I have the following numbers:
100, 200, 300, 400 ... 20000
And I would like to pick a random number within that range. Again, that range is defined as 100:100:20000. Furthermore, by saying 'within that range', I don't mean randomly picking a number from 100->20000, such as 105. I mean randomly choosing a number from the list of numbers available, and that list is defined as 100:100:20000.
How would I do that in Python?
Use random.randrange :
random.randrange(100, 20001, 100)
This is equivalent to choice(range(start, stop, step)), but doesn’t actually build a range object.random.randint(1, 2001) * 100 which it turns out is pretty much the exact same. A difference of about 1 - 2 hundredths of a second for 1M iterations.
random.randrange(100, 20001, 100) is clearer than random.randint(1, 2001) * 100. Performance should be similar because neither generates a sequence.For Python 3:
import random
random.choice(range(100, 20100, 100))
xrange would not avoid this problem. random.choice must evaluate the entire iterator from xrange for its choice to be random.
random.choice just generates a single random index and uses that to access the appropriate element of the sequence. Note that the argument to random.choice must support len and item access by index: random.choice doesn't accept arbitrary iterables.range.
randrange, though.range supports len and access by index.from python manual:
random.randrange([start], stop[, step])Return a randomly selected element from range(start, stop, step). This is equivalent to choice(range(start, stop, step)), but doesn’t actually build a range object.
This would do it
print random.choice(xrange(100, 20100, 100);
But this is probably better:
print int(round(random.randint(100, 200001),-2))
@mouad's answer looks the best to me.
EDIT
Out of curiosity I tried a benchmark of this answer vs. random.randint(1, 2001) * 100 which it turns out is pretty much the exact same. A difference of about 1 - 2 hundredths of a second for 1M iterations.