When I was looking at answers to this question, I found I didn't understand my own answer.
I don't really understand how this is being parsed. Why does the second example return False?
>>> 1 in [1,0] # This is expected
True
>>> 1 in [1,0] == True # This is strange
False
>>> (1 in [1,0]) == True # This is what I wanted it to be
True
>>> 1 in ([1,0] == True) # But it's not just a precedence issue!
# It did not raise an exception on the second example.
Traceback (most recent call last):
File "<pyshell#4>", line 1, in <module>
1 in ([1,0] == True)
TypeError: argument of type 'bool' is not iterable
Thanks for any help. I think I must be missing something really obvious.
I think this is subtly different to the linked duplicate:
Why does the expression 0 < 0 == 0 return False in Python?.
Both questions are to do with human comprehension of the expression. There seemed to be two ways (to my mind) of evaluating the expression. Of course neither were correct, but in my example, the last interpretation is impossible.
Looking at 0 < 0 == 0 you could imagine each half being evaluated and making sense as an expression:
>>> (0 < 0) == 0
True
>>> 0 < (0 == 0)
True
So the link answers why this evaluates False:
>>> 0 < 0 == 0
False
But with my example 1 in ([1,0] == True) doesn't make sense as an expression, so instead of there being two (admittedly wrong) possible interpretations, only one seems possible:
>>> (1 in [1,0]) == True
Python actually applies comparison operator chaining here. The expression is translated to
(1 in [1, 0]) and ([1, 0] == True)
which is obviously False.
This also happens for expressions like
a < b < c
which translate to
(a < b) and (b < c)
(without evaluating b twice).
See the Python language documentation for further details.
Related
When I was looking at answers to this question, I found I didn't understand my own answer.
I don't really understand how this is being parsed. Why does the second example return False?
>>> 1 in [1,0] # This is expected
True
>>> 1 in [1,0] == True # This is strange
False
>>> (1 in [1,0]) == True # This is what I wanted it to be
True
>>> 1 in ([1,0] == True) # But it's not just a precedence issue!
# It did not raise an exception on the second example.
Traceback (most recent call last):
File "<pyshell#4>", line 1, in <module>
1 in ([1,0] == True)
TypeError: argument of type 'bool' is not iterable
Thanks for any help. I think I must be missing something really obvious.
I think this is subtly different to the linked duplicate:
Why does the expression 0 < 0 == 0 return False in Python?.
Both questions are to do with human comprehension of the expression. There seemed to be two ways (to my mind) of evaluating the expression. Of course neither were correct, but in my example, the last interpretation is impossible.
Looking at 0 < 0 == 0 you could imagine each half being evaluated and making sense as an expression:
>>> (0 < 0) == 0
True
>>> 0 < (0 == 0)
True
So the link answers why this evaluates False:
>>> 0 < 0 == 0
False
But with my example 1 in ([1,0] == True) doesn't make sense as an expression, so instead of there being two (admittedly wrong) possible interpretations, only one seems possible:
>>> (1 in [1,0]) == True
Python actually applies comparison operator chaining here. The expression is translated to
(1 in [1, 0]) and ([1, 0] == True)
which is obviously False.
This also happens for expressions like
a < b < c
which translate to
(a < b) and (b < c)
(without evaluating b twice).
See the Python language documentation for further details.
When I was looking at answers to this question, I found I didn't understand my own answer.
I don't really understand how this is being parsed. Why does the second example return False?
>>> 1 in [1,0] # This is expected
True
>>> 1 in [1,0] == True # This is strange
False
>>> (1 in [1,0]) == True # This is what I wanted it to be
True
>>> 1 in ([1,0] == True) # But it's not just a precedence issue!
# It did not raise an exception on the second example.
Traceback (most recent call last):
File "<pyshell#4>", line 1, in <module>
1 in ([1,0] == True)
TypeError: argument of type 'bool' is not iterable
Thanks for any help. I think I must be missing something really obvious.
I think this is subtly different to the linked duplicate:
Why does the expression 0 < 0 == 0 return False in Python?.
Both questions are to do with human comprehension of the expression. There seemed to be two ways (to my mind) of evaluating the expression. Of course neither were correct, but in my example, the last interpretation is impossible.
Looking at 0 < 0 == 0 you could imagine each half being evaluated and making sense as an expression:
>>> (0 < 0) == 0
True
>>> 0 < (0 == 0)
True
So the link answers why this evaluates False:
>>> 0 < 0 == 0
False
But with my example 1 in ([1,0] == True) doesn't make sense as an expression, so instead of there being two (admittedly wrong) possible interpretations, only one seems possible:
>>> (1 in [1,0]) == True
Python actually applies comparison operator chaining here. The expression is translated to
(1 in [1, 0]) and ([1, 0] == True)
which is obviously False.
This also happens for expressions like
a < b < c
which translate to
(a < b) and (b < c)
(without evaluating b twice).
See the Python language documentation for further details.
So I was looking at some code online and I came across a line (at line 286):
if depth > 0 and best <= -MATE_VALUE is None and nullscore > -MATE_VALUE:
The part I had trouble understanding was the best <= -MATE_VALUE is None.
So I fired up the interpreter to see how a statement such as value1 > value2 is value3 work. So I tried
>>> 5 > 2 is True
False
>>> (5 > 2) is True
True
>>> 5 > (2 is True)
True
My Question
Why is 5 > 2 is True not True? And how do these things generally work?
Thanks.
You're seeing python's operator chaining working
5 > 2 is True
Is equivalent to
5>2 and 2 is True
You can see this in that
>>> 5>2 is 2
Returns True.
First, 5 > 2 is True is equivalent to (5 > 2) and (2 is True) because of operator chaining in python (section 5.9 here).
It's clear that 5 > 2 evaluates to True. However, 2 is True will evaluate to False because it is not implicitly converted to bool. If you force the conversion, you will find that bool(2) is True yields True. Other statements such as the if-statement will do this conversion for you, so if 2: will work.
Second, there is an important difference between the is operator and the == operator (taken from here):
Use is when you want to check against an object's identity (e.g.
checking to see if var is None). Use == when you want to check
equality (e.g. Is var equal to 3?).
>> [1,2] is [1,2]
False
>> [1,2] == [1,2]
True
While this does not have an immediate impact on this example, you should keep it in mind for the future.
I've come across a surprising situation when using numpy's arrays. The following code
(True==True)+(True==True)
returns 2, as one would expect. While
import numpy
Array=numpy.zeros((2,2),dtype=bool)
(Array[0][0]==Array[0][0])+(Array[1][0]==Array[1][0])
returns True. This leads to:
(Array[0][0]==Array[0][0])+(Array[1][0]==Array[1][0])-1
returning 0, while
(Array[0][0]==Array[0][0])-1+(Array[1][0]==Array[1][0])
returns 1, making the sum not commutative!
Is this intended? If so, why?
It would appear that numpy.bool_ behaves slightly differently to vanilla Python bool:
>>> int(True+True) == int(True) + int(True)
True
>>> int(numpy.bool_(1)+numpy.bool_(1)) == int(numpy.bool_(1)) + int(numpy.bool_(1))
False
This is because:
>>> True+True
2
>>> numpy.bool_(1)+numpy.bool_(1)
True
>>> int(numpy.bool_(1)+numpy.bool_(1))
1
Basically, the addition operation for numpy.bool_ is logical, rather than numerical; to get the same behaviour with bool:
>>> int(True and True)
1
This is fine if you only use it for truthiness, as intended, but if you try to use it in an integer context without being explicit about that, you end up surprised. As soon as you're explicit, expected behaviour is restored:
>>> int(numpy.bool_(1)) + int(numpy.bool_(1))
2
I think the problem is th autocasting.
in this case:
(Array[0][0]==Array[0][0])+(Array[1][0]==Array[1][0])-1
Python do:
(Array[0][0]==Array[0][0])+(Array[1][0]==Array[1][0]) = True
True -1 =cast= 1 -1 = 0
In the second case the cast is do it before:
(Array[0][0]==Array[0][0])-1+(Array[1][0]==Array[1][0])
True - 1 + True
(True - 1 =cast= 0)
0 + True =cast again= 0+ 1 = 1
So, it's not a bug. It's a autocasting in diferent parts.
As expected, 1 is not contained by the empty tuple
>>> 1 in ()
False
but the False value returned is not equal to False
>>> 1 in () == False
False
Looking at it another way, the in operator returns a bool which is neither True nor False:
>>> type(1 in ())
<type 'bool'>
>>> 1 in () == True, 1 in () == False
(False, False)
However, normal behaviour resumes if the original expression is parenthesized
>>> (1 in ()) == False
True
or its value is stored in a variable
>>> value = 1 in ()
>>> value == False
True
This behaviour is observed in both Python 2 and Python 3.
Can you explain what is going on?
You are running into comparison operator chaining; 1 in () == False does not mean (1 in ()) == False.
Rather, comparisons are chained and the expression really means:
(1 in ()) and (() == False)
Because (1 in ()) is already false, the second half of the chained expression is ignored altogether (since False and something_else returns False whatever the value of something_else would be).
See the comparisons expressions documentation:
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).
For the record, <, >, ==, >=, <=, !=, is, is not, in and not in are all comparison operators (as is the deprecated <>).
In general, don't compare against booleans; just test the expression itself. If you have to test against a boolean literal, at least use parenthesis and the is operator, True and False are singletons, just like None:
>>> (1 in ()) is False
True
This gets more confusing still when integers are involved. The Python bool type is a subclass of int1. As such, False == 0 is true, as is True == 1. You therefor can conceivably create chained operations that almost look sane:
3 > 1 == True
is true because 3 > 1 and 1 == True are both true. But the expression:
3 > 2 == True
is false, because 2 == True is false.
1 bool is a subclass of int for historic reasons; Python didn't always have a bool type and overloaded integers with boolean meaning just like C does. Making bool a subclass kept older code working.