CODEWITHSUNDEEP
pythonscipyfilteringsignal-processing
Astrid
Astrid

Reputation: 1936

What filter do I need? Want to keep high frequency values only

I am asking this question because I am not quite sure which filter I should be using.

Mine is simply a signal made up of discrete values as such s = [1 2 2 2 3 4 2 4 3 4 5 3 2 3 3]. Then I would like, per window size, a filtered signal. So e.g. if I use a window size of 5 for s then I would get; s_filtered = [2 2 2 2 2 4 4 4 4 4 3 3 3 3 3]. Hence I want to keep the values that are of the highest frequency in each block. For indices 0:4 (window size 5) the values of highest frequency is 2, so I want my "filtered" signal (if that is indeed the correct terminology) to have 2 in all indices 0:4 for the "filtered" signal.

Currently I am using just a median filter, but I do no think this is the correct approach.

Here is some python code to demonstrate what I am doing (but as said, which I think is wrong).

import numpy as np
import pylab *
from scipy.signal import medfilt

test = np.random.randint(10, size=1000)

fig, ax1 = plt.subplots(1,sharey=True, sharex=True, figsize=(15,5))
ax1.plot(test)
ax1.plot(medfilt(test,[99]),'r')
plt.show()

Random data with median filtered signal in red.

Where the red line is the filtered signal for a window size of 99.

SOLUTION:

import itertools
import collections

def grouper(iterable, n, fillvalue=None):
    args = [iter(iterable)] * n
    return itertools.izip_longest(*args, fillvalue=fillvalue)

s = [1, 2, 2, 2, 3, 4, 2, 4, 3, 4, 5, 3, 2, 3, 3]

list(chain.from_iterable(repeat(collections.Counter(x).most_common(1)[0][0],5) for x in grouper(s,5)))

Upvotes: 2

Views: 211

Answers (1)

Kasravnd
Kasravnd

Reputation: 107287

You can use grouper function from itertools recipes to group your array based on a specified length and then find the most common item using collections.Counter.most_common() method and use itertools.repeat to repeat your item 5 time and at last chain the repeated objects with itertools.chain.from_iterable : 

def grouper(iterable, n, fillvalue=None):
    args = [iter(iterable)] * n
    return zip_longest(*args, fillvalue=fillvalue)

Demo :

>>> list(chain.from_iterable(repeat(Counter(x).most_common(1)[0][0],5) for x in grouper(s,5)))
[2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3]

Upvotes: 1

Related Questions