63

It has always seemed strange to me that random.randint(a, b) would return an integer in the range [a, b], instead of [a, b-1] like range(...).

Is there any reason for this apparent inconsistency?

dreftymac
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David Wolever
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    (Not answering the question:) Well, you could use `randrange`. – kennytm Apr 02 '10 at 19:40
  • Maybe I'm just not as savvy as you, but I don't see it as an inconsistency. I see it as just providing a method for for programmers to call an inclusively random number without needing to use operators and magic numbers. The better languages give their users avenues through which they can be creative. – Musixauce3000 Apr 29 '16 at 13:45
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    @Musixauce3000 it's an inconsistency because both methods – `range` and `randint` – accept the same logical value (ie, a range of numbers), but that value is specified in two different – inconsistent – ways (ie, one is inclusive of the endpoint, the other is exclusive). – David Wolever Apr 29 '16 at 20:20
  • Why would you want it to be *less* intuitive? – endolith Oct 12 '19 at 01:49

4 Answers4

97

I tried to get to the bottom of this by examining some old sources. I suspected that randint was implemented before Python's long integer: meaning that if you wanted a random number that included INT_MAX, you would have needed to call random.randrange(0, INT_MAX + 1) which would have overflowed and resulted in arguments of (0, 0) or (0, INT_MIN) depending.

However, looking as far back as even the Python 1.5.2 sources, in Lib/whrandom.py we see:

#
# Get a random integer in the range [a, b] including both end points.
# (Deprecated; use randrange below.)
#
def randint(self, a, b):
    return self.randrange(a, b+1)

whrandom.randint was continued to be deprecated in 2.0, 2.1, 2.2, and 2.3; but random.randint was marked as deprecated in 2.1, although no longer marked as deprecated in 2.2.

Also, random.py from version 2.1 is the first to note in random.randint's docstring:

def randrange(self, start, stop=None, step=1, int=int, default=None):
    """Choose a random item from range(start, stop[, step]).

    This fixes the problem with randint() which includes the
    endpoint; in Python this is usually not what you want.
    Do not supply the 'int' and 'default' arguments.
    """

The only available source older than that is the 0.9.1 source, and as far as I can tell, randint was not implemented at that point.

Thus, I conclude that the reasoning for randint including the endpoint is known to only Guido himself at this point; given the docstring from Python 2.1, it sounds like the reason may have been a simple mistake.

Mark Rushakoff
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16

I guess random.randint was just the first attempt at implementing this feature. It seems that the Python developers also felt that this was a problem, which is why in v1.5.2 they added another method randrange with more standard parameters:

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.

You can use randrange instead of randint to avoid surprising people.

On the other hand, in many situations where the problem is phrased as 'choose a random number between 1 and 6' it might be more natural to use randint(1, 6) instead of writing randrange(1, 7) or randrange(min, max + 1).

Community
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Mark Byers
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10

This is speculation, but normal human usage of 'give me a random number from a to b' is inclusive. Implementing it that way sort of makes sense, given Python's general philosophy of being a more human-readable language.

DNS
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3

I don't think there's a reason for that. But at least it's documented.

Wieland
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