Given a list x = [1,0,0,1,1]I can use random.shuffle(x) repeatedly to shuffle this list, but if I try to do this a for loop the list doesn't shuffle.
For example:
x = [1,0,0,1,1]
k = []
for i in range(10):
random.shuffle(x)
k.append(x)
return x
Basically, kcontains the same sequence of x unshuffled? Any work around?
One pythonic way to create new random-orderings of a list is not to shuffle in place at all. Here is one implementation:
[random.sample(x, len(x)) for _ in range(10)]
Explanation
random.sample creates a new list, rather than shuffling in place.len(x) is the size of the sample. In this case, we wish to output lists of the same length as the original list.As mentioned by @jonrsharpe, random.shuffle acts on the list in place. When you append x, you are appending the reference to that specific object. As such, at the end of the loop, k contains ten pointers to the same object!
To correct this, simply create a new copy of the list each iteration, as follows. This is done by calling list() when appending.
import random
x = [1,0,0,1,1]
k = []
for i in range(10):
random.shuffle(x)
k.append(list(x))
Try this:
x = [1,0,0,1,1]
k = []
for i in range(10):
random.shuffle(x)
k.append(x.copy())
return x
By replacing x with x.copy() you append to k a new list that looks like x at that moment instead of x itself.
random.shuffleis in place. You're fillingkwith references to the same list, so all ten lists in the result will be in the same random order. Maybe try a shallow copy,k.append(x[:])?kactually contains the same sequence ofxshuffled, but there are only so many unique orders so sometimes it will be the same as your starting position.