This question already has answers here:
What does it mean that Python comparison operators chain/group left to right?
(2 answers)
Why does the expression 0 < 0 == 0 return False in Python?
(9 answers)
Where in the python docs does it allow the `in` operator to be chained?
(1 answer)
Closed 5 years ago.
I just stumbled upon the following line in Python 3.
1 in range(2) == True
I was expecting this to be True since 1 in range(2) is True and True == True is True.
But this outputs False. So it does not mean the same as (1 in range(2)) == True. Furthermore it does not mean the same as 1 in (range(2) == True) which raises an error.
Despite years of experience in Python, I am taken off guard. What is going on?
This is due to the fact that both operators are comparison operators, so it is being interpreted as operator chaining:
https://docs.python.org/3.6/reference/expressions.html#comparisons
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).
So it is equivalent to:
>>> (1 in range(2)) and (range(2) == True)
False
Related
This question already has answers here:
Why does the expression 0 < 0 == 0 return False in Python?
(9 answers)
Closed 1 year ago.
why python 3 evaluate 3 in [] == False to be False?
But I know (3 in []) == False is True.
This is caused by chained comparisons
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).
Consider the case 1 < 2 < 3. This really breaks out to (1 < 2) and (2 < 3)
So the statement 3 in [] == False breaks out to (3 in []) and ([] == False)
This question already has answers here:
Is False == 0 and True == 1 an implementation detail or is it guaranteed by the language?
(3 answers)
Closed 2 years ago.
I started learning python from a site, and it gave me a question that was kind of weird, I had to add and multiply a number to True and False. I've never done that before, so could someone explain what that means? (Can you replace True with 1 and False with 0?)
This is the question:
What is the value of result at the end of the following code?
x = 20
y = 5
result = (x + True) / (4 - y * False)
A)
False
B)
-21
C)
5
D)
5.25
The answer is 5.25 because in Python, True is equal to 1 while False is equal to 0.
This question already has answers here:
Python boolean expression and or
(4 answers)
Closed 2 years ago.
I came across the below expression in python to evaluate if a number is odd or even. While it looks elegant I was surprised to see it work. I guess for the input 3:
x%2 equals 1 which evaluates to True and non-empty strings (odd) evaulate to True. So we have True and True or <don't have to check because it doesn't matter if True or False as the overall statement will be True>. But why is the output in that case not True but odd - what concept do I miss?
>>> (lambda x:
... (x % 2 and 'odd' or 'even'))(3)
'odd'
The and and or operators are not limited to True and False; they return one of their operands.
>>> 1 and "odd"
'odd'
>>> 0 or "even"
'even'
This question already has answers here:
python operator precedence of in and comparison [duplicate]
(3 answers)
Closed 2 years ago.
Suppose I have the following line of code:
print("valley" in "hillside" == False)
Since the precedence of in and == is equivalent in Python, I expected the operations to be performed from left to right, producing True as the output.
However, in actuality, when I run this line of code, I get False.
I have noticed that adding brackets around "valley" in "hillside" results in True as the output but I don't seem to understand why it's necessary in the first place...
Both in and == are comparison operators, so the parser treats
print("valley" in "hillside" == False)
the same as
print("valley" in "hillside" and "hillside" == False)
See the section on Comparisons in the Python language reference for more details, in particular this note:
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).
This question already has an answer here:
Python's in (__contains__) operator returns a bool whose value is neither True nor False
(1 answer)
Closed 7 years ago.
How does Python parse the first of the following three expressions? (I expected it to be the same as the second, since == and in have the same precedence.)
>>> 1 == 0 in (0,1), (1==0) in (0,1), 1 == (0 in (0,1))
(False, True, True)
See the documentation of comparison operators: they are chained rather than grouped. So 1 == 0 in (0,1) is equivalent to (1==0) and (0 in (0,1)), which is obviously false.