Populate list or tuple from callable or lambda in python

Question

This is a problem I've come across a lot lately. Google doesn't seem to have an answer so I bring it to the good people of stack overflow.

I am looking for a simple way to populate a list with the output of a function. Something like this:

fill(random.random(), 3) #=> [0.04095623, 0.39761869, 0.46227642]

Here are other ways I've found to do this. But I'm not really happy with them, as they seem inefficient.

results = []
for x in xrange(3): results.append(random.random())
#results => [0.04095623, 0.39761869, 0.46227642]

and

map(lambda x: random.random(), [None] * 3)
#=> [0.04095623, 0.39761869, 0.46227642]

Suggestions?

---

Thanks for all the answers. I knew there was a more python-esque way.

And to the efficiency questions...

$ python --version
Python 2.7.1+
$ python -m timeit "import random" "map(lambda x: random.random(), [None] * 3)"
1000000 loops, best of 3: 1.65 usec per loop
$ python -m timeit "import random" "results = []" "for x in xrange(3): results.append(random.random())"
1000000 loops, best of 3: 1.41 usec per loop
$ python -m timeit "import random" "[random.random() for x in xrange(3)]"
1000000 loops, best of 3: 1.09 usec per loop

Answer 1

How about a list comprehension?

[random.random() for x in xrange(3)]

Also, in many cases, you need the values just once. In these cases, a generator expression which computes the values just-in-time and does not require a memory allocation is preferable:

results = (random.random() for x in xrange(3))
for r in results:
   ...
# results is "used up" now.
# We could have used results_list = list(results) to convert the generator

By the way, in Python 3.x, xrange has been replaced by range. In Python 2.x, range allocates the memory and calculates all values beforehand (like a list comprehension), whereas xrange calculates the values just-in-time and does not allocate memory (it's a generator).