The // "integer division" operator of Python surprised me, today:
>>> math.floor(11/1.1)
10.0
>>> 11//1.1
9.0
The documentation reads "(floored) quotient of x and y". So, why is math.floor(11/1.1) equal to 10, but 11//1.1 equal to 9?
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209answers:
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A:
Because 1.1 can't be represented in binary form exactly; the approximation is a littler higher than 1.1 - therefore the division result is a bit too small.
Try the following:
Under Python 2, type at the console:
>>> 1.1
1.1000000000000001
In Python 3.1, the console will display 1.1, but internally, it's still the same number.
But:
>>> 11/1.1
10.0
As gnibbler points out, this is the result of "internal rounding" within the available precision limits of floats. And as The MYYN points out in his comment, // uses a different algorithm to calculate the floor division result than math.floor() in order to preserve a == (a//b)*b + a%b as well as possible.
Use the Decimal type if you need this precision.
Tim Pietzcker
2010-01-07 10:36:04
But according to Python `11.0/1.1 == 10.0` is true
sth
2010-01-07 10:40:33
As we know 1.1 is stored as 1.1000000000000001, so the result 11.0/1.1 in Python is 9.999999999999999090909090909 but the closest float to this number is infact 10.0, so it happens that the result is exactly 10.0
gnibbler
2010-01-07 11:22:00
@Tim I thought of that, and I believe this is the key to the difference between math.floor(11/1.1) and 11//1.1. However, as gnibbler pointed out, the fact that 1.1 is represented by a slightly larger number only indicates that math.floor(11/1.1) should really be 9.0, which leaves the original question still open.
EOL
2010-01-07 12:45:15
@EOL, since 11/1.1 evaluates to exactly 10.0, math.floor(11/1.1)=math.floor(10.0), however this is only due to some luck with the rounding
gnibbler
2010-01-07 22:36:00
@gnibbler: Yeah, it's the "luck" aspect that surprised me: two "identical" operations behaving in a different way. :) The explanation is in the implementation, as pointed out by The MYYN.
EOL
2010-01-08 12:58:33