I was recently learning how to use random.shuffle in Python and I was surprised to see that the function shuffles the variable in place without returning anything. How is this acheived in the function? Looking at the random library's source code yielded no answer.
How can I write my own function that changes the variable in place without reassignment?
I is works because of lists are mutable. You cant reassign any variable because it will be new variable. You cant modify immutable typed variable. You can modify mutable variable.
So:
>>> def addone(x):
... x += 1
>>> a = 2
>>> addone(a)
>>> a
2
>>> def addone(x):
... x.append(1)
...
>>> l=[2]
>>> addone(l)
>>> l
[2, 1]
>>> def addone(x):
... x = x + [1]
...
>>> li=[2]
>>> addone(li)
>>> li
[2]
Well, just look at the implementation of random.shuffle. Find it in a file called random.py, it's a pure python implementation and quite simple - just using a loop of assignments (with tuple unpacking).
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.
Do not supply the 'int' argument.
"""
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]
random.shuffledoesn't replace the list with a new list; it changes the list's state.