Averaging over n elements

Question

I have a numpy array like this [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Let's assume I want the average of 3 elements, my target array should then look like this:

[2, 2, 2, 5, 5, 5, 8, 8, 8, 10]

Notice that when there is no triplet available I want to calculate the average over the remaining elements

Is there a neat way to do that with array operations?

have a look at [np.array_split()](https://docs.scipy.org/doc/numpy/reference/generated/numpy.array_split.html) — DrBwts, Oct 20 '19 at 13:56

Demi-Lune · Accepted Answer · 2019-10-20T14:58:46.287

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You could reshape the array to use the mean function, for example:

a = np.arange(1,11)
b = a[:a.size//3*3]
b.shape = (-1,3)
c = np.mean(b, axis=1)
# c == array([2., 5., 8.])

Then reassign the results in the original array:

c.shape = (-1,1)  # i.e. (len(b), 1)
b[:] = c
print(a)
# array([ 2,  2,  2,  5,  5,  5,  8,  8,  8, 10])

Note that this works because b is a sub-view of a. Also the last element is not as you asked the average (I left it untouched), but it'll be easy to fix, with e.g.:

a[9:] = np.mean(a[9:])

edited Oct 20 '19 at 14:58

answered Oct 20 '19 at 13:57

Demi-Lune

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Well the problem is though I don't know how large my array a is, so I won't be able to create an array be which works fine with reshape. – sebko_iic Oct 20 '19 at 14:00
This is why `b` is introduced: it **is** reshapable. – Demi-Lune Oct 20 '19 at 14:02
Yes I know. But you're setting a fixed size for b with your knowledge of a, but I don't know the size of a – sebko_iic Oct 20 '19 at 14:06
Isn't the comment `(a.size//3)*3` sufficient? Let me put it in the example directly. – Demi-Lune Oct 20 '19 at 14:15
Well sorry I didn't look properly. Yes alright everything;s fine xD – sebko_iic Oct 20 '19 at 14:17
I really like this reshaping solution. But if I'm not mistaken, there is a problem if the array has e.g. 2 more elements than `(a.size//3)*3` (`a = np.arange(1, 12)`). It would not calculate the average over the *remaining* elements as the OP asked. – FObersteiner Oct 20 '19 at 14:38
But the problem now is that when c is not divisible by 3 it won't work either – sebko_iic Oct 20 '19 at 14:45
@sebko98: no, I think that won't happen since it is dependent on the shape of `b` – FObersteiner Oct 20 '19 at 14:47
well it does when a=251 then c=83 which is not divisible by 3 – sebko_iic Oct 20 '19 at 14:49
@sebko98, yes, c should be reshaped the same size as b. Just fixed it with the "magical" -1 – Demi-Lune Oct 20 '19 at 14:52
@MrFuppes, I admit that the last items are not averaged. I'm still wondering if there's an elegant solution (maybe with padding?), or if we just have to `a[10:] = np.mean(a[10:])` – Demi-Lune Oct 20 '19 at 14:55
well, the OP doesn't seem to care ;-) – FObersteiner Oct 20 '19 at 14:58

score 0 · Answer 2 · answered Oct 20 '19 at 15:00

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I have done most of it in a one liner just for fun :D *Notice I'm using sum() to flatten the list.. (that's some weird python trick)

def custom_avg(group: int, arr):
    out = list()
    [out.append(list(np.full( (1, group), np.sum(arr[i:i+group])/ (1 if i + group > len(arr) else group), dtype=int))) for i in range(0, len(arr), group) ]
    return sum(out,[])

Enjoy! good luck.

answered Oct 20 '19 at 15:00

Ori V Agmon

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that's... some 1-liner... just because you *can* write it in one line. anyway, check the last element of your result, doesn't seem to be correct to me. – FObersteiner Oct 20 '19 at 15:10

Averaging over n elements

2 Answers2