269
7   5.990105  0.667533
6   6.410582  0.193027
9   6.881429  0.041691
15  7.522877  0.268144
1   8.093155  0.130559
0   8.699004  0.996624
8   8.755095  0.495984
4   9.135271  0.792966
18  9.440045  0.477514
2   9.654226  0.509812

是否可以有效地计算t他在 A 列间隔中的 B 列的平均值是什么?

Is it possible to efficiently calculate the mean of column B values in intervals of column A?

例如,一个人可能想计算列 B 中的值的均值,这些值落入bin范围 [0,1,2,3 ,$ 4,5,6,7,8,9,10] A 。因此,对于bin范围 A = {0-1} ,落入该bin的 B 值的平均值为 0.465408 ,对于bin范围 A = {1-2} ,落入该bin的B值的平均值将为 0.522665 等。

For example one might want to calculate the mean of values in column B which fall into the bin ranges [0,1,2,3,4,5,6,7,8,9,10] of column A. So for the bin range A = {0-1} the mean of B values falling within this bin would be 0.465408, for the bin range A = {1-2} the mean of B values falling within this bin would be 0.522665, etc.

我发现 pandas.core.window.Rolling.mean (请参见),但它似乎是在计算窗口的平均值具有指定的长度,而不是超过另一列的合并宽度。

I've found pandas.core.window.Rolling.mean (see https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.core.window.Rolling.mean.html) but it appears to calculate the mean values over a window of specified length rather than over bin widths of another column.

推荐答案

使用 cut 细分 A 列放入垃圾箱,然后在这些段上应用 groupby 并计算平均值的值 B

Using cut to segment A column into bins, and then applying groupby on these segments and calculating the mean value of B:

df.groupby(pd.cut(df['A'], bins=np.arange(11)))['B'].mean()

输出:

A
(0, 1]     0.465408
(1, 2]     0.522665
(2, 3]          NaN
(3, 4]     0.571255
(4, 5]     0.659662
(5, 6]     0.394901
(6, 7]     0.117359
(7, 8]     0.268144
(8, 9]     0.541056
(9, 10]    0.593431

更新::您可以使用 agg 应用一组不同的聚合函数,例如平均值 std 大小,例如:

Update: you can use agg to apply a set of different aggregation functions, such as mean, std and size for example:

df.groupby(pd.cut(df['A'], bins=np.arange(11)))['B'].agg(['mean', 'std', 'size'])

输出:

             mean       std  size
A
(0, 1]   0.465408       NaN     1
(1, 2]   0.522665  0.149038     4
(2, 3]        NaN       NaN     0
(3, 4]   0.571255  0.441983     3
(4, 5]   0.659662       NaN     1
(5, 6]   0.394901  0.385560     2
(6, 7]   0.117359  0.107011     2
(7, 8]   0.268144       NaN     1
(8, 9]   0.541056  0.434788     3
(9, 10]  0.593431  0.173556     3

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08-11 15:23