Selecting element in a list with constraint in Python

Question

I have a list [1,2,3,3], and I want to select a random element from this list multiple times. However, I don't want my random generator to select the same element from the same index more than once.

What I'm currently have:

[1,2,3,3] --> [2] with index 1
[1,2,3,3] --> [1] with index 0       
[1,2,3,3] --> [2] with index 1       (this is wrong because chose the same index)

What I want is:

[1,2,3,3] --> [2] with index 1
[1,2,3,3] --> [1] with index 0       
[1,2,3,3] --> [3] with index 3       
[1,2,3,3] --> [3] with index 4       (this is perfect, no repeats!)

What should I do to solve this issue? The function random.choice(...) itself doesn't solve this.

UPDATE: I noticed some of you recommended me to use shuffle. It was a really good idea. But, what if I want to keep track of the original index too at later time? I don't think shuffle and pop could do that, right?

please see this post; https://stackoverflow.com/questions/9755538/how-do-i-create-a-list-of-unique-random-numbers — SteveJ, Dec 01 '17 at 05:16
I would use `shuffle` or something, then pop the list until it's empty. — Shadow, Dec 01 '17 at 05:17

user2390182 · Answer 1 · 2017-12-01T05:46:31.687

2

The following generator produces (index, element) pairs in random order while keeping the original list as it is:

def gen(lst):
    lst2 = list(enumerate(lst))
    random.shuffle(lst2)
    for x in lst2:
        yield x

l = [1, 2, 3, 3]
for index, elmnt in gen(l):
    # do stuff
    print(index, elmnt)
# 0 1
# 2 3
# 3 3
# 1 2

edited Dec 01 '17 at 05:46

answered Dec 01 '17 at 05:33

user2390182

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You are using `enumeration` just what I need! Thanks. – ThomasWest Dec 01 '17 at 05:39

thanasisp · Accepted Answer · 2017-12-01T05:47:57.833

1

shuffle the list once and then pop() returns the last element, reducing the list.

>>> from random import shuffle
>>> a=[1,2,3,3]
>>> shuffle(a)
>>> a.pop()
3
>>> a.pop()
2
>>> a.pop()
3
>>> a.pop()
1

edit: to keep track of the index using enumerate

from random import shuffle
a=[1,2,3,3]
b=list(enumerate(a))
shuffle(b)
b.pop()

(3, 3)

docs: enumerate and random.shuffle

edited Dec 01 '17 at 05:47

answered Dec 01 '17 at 05:25

thanasisp

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What if I want to keep track of the original index too tho? – ThomasWest Dec 01 '17 at 05:28
you can use `enumerate`. https://docs.python.org/2/library/functions.html#enumerate – thanasisp Dec 01 '17 at 05:36
@thanasisp If you add the enumerate solution I will give +1 vote. Also because the `enumerate` zip will be non-destructive. – Jan Christoph Terasa Dec 01 '17 at 05:38

score 0 · Answer 3 · answered Dec 01 '17 at 05:20

0

You can use random.sample.

random.sample(population, k)

Returns a k length list of unique elements chosen from the population sequence or set. Used for random sampling without replacement.

answered Dec 01 '17 at 05:20

kchawla-pi

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score 0 · Answer 4 · answered Dec 01 '17 at 05:34

0

Use @thanasisp's solution and use shuffle along with zip to keep track of the indices:

a = range(10, 15)
b = zip(a, range(len(a))) # pairs of values and indices

import random
random.shuffle(list(b)) # list() evaluates the zip

a_shuffle, idxes = list(zip(*b))
# e.g. [(13, 11, 10, 14, 12), (3, 1, 0, 4, 2)]

answered Dec 01 '17 at 05:34

Jan Christoph Terasa

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@thanasisp suggestion ot use `enumerate` is even better, because that bascially is what this zip does. :-) – Jan Christoph Terasa Dec 01 '17 at 05:37
I see, Im going to give it a try. Thanks! – ThomasWest Dec 01 '17 at 05:37

Selecting element in a list with constraint in Python

4 Answers4