Splitting values in an array 'logarithmically' / based on another array

user_3696204 0
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I have a 2d array, where each element is a fourier transform. I'd like to split transform 'logarithmically'. For example, let's take a single one of those arrays and call it a:

a = np.arange(0, 512)

# I want to split a into 'bins' defined by b, below:
b = np.array([0] + [10 * 2**i for i in range(6)]) # [0, 10, 20, 40, 80, 160, 320, 640]

What I'm looking to do is something like using np.split, except I would like to split values into 'bins' based on array b such that all values of a between [0, 10) are in one bin, all values between [10, 20) in another, etc.

I could do this in some sort of convoluted for loop:

split_arr = []
for i in range(1, len(b)):
    fbin = []
    for amp in a:
        if (amp >= b[i-1]) and (amp < b[i]):
            fbin.append(amp)
    split_arr.append(fbin)

I have many arrays to split, and also this is ugly (just my opinion). Is there a better way?

numpy arrays python
2 Answers

Here is how you can do it, using np.split:

np.split(a, np.searchsorted(a,b))

If your array a is not sorted, sort it before the above command:

a = np.sort(a)

np.searchsorted finds the locations of values in b that would be inserted in the sorted array a. In other words, np.searchsorted finds the locations where you want to split your array. And if you do not want the empty array at the beginning, simply remove 0 from b.

user_49759810

First you can reduce the 'ugliness' by using list comprehension:

split_arr = [[amp for amp in a if (amp >= b[i-1]) and (amp < b[i])] for i in range(1, len(b))]

Then you can apply the same logic using numpy fast parallelized functionalities (which has the bonus of looking even cleaner):

split_arr = [a[(a >= b[i-1]) & (a < b[i])] for i in range(1, len(b))]

Comparison:

%timeit [[amp for amp in a if (amp >= b[i-1]) and (amp < b[i])] for i in range(1, len(b))]
1.29 ms ± 109 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit [a[(a >= b[i-1]) & (a < b[i])] for i in range(1, len(b))]
35.9 µs ± 4.52 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
user_45659470
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