[SciPy-User] Identify unique sequence data from array

Skipper Seabold jsseabold@gmail....
Wed Dec 22 22:49:02 CST 2010


On Wed, Dec 22, 2010 at 6:00 PM, Charles R Harris
<charlesr.harris@gmail.com> wrote:
>
>
> On Wed, Dec 22, 2010 at 2:51 PM, Robert Kern <robert.kern@gmail.com> wrote:
>>
>> On Wed, Dec 22, 2010 at 16:47, Paul Anton Letnes
>> <paul.anton.letnes@gmail.com> wrote:
>> >
>> > On 22. des. 2010, at 21.18, otrov wrote:
>> >
>> >>>> The problem:
>> >>
>> >>>> I have 2D data sets (scipy/numpy arrays) of 10^7 to 10^8 rows, which
>> >>>> consists of repeated sequences of one unique sequence, usually ~10^5 rows,
>> >>>> but may differ in scale. Period is same for both columns, so there is not
>> >>>> really difference if we consider 2D or 1D array.
>> >>>> I want to track this data block.
>> >>
>> >>> for i in range(1, len(X)-1):
>> >>>    if (X[i:] == X[:-i]).all():
>> >>>        break
>> >>
>> >> Just look at that python beauty! Such a great language when in hand of
>> >> a smart user.
>> >> Thanks for you snippet, but unfortunately it takes forever to finish
>> >> the task
>> >
>> > You could also check one element at a time. I think it will be faster,
>> > because it will break if comparison of the first element doesn't hold. Then,
>> > if you find such an occurrence, use Robert's method to double check that you
>> > found the true repetition period.
>>
>> Excellent point.
>>
>
> Why not do an FFT and look at the shape around the carrier frequency? The DC
> level should probably be subtracted first. It shoud also be possible to
> construct a Weiner filter to extract the sequences if they don't occur with
> strict periods.
>

Could you give an example?  I've used a convolution to find the number
of successive discrete events, but I'm not sure how to generalize it
or if your suggestion is similar.

For example, to count the number of three successes in a row for
Bernoulli trials and to find where

In [1]: import numpy as np

In [2]: x = np.array([1,1,1,0,0,1,0,1,0,1,1,1,0,1,0,1])

In [3]: y = np.array([1,1,1])

In [4]: idx = np.convolve(x,y)

In [5]: num_runs = len(np.where(idx==len(y))[0])

In [6]: # Extract runs from original array

In [6]: idx = np.where(idx==len(y))[0]-(len(y)-1)

In [7]: idx = np.hstack([np.arange(i,i+3) for i in idx])

In [8]: x[idx]
Out[8]: array([1, 1, 1, 1, 1, 1])

Skipper


More information about the SciPy-User mailing list