Calculate mean across dimension in a 2D array

Question

I have an array a like this:

a = [[40, 10], [50, 11]]

I need to calculate the mean for each dimension separately, the result should be this:

[45, 10.5]

45 being the mean of a[*][0] and 10.5 the mean of a[*][1].

What is the most elegant way of solving this without using a loop?

score 161 · Answer 1 · answered Apr 04 '13 at 19:26

161

a.mean() takes an axis argument:

In [1]: import numpy as np

In [2]: a = np.array([[40, 10], [50, 11]])

In [3]: a.mean(axis=1)     # to take the mean of each row
Out[3]: array([ 25. ,  30.5])

In [4]: a.mean(axis=0)     # to take the mean of each col
Out[4]: array([ 45. ,  10.5])

Or, as a standalone function:

In [5]: np.mean(a, axis=1)
Out[5]: array([ 25. ,  30.5])

The reason your slicing wasn't working is because this is the syntax for slicing:

In [6]: a[:,0].mean() # first column
Out[6]: 45.0

In [7]: a[:,1].mean() # second column
Out[7]: 10.5

answered Apr 04 '13 at 19:26

askewchan

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thanks for your quick response. What does `In [n]:` means? is this part of the code? – otmezger Apr 04 '13 at 19:31
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I'm using numpy, so line 2 and 3 works great, but with `axis=0` instead of `axis=1` – otmezger Apr 04 '13 at 19:32
@otmezger `axis=0` is on the next line. I edited to show more info, refresh, perhaps? – askewchan Apr 04 '13 at 19:34
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@otmezger You're welcome. Take note that many numpy array methods take an axis argument just like this. – askewchan Apr 04 '13 at 19:38
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@askewchan: what does `mean = np.mean(a, axis=(0,2,3)) mean? `knowing that the input tensor (a) is of shape (batch,channel,width,height) ? – Hossein Jan 12 '17 at 11:14

score 16 · Answer 2 · answered Apr 04 '13 at 19:28

16

Here is a non-numpy solution:

>>> a = [[40, 10], [50, 11]]
>>> [float(sum(l))/len(l) for l in zip(*a)]
[45.0, 10.5]

answered Apr 04 '13 at 19:28

Andrew Clark

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score 6 · Answer 3 · answered Apr 04 '13 at 19:28

6

If you do this a lot, NumPy is the way to go.

If for some reason you can't use NumPy:

>>> map(lambda x:sum(x)/float(len(x)), zip(*a))
[45.0, 10.5]

answered Apr 04 '13 at 19:28

NPE

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Calculate mean across dimension in a 2D array

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