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Possible Duplicate:
Why does the expression 0 < 0 == 0 return False in Python?

The following output from the Python 2 REPL confuses me:

>>> 15>10==True
False
>>> 15>1==True
True
>>> 15>2==True
False
>>> 15>False
True

If 15>10==True is evaluated as (15>10)==True the expression would simplify to print True==True, which obviously evaluates to True. If 15>10==True is evaluated as 15>(10==True) the expression simplifies to 15>False which also evaluates to True. Both of these interpretations contradict the actual value of the expression (False).

I can understand 15>1==True evaluating to True since 1==True is true, but no interpretation of 15>10==True makes sense to me.

Summary: In Python 2, why does 15>10==True evaluate to False?

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Viktor Dahl
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2 Answers2

5

This is because chained comparison in Python. Namely, 15>10==True is actually evaluated as: 

15 > 10 and 10 == True

which is False.

On the other hand, 15>1==True is the same as

15 > 1 and 1 == True

which evaluates to True.

To quote from the docs:

Unlike C, all comparison operations in Python have the same priority, which is lower than that of any arithmetic, shifting or bitwise operation. Also unlike C, expressions like a < b < c have the interpretation that is conventional in mathematics:

comparison    ::=  or_expr ( comp_operator or_expr )*
comp_operator ::=  "<" | ">" | "==" | ">=" | "<=" | "<>" | "!="
                   | "is" ["not"] | ["not"] "in"

Comparisons yield boolean values: True or False.

Comparisons can be chained arbitrarily, e.g., x < y <= z is equivalent to x < y and y <= z, except that y is evaluated only once (but in both cases z is not evaluated at all when x < y is found to be false).

Formally, if a, b, c, ..., y, z are expressions and op1, op2, ..., opN are comparison operators, then a op1 b op2 c ... y opN z is equivalent to a op1 b and b op2 c and ... y opN z, except that each expression is evaluated at most once.

Note that a op1 b op2 c doesn’t imply any kind of comparison between a and c, so that, e.g., x < y > z is perfectly legal (though perhaps not pretty).

K Z
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2

I beleive you're unknowingly running into a "neat" feature of python operators:

Unlike C, expressions like a < b < c have the interpretation that is conventional in mathematics:

In [12]: 15>10==True
Out[12]: False

True happens to be treated as 1 in this expression, ie:

In [13]: 15>10==1
Out[13]: False

But then replacing == with in a > operator, we see how the new expression is true as per the rule quoted above:

In [14]: 15>10>1
Out[14]: True
Doug T.
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