How to replace a list comprehension with a numpy command?

Question

Is there a way to replace the following python list comprehension with a numpy function that doesn't work with loops?

a = np.array([0, 1, 1, 1, 0, 3])

bins = np.bincount(a)
>>> bins: [2 3 0 1]


a_counts = [bins[val] for val in y_true]
>>> a_counts: [2, 3, 3, 3, 2, 1]

So the basic idea is to generate an array where the actual values are replaced by the number of occurrences of that specific value in the array.

I want to do this calculation in a custom keras loss function which, to my knowledge, doesn't work with loops or list comprehensions.

yatu · Accepted Answer · 2019-05-28 13:13:00Z

3

You just need to index the result from np.bincount with a:

a = np.array([0, 1, 1, 1, 0, 3])
bins = np.bincount(a)
a_counts = bins[a]

print(a_counts)
# array([2, 3, 3, 3, 2, 1], dtype=int64)

edited May 28, 2019 at 13:13

answered May 28, 2019 at 12:28

yatu

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1 Comment

LuckerNo1 Over a year ago

Little follow up question: I tried doing the same thing in keras using tf backend with the same arrays: python bins = k.tf.bincount(y_true) y_true_counts = bins[y_true] Unfortunately I get the following error: Shape must be rank 1 but is rank 2 for 'strided_slice_10' (op: 'StridedSlice') with input shapes: [?], [1,6], [1,6], [1].

U13-Forward · Accepted Answer · 2019-05-28 12:27:33Z

0

Or use collections.Counter:

from collections import Counter
l = [0, 1, 1, 1, 0, 3]
print(Counter(l))

Which Outputs:

Counter({1: 3, 0: 2, 3: 1})

answered May 28, 2019 at 12:27

U13-Forward

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Comments

Dilshat · Accepted Answer · 2019-05-28 12:34:00Z

0

If you want to avoid loops, you may use pandas library:

import pandas as pd
import numpy as np
a = np.array([0, 1, 1, 1, 0, 3])
a_counts = pd.value_counts(a)[a].values
>>> a_counts: array([2, 3, 3, 3, 2, 1], dtype=int64)

answered May 28, 2019 at 12:34

Dilshat

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