Shuffling a list

Question

in this program I'm trying to shuffle a list by randomly choosing two items from a list and swapping them round, and then repeating this process several times.

The problem I have encountered is I don't know how I can swap the items round and print the shuffled list.

For instance if my two random values were a and b, if I were to just put:

a = b
b = a

then that would change the value of a to b, but when it tries to change b to a, no change would occur as a has already been changed to b.

The only way I can think that this would work is swapping them at the same time, but I do not know of a function/ way to swap them round.

Also if a, b were items of a list L, after I swapped them round if I used

print L

should it print the altered version? I only ask because from what I have tried it is not doing that.

NB I am trying to shuffle this list stage by stage by swapping, instead of using the shuffle function imported from random.

The classic way of swapping back when I were a lad, was to use a temporary variable: tmp=a, a=b, b=tmp - but python is smarter than BASIC was then! [too lame to deserve to be an answer] — Spacedman, Nov 16 '11 at 17:25

score 4 · Answer 1 · edited May 23 '17 at 11:47

4

In Python, you can swap two variables like this:

a, b = b, a

This is called multiple assignment, you can find more information about it here.

In other languages this is usually done by assigning a temporary variable:

tmp = a
a = b
b = tmp

Isn't Python great?

edited May 23 '17 at 11:47

Community

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answered Nov 16 '11 at 17:22

Andrew Clark

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Raymond Hettinger · Answer 2 · 2011-11-16T17:29:39.697

The random.shuffle function uses swapping too. It would be worthwhile to look at its source code:

def shuffle(self, x, random=None, int=int):
    """x, random=random.random -> shuffle list x in place; return None.

    Optional arg random is a 0-argument function returning a random
    float in [0.0, 1.0); by default, the standard random.random.
    """

    if random is None:
        random = self.random
    for i in reversed(xrange(1, len(x))):
        # pick an element in x[:i+1] with which to exchange x[i]
        j = int(random() * (i+1))
        x[i], x[j] = x[j], x[i]

Observe how the last line performs a swap using tuple packing and unpacking.

As an alternative to packing and unpacking, the traditional way to swap variables is to use a temporary variable:

t    = x[i]
x[i] = x[j]
x[j] = t

score 3 · Accepted Answer · answered Nov 16 '11 at 17:24

Use a temp variable for your first problem:

temp = a
a = b
b = temp

In Python you can also do this:

a, b = b, a

I suspect your second problem is because you're changing things you got out of the list, instead of changing the list. Try this:

i, j = # two indexes to swap in the list
L[i], L[j] = L[j], L[i]

score 1 · Answer 4 · answered Nov 16 '11 at 17:24

1

Use a temporary variable:

temp = a
a = b
b = temp

answered Nov 16 '11 at 17:24

Oliver

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score 0 · Answer 5 · answered Nov 16 '11 at 20:32

http://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle

Just swapping items is known to be bad.

Don't forget that if you have n items, there are n! arrangements. If your random number is 32 bits, there are 2^32 numbers.

It's hard then to shuffle a pack of cards with a 32 bit number as 52! is very much bigger than 2^32

Shuffling a list

5 Answers5