标签归档:scientific-computing

使用scipy / numpy在python中合并数据

问题:使用scipy / numpy在python中合并数据

有没有更有效的方法来对预先指定的bin中的数组取平均值?例如,我有一个数字数组以及一个与该数组中bin的开始和结束位置相对应的数组,我只想取这些bin中的均值?我下面有执行此操作的代码,但我想知道如何减少和改进它。谢谢。

from scipy import *
from numpy import *

def get_bin_mean(a, b_start, b_end):
    ind_upper = nonzero(a >= b_start)[0]
    a_upper = a[ind_upper]
    a_range = a_upper[nonzero(a_upper < b_end)[0]]
    mean_val = mean(a_range)
    return mean_val


data = rand(100)
bins = linspace(0, 1, 10)
binned_data = []

n = 0
for n in range(0, len(bins)-1):
    b_start = bins[n]
    b_end = bins[n+1]
    binned_data.append(get_bin_mean(data, b_start, b_end))

print binned_data

is there a more efficient way to take an average of an array in prespecified bins? for example, i have an array of numbers and an array corresponding to bin start and end positions in that array, and I want to just take the mean in those bins? I have code that does it below but i am wondering how it can be cut down and improved. thanks.

from scipy import *
from numpy import *

def get_bin_mean(a, b_start, b_end):
    ind_upper = nonzero(a >= b_start)[0]
    a_upper = a[ind_upper]
    a_range = a_upper[nonzero(a_upper < b_end)[0]]
    mean_val = mean(a_range)
    return mean_val


data = rand(100)
bins = linspace(0, 1, 10)
binned_data = []

n = 0
for n in range(0, len(bins)-1):
    b_start = bins[n]
    b_end = bins[n+1]
    binned_data.append(get_bin_mean(data, b_start, b_end))

print binned_data

回答 0

它可能更快,更容易使用numpy.digitize()

import numpy
data = numpy.random.random(100)
bins = numpy.linspace(0, 1, 10)
digitized = numpy.digitize(data, bins)
bin_means = [data[digitized == i].mean() for i in range(1, len(bins))]

替代方法是使用numpy.histogram()

bin_means = (numpy.histogram(data, bins, weights=data)[0] /
             numpy.histogram(data, bins)[0])

自己尝试哪个更快… :)

It’s probably faster and easier to use numpy.digitize():

import numpy
data = numpy.random.random(100)
bins = numpy.linspace(0, 1, 10)
digitized = numpy.digitize(data, bins)
bin_means = [data[digitized == i].mean() for i in range(1, len(bins))]

An alternative to this is to use numpy.histogram():

bin_means = (numpy.histogram(data, bins, weights=data)[0] /
             numpy.histogram(data, bins)[0])

Try for yourself which one is faster… :)


回答 1

Scipy(> = 0.11)函数scipy.stats.binned_statistic特别解决了上述问题。

对于与先前答案相同的示例,Scipy解决方案将是

import numpy as np
from scipy.stats import binned_statistic

data = np.random.rand(100)
bin_means = binned_statistic(data, data, bins=10, range=(0, 1))[0]

The Scipy (>=0.11) function scipy.stats.binned_statistic specifically addresses the above question.

For the same example as in the previous answers, the Scipy solution would be

import numpy as np
from scipy.stats import binned_statistic

data = np.random.rand(100)
bin_means = binned_statistic(data, data, bins=10, range=(0, 1))[0]

回答 2

不知道为什么这个线程坏掉了;但是这是2014年批准的答案,应该更快一些:

import numpy as np

data = np.random.rand(100)
bins = 10
slices = np.linspace(0, 100, bins+1, True).astype(np.int)
counts = np.diff(slices)

mean = np.add.reduceat(data, slices[:-1]) / counts
print mean

Not sure why this thread got necroed; but here is a 2014 approved answer, which should be far faster:

import numpy as np

data = np.random.rand(100)
bins = 10
slices = np.linspace(0, 100, bins+1, True).astype(np.int)
counts = np.diff(slices)

mean = np.add.reduceat(data, slices[:-1]) / counts
print mean

回答 3

numpy_indexed包(免责声明:我是它的作者)包含的功能有效地执行这种类型的操作:

import numpy_indexed as npi
print(npi.group_by(np.digitize(data, bins)).mean(data))

这基本上与我之前发布的解决方案相同;但现在包装在一个不错的界面中,包括测试和所有功能:)

The numpy_indexed package (disclaimer: I am its author) contains functionality to efficiently perform operations of this type:

import numpy_indexed as npi
print(npi.group_by(np.digitize(data, bins)).mean(data))

This is essentially the same solution as the one I posted earlier; but now wrapped in a nice interface, with tests and all :)


回答 4

我将添加并回答这个问题,即使用histogram2d python查找均值bin值,即scipy也具有专门设计用于计算一个或多个数据集的二维合并统计量的功能

import numpy as np
from scipy.stats import binned_statistic_2d

x = np.random.rand(100)
y = np.random.rand(100)
values = np.random.rand(100)
bin_means = binned_statistic_2d(x, y, values, bins=10).statistic

函数scipy.stats.binned_statistic_dd是此函数对更高维度数据集的概括

I would add, and also to answer the question find mean bin values using histogram2d python that the scipy also have a function specially designed to compute a bidimensional binned statistic for one or more sets of data

import numpy as np
from scipy.stats import binned_statistic_2d

x = np.random.rand(100)
y = np.random.rand(100)
values = np.random.rand(100)
bin_means = binned_statistic_2d(x, y, values, bins=10).statistic

the function scipy.stats.binned_statistic_dd is a generalization of this funcion for higher dimensions datasets


回答 5

另一种选择是使用ufunc.at。此方法在指定索引处就地应用所需的操作。我们可以使用searchsorted方法获取每个数据点的bin位置。然后,每次在bin_indexes遇到索引时,我们就可以使用at将bin_indexes给定的索引处的直方图位置增加1。

np.random.seed(1)
data = np.random.random(100) * 100
bins = np.linspace(0, 100, 10)

histogram = np.zeros_like(bins)

bin_indexes = np.searchsorted(bins, data)
np.add.at(histogram, bin_indexes, 1)

Another alternative is to use the ufunc.at. This method applies in-place a desired operation at specified indices. We can get the bin position for each datapoint using the searchsorted method. Then we can use at to increment by 1 the position of histogram at the index given by bin_indexes, every time we encounter an index at bin_indexes.

np.random.seed(1)
data = np.random.random(100) * 100
bins = np.linspace(0, 100, 10)

histogram = np.zeros_like(bins)

bin_indexes = np.searchsorted(bins, data)
np.add.at(histogram, bin_indexes, 1)