blocks|key|884681|text|使用numpy.digitize()可能更快、更容易|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|884682|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))]|code-block|syntax|javascript|884683|另一种方法是使用numpy.histogram()|884684|bin_means+=+(numpy.histogram(data,+bins,+weights=data)[0]+/
+++++++++++++numpy.histogram(data,+bins)[0])|884685|你自己试试哪一个更快...:)|884686|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.digitize.html|1|http://docs.scipy.org/doc/numpy/reference/generated/numpy.histogram.html^0|2|G|2|G|0|0|0|8|H|8|H|1|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@$9|1A|A|1B|1|1C]]|E|$]]|$1|M|3|N|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1E|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]|Y|$5|T|U|V|E|$W|Z]]]]

It's probably faster and easier to use <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.digitize.html" rel="noreferrer"><code>numpy.digitize()</code></a>:

<pre><code>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))]
</code></pre>

An alternative to this is to use <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.histogram.html" rel="noreferrer"><code>numpy.histogram()</code></a>:

<pre><code>bin_means = (numpy.histogram(data, bins, weights=data)[0] /
 numpy.histogram(data, bins)[0])
</code></pre>

Try for yourself which one is faster... :)

blocks|key|667549|text|不确定为什么这个线程会被破坏；但这是2014年批准的答案，它应该快得多：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|667550|import+numpy+as+np

data+=+np.random.rand(100)
bins+=+10
slices+=+np.linspace(0,+100,+bins%2B1,+True).astype(np.int)
counts+=+np.diff(slices)

mean+=+np.add.reduceat(data,+slices[:-1])+/+counts
print+mean|code-block|syntax|javascript|667551|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

<pre><code>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
</code></pre>

blocks|key|884798|text|Scipy+(>=0.11)函数scipy.stats.binned_statistic专门解决了上述问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|884799|对于与前面答案相同的示例，Scipy解决方案为|884800|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]|code-block|syntax|javascript|884801|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.binned_statistic.html^0|G|S|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

The Scipy (&gt;=0.11) function <a href="https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.binned_statistic.html" rel="nofollow noreferrer">scipy.stats.binned_statistic</a> specifically addresses the above question.
For the same example as in the previous answers, the Scipy solution would be
<pre><code>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]
</code></pre>

blocks|key|667674|text|numpy_indexed包(免责声明:我是它的作者)包含有效执行此类型操作的功能：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|667675|import+numpy_indexed+as+npi
print(npi.group_by(np.digitize(data,+bins)).mean(data))|code-block|syntax|javascript|667676|这基本上与我之前发布的解决方案相同；但现在包装在一个很好的界面中，包含测试和所有内容:)|667677|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/EelcoHoogendoorn/Numpy_arraysetops_EP^0|0|D|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

The <a href="https://github.com/EelcoHoogendoorn/Numpy_arraysetops_EP" rel="noreferrer">numpy_indexed</a> package (disclaimer: I am its author) contains functionality to efficiently perform operations of this type:

<pre><code>import numpy_indexed as npi
print(npi.group_by(np.digitize(data, bins)).mean(data))
</code></pre>

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

blocks|key|667737|text|我想补充一点，也是为了回答find+mean+bin+values+using+histogram2d+python的问题，scipy还具有一个专门为compute+a+bidimensional+binned+statistic+for+one+or+more+sets+of+data设计的函数|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|667738|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|code-block|syntax|javascript|667739|函数scipy.stats.binned_statistic_dd是该函数在高维数据集上的推广|667740|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/24917685|1|http://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.binned_statistic_2d.html|2|http://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.binned_statistic_dd.html^0|D|19|0|23|1X|1|0|0|2|V|2|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]|$A|10|B|11|1|12]]|C|$]]|$1|D|3|E|5|F|7|13|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|14|8|@]|9|@$A|15|B|16|1|17]]|C|$]]|$1|K|3|-4|5|6|7|18|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]|U|$5|N|O|P|C|$Q|V]]]]

I would add, and also to answer the question <a href="https://stackoverflow.com/questions/24917685">find mean bin values using histogram2d python</a> that the scipy also have a function specially designed to <a href="http://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.binned_statistic_2d.html" rel="nofollow noreferrer" title="scipy.stats.binned_statistic_2d">compute a bidimensional binned statistic for one or more sets of data</a>

<pre><code>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
</code></pre>

the function <a href="http://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.binned_statistic_dd.html" rel="nofollow noreferrer" title="scipy.stats.binned_statistic_dd">scipy.stats.binned_statistic_dd</a> is a generalization of this funcion for higher dimensions datasets

blocks|key|667809|text|另一种选择是使用ufunc.at。此方法在指定的索引处就地应用所需的操作。我们可以使用searchsorted方法获得每个数据点的bin位置。然后，每当我们在bin_indexes遇到索引时，我们可以使用at将直方图在bin_indexes给出的索引处的位置递增1。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|667810|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)|code-block|syntax|javascript|667811|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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. 

<pre><code>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)
</code></pre>

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.

<pre><code>from scipy import *
from numpy import *

def get_bin_mean(a, b_start, b_end):
 ind_upper = nonzero(a &gt;= b_start)[0]
 a_upper = a[ind_upper]
 a_range = a_upper[nonzero(a_upper &lt; 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
</code></pre>

binning data in python with scipy/numpy

有没有一种更有效的方法来在预先指定的存储箱中取一个数组的平均值？例如，我有一个数字数组和一个对应于该数组中bin开始和结束位置的数组，我想只取这些bin中的平均值？我有下面的代码可以做到这一点，但我想知道如何减少和改进它。谢谢。from scipy import *from numpy import *def get_...

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