Looking for module for shrinking a list with n-point means
Robert Kern
robert.kern at gmail.com
Fri May 22 21:06:44 CEST 2009
On 2009-05-22 08:50, Scott David Daniels wrote:
> Yash Ganthe wrote:
>> I would like to shrink a large list in the following way:
>> If the List has 1000 integers, we need only 100 averages such that the
>> 1000 points are average for every 10 consecutive values. So p0 to p9
>> will be averaged to obtain t0. p10 to p19 will be averaged to obtain
>> t1 and so on. This is a 10-point mean.
>>
>> We are doing this as we collect a lot of data and plot it on a graph.
>> Too many samples makes the graph cluttered. So we need to reduce the
>> number of values in the way described above.
>
> Does this give you a clue?
>
> import numpy as np
>
> v = np.arange(128)
> v.shape = (16, 8)
> sum(v.transpose()) / 8.
Or even:
import numpy as np
v = np.arange(1000).reshape((-1, 10))
ten_point_mean = v.mean(axis=1)
--
Robert Kern
"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco
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