Numpy: Find maximum value and corresponding location along one axis in 3D array

Question

I am looking for an efficient way to find the maximum value and corresponding index along the first axis of a 3D array using Numpy.

Context: I am working with data stored in a 3D array z.shape() = (t, x, y) and have taken a Fourier transform along the t axis using

spectrum = np.fft.fft(z, axis=0)
spectrum.shape(f, x, y)

Here, f is the frequency space, having the same size as t.

I need to construct an array peak_vals.shape() = (x, y) which holds the highest value of the transformed data for each x- and y-coordinate. Moreover, I need to store the frequency corresponding to each peak: peak_freqs.shape() = (x, y)

I already have a function that computes the frequencies corresponding to the Fourier transform:

def freq_arr(t):
    """Determine the frequency space that a DFT of vector t would have"""
    samp_freq = 1/(t[1]-t[0])
    return np.arange(len(t)) * samp_freq / len(t)

To find the peak along the f axis, I came up with:

peak_vals = np.amax(spectrum, axis=0)

How can I find the frequency corresponding to these peaks? I would like to construct a function

def peaks(freqs, spectrum):
    peak_vals = np.amax(spectrum, axis=0)
    # do something
    return peak_vals, peak_freqs

I want to avoid for-loops since I will be working with large arrays.

np.argmax returns the index where np.amax occurs.

hpaulj
– hpaulj

2019-02-15 22:35:52 +00:00
Commented Feb 15, 2019 at 22:35 — hpaulj
– hpaulj, Commented Feb 15, 2019 at 22:35

p13rr0m · Accepted Answer · 2019-02-16 00:04:06Z

1

def peaks(freqs, spectrum):
    peak_vals = np.amax(spectrum, axis=0)

    # get the indices with the max value
    peak_idx = np.argmax(spectrum, axis=0)

    # now use the indices to get the corresponding frequencies
    # peak_freqs is now 2d even though freqs is 1d
    peak_freqs = freqs[peak_idx]

    return peak_vals, peak_freqs

edited Feb 16, 2019 at 0:04

answered Feb 15, 2019 at 22:56

p13rr0m

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3 Comments

falidoro Over a year ago

freqs is a 1D-array, since it is just a transformed time axis., so in can't be unpacked into f, x, y. I also don't fully understand the flatten() part.

p13rr0m Over a year ago

Sorry, my bad. I must have overread that the freqs is just a 1D-array. Then the problem becomes even simpler, without using the flatten part.

falidoro Over a year ago

Indeed seems to do the trick. Thank you. I wasn't aware of this manner of passing indices.

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