Question

您是否知道针对以下问题的快速/优雅的Python / Scipy / Numpy解决方案：你有一组x，y坐标和相关的值w（所有1D数组）。现在将bin x和y放到2D网格（大小为BINSxBINS）上，并计算每个bin的w值的分位数（如中位数），最终应该生成具有所需分位数的BINSxBINS 2D数组。

这对于一些嵌套循环很容易，但我确信有更优雅的解决方案。

谢谢，标记

Answer 1

这是我提出的，我希望它有用。它不一定比使用循环更清洁或更好，但也许它会让你开始更好地开始。

import numpy as np
bins_x, bins_y = 1., 1.
x = np.array([1,1,2,2,3,3,3])
y = np.array([1,1,2,2,3,3,3])
w = np.array([1,2,3,4,5,6,7], 'float')

# You can get a bin number for each point like this
x = (x // bins_x).astype('int')
y = (y // bins_y).astype('int')
shape = [x.max()+1, y.max()+1]
bin = np.ravel_multi_index([x, y], shape)

# You could get the mean by doing something like:
mean = np.bincount(bin, w) / np.bincount(bin)

# Median is a bit harder
order = bin.argsort()
bin = bin[order]
w = w[order]
edges = (bin[1:] != bin[:-1]).nonzero()[0] + 1
med_index = (np.r_[0, edges] + np.r_[edges, len(w)]) // 2
median = w[med_index]

# But that's not quite right, so maybe
median2 = [np.median(i) for i in np.split(w, edges)]

另请查看numpy.histogram2d

Answer 2

我只是想自己这样做，听起来你想要你的命令“scipy.stats.binned_statistic_2d”可以找到给定箱子的第三个参数的平均值，中位数，标准偏差或任何定义的函数。

我意识到这个问题已经得到解答，但我相信这是一个很好的内置解决方案。

Answer 3

非常感谢您的代码。基于此，我找到了我的问题的以下解决方案（只对代码进行了一些小修改）：

import numpy as np
BINS=10
boxsize=10.0
bins_x, bins_y = boxsize/BINS, boxsize/BINS
x = np.array([0,0,0,1,1,1,2,2,2,3,3,3])
y = np.array([0,0,0,1,1,1,2,2,2,3,3,3])
w = np.array([0,1,2,0,1,2,0,1,2,0,1,2], 'float')

# You can get a bin number for each point like this
x = (x // bins_x).astype('int')
y = (y // bins_y).astype('int')
shape = [BINS, BINS]
bin = np.ravel_multi_index([x, y], shape)


# Median 
order = bin.argsort()
bin = bin[order]
w = w[order]
edges = (bin[1:] != bin[:-1]).nonzero()[0] + 1
median = [np.median(i) for i in np.split(w, edges)]

#construct BINSxBINS matrix with median values
binvals=np.unique(bin)
medvals=np.zeros([BINS*BINS])
medvals[binvals]=median
medvals=medvals.reshape([BINS,BINS])

print medvals

Answer 4

numpy / scipy就像这样：

    import numpy as np
    import scipy.stats as stats

    x = np.random.uniform(0,200,100)
    y = np.random.uniform(0,200,100)
    w = np.random.uniform(1,10,100)

    h = np.histogram2d(x,y,bins=[10,10], weights=w,range=[[0,200],[0,200]])
    hist, bins_x, bins_y = h
    q = stats.mstats.mquantiles(hist,prob=[0.25, 0.5, 0.75])

    >>> q.round(2)
    array([ 512.8 ,  555.41,  592.73])

    q1 = np.where(hist<q[0],1,0)
    q2 = np.where(np.logical_and(q[0]<=hist,hist<q[1]),2,0)
    q3 = np.where(np.logical_and(q[1]<=hist,hist<=q[2]),3,0)
    q4 = np.where(q[2]<hist,4,0)

    >>>q1 + q2 + q3 + q4
    array([[4, 3, 4, 3, 1, 1, 4, 3, 1, 2],
   [1, 1, 4, 4, 2, 3, 1, 3, 3, 3],
   [2, 3, 3, 2, 2, 2, 3, 2, 4, 2],
   [2, 2, 3, 3, 3, 1, 2, 2, 1, 4],
   [1, 3, 1, 4, 2, 1, 3, 1, 1, 3],
   [4, 2, 2, 1, 2, 1, 3, 2, 1, 1],
   [4, 1, 1, 3, 1, 3, 4, 3, 2, 1],
   [4, 3, 1, 4, 4, 4, 1, 1, 2, 4],
   [2, 4, 4, 4, 3, 4, 2, 2, 2, 4],
   [2, 2, 4, 4, 3, 3, 1, 3, 4, 4]])

prob = [0.25,0.5,0.75]是分位数设置的默认值，您可以更改它或将其保留。

Python中的分位数/中值/ 2D分级

4 个答案: