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I have an array of varying size from which I would like to average each consecutive n numbers and build another array as a result.

I have come up with two different ways but each have their problems and I am not sure if this is the best way to solve this:

  1. Using numpy.array_split() function:

    import numpy as np
no_splits = 3 #Or any number user defines
no_items = int(np.random.random(1)*100) # To get a variable number of items
pre_array = np.random.random(no_items)
mean_array = np.mean(np.array_split(pre_array,no_splits)) 
#This is efficient but gives an error if len(pre_array)%no_splits != 0

  2. enumerate(pre_array) alternative:

    mean_array = [np.mean(pre_array[i-no_splits+1:i]) for i, x in enumerate(pre_array) if i%no_splits == 0 and i != 0]

This is fine but clips the last values if i%no_splits != 0. Ideally, I would create a last value that is the mean of the remaining ones whilst keeping the code compact.

Each of this works for my purposes but I am not sure if they are the most efficient for larger arrays.

Thank you in advance!

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  • Do you mean you want to average 0 1 2 and 1 2 3 ... or 0 1 2 and 3 4 5 ... from 0 1 2 3 4...? Commented Aug 30, 2013 at 17:39
  • Hi @Ophion I need the latter, your posted answer if I understood and tried it correctly uses a moving window average method with length of the array being the same as the original array. I would want the length to equal len(new_array) = original_array/no_splits rounded up. Commented Aug 31, 2013 at 9:49
  • I have updated my answer, please let me know if this is what you are looking for. Commented Aug 31, 2013 at 14:10
  • Yes it's great thanks a lot! I was hoping for a one-liner but making a new function works as well. Thanks for help! Commented Sep 1, 2013 at 13:20

2 Answers 2

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Use uniform_filter:

>>> import scipy.ndimage.filters as filter

>>> a=np.arange(5,dtype=np.double)
>>> filter.uniform_filter(a,size=3)
array([ 0.33333333,  1.        ,  2.        ,  3.        ,  3.66666667])

#What this is actually doing
>>> np.mean([0,0,1]) #ind0
0.33333333333333331
>>> np.mean([0,1,2]) #ind1
1.0
>>> np.mean([1,2,3]) #ind2
2.0

Can be used with any size window.

>>> filter.uniform_filter(a,size=5)
array([ 0.8,  1.2,  2. ,  2.8,  3.2])

The caveat here is that the accumulator will be whatever the dtype of the array is.

Group by three then take mean:

def stride_mean(arr,stride):
    extra = arr.shape[0]%stride
    if extra==0:
        return np.mean(arr.reshape(-1,stride),axis=1)
    else:
        toslice = arr.shape[0]-extra
        first = np.mean(arr[:toslice].reshape(-1,stride),axis=1)
        rest = np.mean(arr[toslice:])
        return np.hstack((first,rest))

print pre_array
[ 0.50712539  0.75062019  0.78681352  0.35659332]

print stride_mean(pre_array,3)
[ 0.6815197   0.35659332]
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1 Comment

The second part seems to be what the OP is asking for.
1 
no_splits = 3
no_items = 100
a = np.random.rand(no_items)

no_bins = no_splits + no_items % no_splits
b = np.empty((no_bins,), dtype=a.dtype)
endpoint = no_items//no_splits

b[:no_splits] = np.mean(a[:endpoint*no_splits].reshape(-1, endpoint),
                       axis=-1)
b[no_splits:] = np.mean(a[endpoint*no_splits:])
>>> b
array([ 0.49898723,  0.49457975,  0.45601632,  0.5316093 ])
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1 Comment

I am not sure this gives me what I wanted - if there are 100 items len(b) = 34 with the last point being average of one item. In array = [0, 1, 2, 3, 4, 5, 6] I want new array to be [mean[0 1 2], mean [3 4 5], mean[6]]. Thanks for helping out!

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