Python Cookbook 2Nd Edition Jun 1002005 [Electronic resources] نسخه متنی

Recipe 18.3. Generating Random Samples with Replacement

Credit: Sean Ross

Problem

You need to generate random samples with replacement out of a
"population" of items that are held
in a sequence.

Solution

A generator for the purpose is quintessentially simple:

import random
def sample_wr(population, _choose=random.choice):
while True: yield _choose(population)

Discussion

random.sample lets you do random sampling without
replacement, and Recipe 18.4, which follows, shows a
generator to perform sampling without replacement with excellent
memory-consumption characteristics. This recipe provides a generator
for sampling with replacement, which is an even
simpler task. Basically all the work gets delegated to
random.choice. The sample_wr
generator shown in this recipe is unbounded: used on its own, it will
keep looping forever. To bound the output of an intrinsically
unbounded generator, you can use it in a for
statement that at some point executes a break, or
use other techniques shown in Recipe 19.2.

For example, to make a random string of 50 lowercase ASCII letters:

import itertools
from string import ascii_lowercase
x = ''.join(itertools.slice(sample_wr(ascii_lowercase), 50))

string.ascii_lowercase is exactly the string
'abcdefghijklmnopqrstuvwxyz'. If you
didn't have the sample_wr
generator, the equivalent code might be something like:

from string import ascii_lowercase
from random import choice
x = ''.join([ random.choice(ascii_lowercase) for i in xrange(50) ])

So, the practical added-value of sample_wr is
modest, when compared to other available building-blocks. It is,
however, preferable to have such a fundamental concept of statistics
as sampling with replacement embodied as its own
function, rather than having to implement it with an explicit loop
over random.choice each time it is needed.

See Also

Library Reference and Python in a
Nutshell docs for module random.