如何为binned_statistic制作用户定义的函数 [英] How to make user defined functions for binned_statistic

问题描述

我正在使用scipy stats软件包沿轴进行统计，但是在使用binned_statistic进行百分位数统计时遇到了麻烦.我在下面对代码进行了概括，尝试在一系列x箱中使用具有x，y值的数据集的第10个百分位数，但失败了.

I am using scipy stats package to take statistics along the an axis, but I am having trouble taking the percentile statistic using binned_statistic. I have generalized the code below, where I am trying taking the 10th percentile of a dataset with x, y values within a series of x bins, and it fails.

我当然可以使用np.std进行函数选择，例如中位数，甚至是numpy标准差.但是，我无法弄清楚如何使用np.percentile，因为它需要2个参数(例如np.percentile(y, 10))，但随后却给我一个ValueError: statistic not understood错误.

I can of course do function options, like median, and even the numpy standard deviation using np.std. However, I cannot figure out how to use np.percentile because it requires 2 arguments (e.g. np.percentile(y, 10)), but then it gives me a ValueError: statistic not understood error.

import numpy as np
import scipy.stats as scist

y_median = scist.binned_statistic(x,y,statistic='median',bins=20,range=[(0,5)])[0]

y_std = scist.binned_statistic(x,y,statistic=np.std,bins=20,range=[(0,5)])[0]

y_10 = scist.binned_statistic(x,y,statistic=np.percentile(10),bins=20,range=[(0,5)])[0]

print y_median
print y_std
print y_10

我很茫然，甚至玩过这样的用户定义的函数，但是都没有运气:

I am at a loss and have even played around with user defined functions like this, but with no luck:

def percentile10():
   return(np.percentile(y,10))

任何帮助，我们将不胜感激.

Any help, is greatly appreciated.

谢谢.

推荐答案

您定义的函数的问题在于它根本不带任何参数！它需要采用与您的示例相对应的y参数，如下所示:

The problem with the function you defined is that it takes no arguments at all! It needs to take a y argument that corresponds to your sample, like this:

def percentile10(y):
   return(np.percentile(y,10))

为简洁起见，您还可以使用lambda函数:

You could also use a lambda function for brevity:

scist.binned_statistic(x, y, statistic=lambda y: np.percentile(y, 10), bins=20,
                       range=[(0, 5)])[0]

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