Is there an operation for not less than or not greater than in python?
Question:
Suppose I have this code:
a = 0
if a == 0 or a > 0:
print(a)
That is: I want to do something when a is not negative.
I know that I can write if a != 0: to check whether a is not equal to 0.
So, I tried using if a !< 0:, following similar logic. However, this is apparently not supported:
>>> if a !< 0:
File "<stdin>", line 1
if a !< 0:
^
SyntaxError: invalid syntax
Why is this syntax invalid? What can I use instead to simplify the conditional?
Answers:
Instead of a == 0 or a > 0, simply use a >= 0.
See https://docs.python.org/library/stdtypes.html#comparisons for a complete list of available comparison operators.
Python does not provide e.g. a !< operator, because it is not needed. if a == 0 or a > 0 means the same thing as if a >= 0.
You can use the equal or greater than operator:
if a >= 0:
print(a)
!< does not work in Python; but not can be placed before a comparison to get the opposite effect, like so:
if not a < 70:
print('The number is Not smaller than 70')
else:
print('The number is DEFINITELY smaller than 70')
You could look into the "not" operator with a > or < expression.
Following worked for me though (not operator does not work, using tilde ~, doesn’t work):
((-1 > 0) or (0 > 0)) == False
Result:
True
To check whether a is not greater than b, use if not (a > b).
an operation for not less than or not greater than in python?
Ahh, pretty old question. But I believe our future programmers are probably searching this one. Here it goes:
Well, if you want the above two operations on the same line: you are looking for == operator [i.e. if a is not less than b and a is also not greater than b, then it ought to be equal to b]. If you meant operations like "not less than", you are looking for not operator.
if not a < b:
# do something here.
The not operator in Python reverts the boolean value, i.e. not True is equivalent to False and not False is equivalent to True.
Suppose I have this code:
a = 0
if a == 0 or a > 0:
print(a)
That is: I want to do something when a is not negative.
I know that I can write if a != 0: to check whether a is not equal to 0.
So, I tried using if a !< 0:, following similar logic. However, this is apparently not supported:
>>> if a !< 0:
File "<stdin>", line 1
if a !< 0:
^
SyntaxError: invalid syntax
Why is this syntax invalid? What can I use instead to simplify the conditional?
Instead of a == 0 or a > 0, simply use a >= 0.
See https://docs.python.org/library/stdtypes.html#comparisons for a complete list of available comparison operators.
Python does not provide e.g. a !< operator, because it is not needed. if a == 0 or a > 0 means the same thing as if a >= 0.
You can use the equal or greater than operator:
if a >= 0:
print(a)
!< does not work in Python; but not can be placed before a comparison to get the opposite effect, like so:
if not a < 70:
print('The number is Not smaller than 70')
else:
print('The number is DEFINITELY smaller than 70')
You could look into the "not" operator with a > or < expression.
Following worked for me though (not operator does not work, using tilde ~, doesn’t work):
((-1 > 0) or (0 > 0)) == False
Result:
True
To check whether a is not greater than b, use if not (a > b).
an operation for not less than or not greater than in python?
Ahh, pretty old question. But I believe our future programmers are probably searching this one. Here it goes:
Well, if you want the above two operations on the same line: you are looking for == operator [i.e. if a is not less than b and a is also not greater than b, then it ought to be equal to b]. If you meant operations like "not less than", you are looking for not operator.
if not a < b:
# do something here.
The not operator in Python reverts the boolean value, i.e. not True is equivalent to False and not False is equivalent to True.