Velocity Reviews > Looking for module for shrinking a list with n-point means

# Looking for module for shrinking a list with n-point means

Yash Ganthe
Guest
Posts: n/a

 05-22-2009
Hi,

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 SciPy or NumPy or any other module have functions for achieving
this?

Which function can be used for doing this?

Thanks,
Yash

John Machin
Guest
Posts: n/a

 05-22-2009
On May 22, 8:03*pm, Yash Ganthe <(E-Mail Removed)> wrote:
> Hi,
>
> 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 SciPy or NumPy

What do their docs say?

> or any other module have functions for achieving
> this?
>
> Which function can be used for doing this?

Perhaps one like this:

| >>> def n_point_means(alist, n):
| ... blist = alist[:]
| ... blist.sort()
| ... size = len(blist)
| ... assert 1 <= n <= size
| ... assert size % n == 0
| ... clist = []
| ... fn = float(n)
| ... for i in xrange(0, size, n):
| ... clist.append(sum(blist[i:i+n]) / fn)
| ... return clist
| ...
| >>> aaa = [9,8,7,6,5,4,3,2,0]
| >>> n_point_means(aaa,2)
| Traceback (most recent call last):
| File "<stdin>", line 1, in <module>
| File "<stdin>", line 6, in n_point_means
| AssertionError
| >>> aaa = [9,8,7,6,5,4,3,2,1,0]
| >>> n_point_means(aaa,2)
| [0.5, 2.5, 4.5, 6.5, 8.5]
| >>> n_point_means(aaa,5)
| [2.0, 7.0]
| >>>

Does that do what you want? If your requirement is so simple, why not
write it yourself?

Robert Kern
Guest
Posts: n/a

 05-22-2009
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(12
> v.shape = (16,
> 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
an underlying truth."
-- Umberto Eco

Yash Ganthe
Guest
Posts: n/a

 05-23-2009
Thanks John,

The code u provided works for me. Indeed it is a simple requirement
and I am a complete novice to Python.

-Yash