CODEWITHSUNDEEP
pythonlistnumpyrandom
bapors
bapors

Reputation: 909

Choosing more than 1 element from the list

I am trying to make a choice for each element in elements, and then I am pairing up the element in elements list with its preferential choice(one,two or three). The choice is mostly done regarding the probabilities (weights) of the elements. The code until here:

from numpy.random import choice
elements = ['one', 'two', 'three']
weights = [0.2, 0.3, 0.5]
chosenones= []
for el in elements:
    chosenones.append(choice(elements,p=weights))
tuples = list(zip(elements,chosenones))

Yields:

[('one', 'two'), ('two', 'two'), ('three', 'two')]

What I need is, for each element to make two choices instead of one.

The expected output should look like:

[('one', 'two'), ('one', 'one'), ('two', 'two'),('two', 'three'), ('three', 'two'), ('three', 'one')]

Do you know how to do have this output?

Upvotes: 0

Views: 147

Answers (2)

Martijn Pieters
Martijn Pieters

Reputation: 1125058

If you need two, just tell numpy.random.choice() to pick two values; include the el value as a tuple as you loop (no need to use zip()):

tuples = []
for el in elements:
    for chosen in choice(elements, size=2, replace=False, p=weights):
        tuples.append((el, chosen))

or by using a list comprehension:

tuples = [(el, chosen) for el in elements
          for chosen in choice(elements, size=2, replace=False, p=weights)]

By setting replace=False, you get unique values; remove it or set it to True explicitly to allow for repetition. See the numpy.random.choice() documentation:

size : int or tuple of ints, optional
Output shape. If the given shape is, e.g., (m, n, k), then m * n * k samples are drawn. Default is None, in which case a single value is returned.

replace : boolean, optional
Whether the sample is with or without replacement

Demo:

>>> from numpy.random import choice
>>> elements = ['one', 'two', 'three']
>>> weights = [0.2, 0.3, 0.5]
>>> tuples = []
>>> for el in elements:
...     for chosen in choice(elements, size=2, replace=False, p=weights):
...         tuples.append((el, chosen))
...
>>> tuples
[('one', 'three'), ('one', 'one'), ('two', 'three'), ('two', 'two'), ('three', 'three'), ('three', 'two')]
>>> [(el, chosen) for el in elements for chosen in choice(elements, size=2, replace=False, p=weights)]
[('one', 'one'), ('one', 'three'), ('two', 'one'), ('two', 'three'), ('three', 'two'), ('three', 'three')]

Upvotes: 1

Right leg
Right leg

Reputation: 16740

If you accept duplicates, random.choices will do the job:

random.choices(population, weights=None, *, cum_weights=None, k=1)

Return a k sized list of elements chosen from the population with replacement. If the population is empty, raises IndexError.

If a weights sequence is specified, selections are made according to the relative weights.

>>> random.choices(['one', 'two', 'three'], weights=[0.2, 0.3, 0.5], k=2)
['one', 'three']

Upvotes: 1

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