rolling mean with increasing window

Question

I have a range

np.arange(1,11) # [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

and for each element, i , in my range I want to compute the average from element i=0 to my current element. the result would be something like:

array([ 1. ,  1.5,  2. ,  2.5,  3. ,  3.5,  4. ,  4.5,  5. ,  5.5])
 # got this result via np.cumsum(np.arange(1,11,dtype=np.float32))/(np.arange(1, 11))

I was wondering if there isn't an out of the box function in numpy / pandas that gives me this result?

Answer 1

You can use expanding() (requires pandas 0.18.0):

ser = pd.Series(np.arange(1, 11))

ser.expanding().mean()
Out: 
0    1.0
1    1.5
2    2.0
3    2.5
4    3.0
5    3.5
6    4.0
7    4.5
8    5.0
9    5.5
dtype: float64

Answer 2

This seems to be the simplest, although it may become inefficient if x is very large:

x = range(1,11)
[np.mean(x[:i+1]) for i in xrange(0,len(x))]

Answer 3

Here's a vectorized approach -

a.cumsum()/(np.arange(a.size)+1)

Please note that to make sure the results are floating pt numbers, we need to add in at the start :

from __future__ import division

Alternatively, we can use np.true_divide for the division -

np.true_divide(a.cumsum(),(np.arange(a.size)+1))

Sample runs -

In [17]: a
Out[17]: array([ 1,  2,  3,  4,  5,  6,  7,  8,  9, 10])

In [18]: a.cumsum()/(np.arange(a.size)+1)
Out[18]: array([ 1. ,  1.5,  2. ,  2.5,  3. ,  3.5,  4. ,  4.5,  5. ,  5.5])

In [20]: a
Out[20]: array([3, 3, 2, 4, 6, 6, 3, 5, 6, 4])

In [21]: a.cumsum()/(np.arange(a.size)+1)
Out[21]: 
array([ 3.        ,  3.        ,  2.66666667,  3.        ,  3.6       ,
        4.        ,  3.85714286,  4.        ,  4.22222222,  4.2       ])

Answer 4

From Pandas 0.18.0 out of the box, as you wanted :)

s = pd.Series([1, 2, 3, 4, 5])
s.rolling(5, min_periods=1).mean()

result is:

0    1.0
1    1.5
2    2.0
3    2.5
4    3.0
dtype: float64