Order of operation evaluation in condition in Python - python

If I have something like a = [1,2,3] and I write a statement like 1 in a == True, that appears to evaluate to false.
However, if I write (1 in a) == True, that evaluates to true.
I am slightly confused about how Python evaluates the first statement to reach false in the end.

Both == and in are considered comparison operators, meaning that operator chaining comes into effect:
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 chaining is what allows you to write:
if 0 <= x < 20: # meaning: if x >= 0 and x < 20:
Hence the expression 1 in a == True is chained as 1 in a and a == True and, since the right side of that and evaluates to false, the entire expression is false.
This chaining does not occur when you "isolate" part of the expression with parentheses, which is why (1 in a) == True acts as you expect.
Having explained that, comparing a boolean value with True (or False for that matter) is not really something you should be doing anyway, since that leads to logically infinite scenarios like:
if (((a == 7) == True) != False) == True # 'a == 7' already a boolean.
Far better just to use one of:
if expression # if comparing with True
if not expression # if comparing with False

Related

How to use conditionals without relational operators

# x = 1
# x = 0
if x:
print(f'x = {x}, and therefore it\'s truthy')
else:
print(f"x = {x}, and therefore it's falsey")
Hi everyone, I am a little confused with this exercise. The code doesn't have any explanation as when the system should print truthy or falsey yet it knows when to print it. Why is that?
I think it's just a way to show that 0 is interpreted as False in a boolean expression like a comparison or an if etc, and 1 (or any other non-zero number) is interpreted as True.
Edit: and like said in some comments, the fact that a value is "truthy" doesn't mean that is strictly equal to True
for example:
print(1 == True) # True
print(0 == False) # True
print(2 == True) # False
print(not not 2) # True
print(not 0) # True
print(not not 0) # False
print(0 and True) # False
print(2 and False) # False
print(2 and True) # True
print(0 or 2) # True
# etc ...
If x is an int and has a value of zero it is equal to False and therefore falsey. If x is an int and has a value of 1 it is equal to True and is therefore truthy. However, if x is an int and is neither 0 nor 1 it is truthy but not equal to True. Therefore, when x is of type int, if x: can be interpreted as 'if x is non-zero'
Its all in the if x: line.
To execute this line, I like to think that python translates it to:
if bool(x):
because bool() will only ever return True or False, then the first option is only ever taken when the result is True.
See this answer for details of a comprehensive list of values which are considered Falsey.
the funny part is if you use
x = 0.00000000000001
it will also consider as true. cause it is not actually equal to zero. in math we consider this as zero for easy calculation. but computer dont do this cause this can create big problem in calculation if we consider a big picture. Example:
y = x ** x
print(y)
so if we consider x as zero, y will be also zero. but its not.
i hope you enjoyed to learn why 0.000000000001 is not equal to zero in computer.Thanks.

Need help understanding Booleans?

I am reading about booleans in my book, and it says:
x and y------------If x is false, return x. Otherwise, return y.
x or y--------------If x is true, return x. Otherwise, return y.
This doesn't make sense to me from the everyday usage of "or" and "and"
To me, it would make sense if it said:
x and y------------If x is false, do nothing. If y is false, do nothing. If x and y are true, return x and y
x or y--------------If x is true, return x. If x is false, check y. If y is false, do nothing. If y is true, return y
Do I just need to blindly accept the actual definitions, or can I understand them because they actually make sense.
"Do nothing" isn't an option. The mathematical expression x or y must have a value (which is true if either x, y, or both is true). And x and y must have a value (which is true if and only if x and y are both true).
The way your book defines them is mathematically correct, but confusing (until you get into things like short-circuit evaluation).
The behavior may seem strange, but consider the following hypothetical examples. Starting with something obvious,
>>> True and False
False
>>> False and True
False
this is really easy to understand since we're dealing with boolean values. Remember this example, because every other example can be thought of this way.
Now consider if the and and or operators were to convert every object to a boolean before comparing. For example, an empty string or empty list would be False, and non-empty ones would be True. It would look like this (obviously this is not what it ACTUALLY looks like)
>>> "vanilla" and ""
False
>>> "" and "vanilla"
False
This makes perfect sense. After all bool("vanilla") and bool("") would be the same as True and False which we already know is False
Instead of actually converting them to True or False, though, it can do the comparison without ever converting them. As a result, you don't really need it to return True or False. It can just return the actual object it tested.
>>> "vanilla" and ""
""
>>> "" and "vanilla"
""
For truth testing purposes, returning "" is the same as returning False, so there's no need to convert it to a boolean. That's why it always returns an object whose truth value is the same as the result of the operator.
Unlike some other languages, you can use any object as operands of boolean operations. All your books says is that you can use boolean operators as a quick "filter" of values.
For instance, say you want to choose a non-empty list between two lists. The following is a valid (and more Pythonic) way to do it:
>>> [] or ['something', 'here']
['something', 'here']
Contrast to (not making use of Python idioms in Python):
if len(l1) != 0:
return l1
else:
return l2
Your book is right - see the documentation. It might be unintuitive at first, but this behavior (called short-circuiting) is extremely useful. In a simple case, it allows you to save a lot of time when checking certain conditions. In this example, where function f takes 10 seconds to evaluate, you can definitely see one use:
if f(foo) or f(bar) or f(baz):
If f(foo) is True, then there is no need to evaluate f(bar) or f(baz), as the whole if statement will be True. Those values are unnecessary and you'll just be wasting your time computing them.
Another extremely common use of this behavior is in null (or for python None) checking. It allows safe usage of functions all within one line:
if obj != None and obj.foo():
If obj is None, the if statement is guaranteed to be False, and so there is no need to check (or even evaluate) obj.foo(), which is good, since that would cause an exception.
Short-circuiting is very common in many programming languages, and is very useful once you fully understand how to use it.
While the book presents it in a slightly confusing way, it is correct. Boolean logic must evaluate to either true or false.
For X and Y in order to return true they both must be true, if X is false then it returns false. If X is true then it returns Y which is either true or false and also the correct answer.
For X or Y to return false they both must be false. If X is true then it can return true (X). If X is false then it returns whatever the value of Y is.
Many (most?) programming languages, including Python, implement short-circuiting in their boolean operators and and or. So they evaluate their operands from left to right, and stop as soon as the final result can be determined.
Since x and y and z and ... is guaranteed be false if any operand is false, it stops evaluating the operands as soon as it encounters a false one. And x or y or z or ... is guaranteed to be true if any operand is true, so it stops as soon as it reaches a true operand. In either case, if they make it all the way to the last operand, they return its value.
In some languages, the boolean operators just return a strict boolean result, either true or false (in some languages these are represented as 1 and 0). But in Python (and some others, such as Javascript and Common Lisp), they return the value of the last operand that was evaluated, which determined the final result. This is often more useful than just the truth value of the expression.
When you put these features together, it allows some succinct idioms, e.g.
quotient = b != 0 && a/b
instead of
if b != 0:
quotient = false
else:
quotient = a/b
The first set of descriptions are shortcuts: following these will give you exactly the same results as the "usual definitions" of true and false.
But your own descriptions don't make much sense. You can't "do nothing"; you have to return some value from a comparison, either true or false. And you can't return both a and b, either: again, the result of a boolean comparison must be a boolean, not a pair of booleans.
Its probably easier if you think about it in each literal case
x and y------------If x is false, return x. Otherwise, return y.
x or y--------------If x is true, return x. Otherwise, return y.
CASE 1: x = true, y = true
"if x is false, return x. Otherwise, return y."
Then this will return y which is true. This makes sense because x and y are both true.
(true and true == true)
"If x is true, return x. Otherwise, return y."
The this will return x which is true. This makes sense because one of x or y is true.
(true or true == true)
CASE 2: x = false, y = true
"if x is false, return x. Otherwise, return y."
Then this will return x which is false. This makes sense because x and y are not both true.
(false and true == false)
"If x is true, return x. Otherwise, return y."
The this will return y. This makes sense because one of x or y is true.
(false or true == true)
CASE 3: x = true, y = false
"if x is false, return x. Otherwise, return y."
Then this will return y which is false. This makes sense because x and y are not both true.
(true and false == false)
"If x is true, return x. Otherwise, return y."
Then this will return x which is true. This makes sense because x or y is true
(true or false == true)
CASE 4: x = false, y = false
"if x is false, return x. Otherwise, return y."
Then this will return x which is false. This makes sense because x and y are not both true.
(false and false == false)
"If x is true, return x. Otherwise, return y."
The this will return y which is false. This makes sense because neither x nor y is true.
(false or false == false)

What should be simple expression (containing less_than, is, and) result? Why? [duplicate]

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.

How does python interpret "True != True != True"? [duplicate]

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.

Python's in (__contains__) operator returns a bool whose value is neither True nor False

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.

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