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pythonnumpyrandom
Kurt Peek
Kurt Peek

Reputation: 57471

Using Numpy's random.choice to randomly remove item from list

According to the http://docs.scipy.org/doc/numpy-dev/reference/generated/numpy.random.choice.html, using the replace = False with Numpy's random.choice method should make the sample without replacement. However, this does not seem to work for me:

In [33]: import numpy as np

In [34]: arr = range(5)

In [35]: number = np.random.choice(arr, replace = False)

In [36]: arr
Out[36]: [0, 1, 2, 3, 4]

The array arr is still range(5) after sampling, and not missing a (random) number as I would expect. How could I sample a number from range(5) without replacement?

Upvotes: 4

Views: 14712

Answers (2)

Jblasco
Jblasco

Reputation: 3967

As mentioned in one of the comments, np.choice selects with or without replacement, a series of numbers from a sequence. But it does not modify the sequence.

Easy alternative

arr = range(5)
# numbers below will never contain repeated numbers (replace=False)
numbers = np.random.choice(arr, 3, replace=False)

The behaviour I think you want would be:

arr = range(5)
all_but_one = np.random.choice(arr, len(arr) -1, replace=False)

so you would select N-1 numbers without replacement (to avoid repetitions), effectively removing a random element from the iterable.

More efficient alternative 

arr = range(5)
random_index = np.random.randint(0, len(arr))
arr.pop(random_index)

Upvotes: 12

Kurt Peek
Kurt Peek

Reputation: 57471

I ended up defining a function using the random library:

import random
def sample_without_replacement(arr):
    random.shuffle(arr)
    return arr.pop()

Its use is shown below:

In [51]: arr = range(5)

In [52]: number = sample_without_replacement(arr)

In [53]: number
Out[53]: 4

In [54]: arr
Out[54]: [2, 0, 1, 3]

Note that the method also shuffles the array in place, but for my purposes that doesn't matter.

Upvotes: 1

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