how multiple OR's and AND evaluate - if-statement

This is very simple question, but I can't get it. I have simple condition:
bool c = true || true || true && false;
Why this evaluation is true? As far as I know it evaluates like this:
true || true || true && false => true || true && false => true && false => false
But guess im wrong.

You just have to learn some few basic rules:
(a OR b) is true IF AND ONLY IF at least one of a or b is true.
(a AND b) is true IF AND ONLY IF both of a and b are true.
ORDER of Operators MATTERS: you can't just do the logic any way that you want. computer calculates the output of the a logical statement by an order. a simplified order is like this: First is Grouping (), Second is And, Third is OR.
so when you say bool c = true || true || true && false;. the computer says ok let's first calculate true && false. it is false! and then it calculates true || true || false which is true.
EDIT
Note 1: a complete list of logical operators and their precedence is heavily dependent on the language. you can refer to the documentation for that.
for C#
Note 2: the best practice is to use GROUPING like parenthesis because GROUPING is always of priority. for example it's better to say:
bool c = (true || true) || (true && false);

Think of OR || like addition and AND && as multiplication. There is a priority in there, in fact, you will sometimes see them written as such:
bool c = true + true + true + true * false
In this case, the first evaluation is true * false, then the rest of the ORs. In this particular case, the true order of evaluation of the ORs will depend on language/compiler.
If you want to force a particular order, you can always use parentheses.

Related

Why is my elif statement always executing in Godot?

I'm new to Godot programming, just to preface this. I'm trying to make it so that when you fall off of a platform without jumping you start accelerating downwards rather than just going down at a fixed speed. I almost have that, but my elif statement is happening every frame rather than the frame you leave the ground.
I've tried changing things, but I'm stuck at a logical dead end where any option I think of or find has resulted in the same or worse behavior.
extends KinematicBody2D
const UP_DIRECTION = Vector2.UP #up is up and down is down
# Declare member variables here.
var speed = 200 #speed of walk
var velocity = Vector2() #movement Vectors
var gravity = 1000 #gravity, it's real
var jump = 0 #jumping speed
var canjump = 0 #jumps left
var jumpcount = 1 #times you can jump
var nojump = false #did you fall or did you jump?
# Called every physics update.
func _physics_process(delta):
#jumping
if (canjump > 0 && Input.is_action_just_pressed("Jump") || is_on_floor() && canjump > 0 && Input.is_action_pressed("Jump")):
jump = -2000
canjump -= 1
nojump = false
#falling without jumping first
elif not (is_on_floor() && Input.is_action_just_pressed("Jump") && nojump == false):
jump = -gravity
canjump -= 1
nojump = false
#decelerating jumps
if (jump < 0):
jump += 45
#grounding jump variables
if(is_on_floor()):
canjump = jumpcount
nojump = true
#setting x and y
velocity.x = (Input.get_action_strength("Right") - Input.get_action_strength("Left")) * speed
velocity.y = gravity + jump
#using found inputs
velocity = move_and_slide(velocity, UP_DIRECTION)
pass
Here's my code, it tried commenting in the tdlr of what things are supposed to do. Again all I'm looking for is the reason why my elif statement is being applied every frame and how to fix that.
If you have a conditional that looks like this:
if condition:
prints("Something")
The execution flow would enter it when the condition is true.
Let us look at this as example:
if is_on_floor() && Input.is_action_just_pressed("Jump") && nojump == false:
prints("Something")
The above conditional requires three things:
is_on_floor() must be true.
Input.is_action_just_pressed("Jump") must be true.
nojump == false must be true. I mean, nojump must be false.
I think we can agree that does not happen all the time.
Now, if I negate the conditional:
if not condition:
prints("Something")
The execution flow would enter when the condition is false.
So, in a case like the one on your code:
if not (is_on_floor() && Input.is_action_just_pressed("Jump") && nojump == false):
prints("Something")
The conditional has one requirement:
not (is_on_floor() && Input.is_action_just_pressed("Jump") && nojump == false)`
Must be true. I mean:
is_on_floor() && Input.is_action_just_pressed("Jump") && nojump == false
Must be false. Double check D'Morgan's Law if you need to.
It is false when either:
is_on_floor() is false.
Or Input.is_action_just_pressed("Jump") is false.
Or nojump == false is false.
In other words, only one of these being false is sufficient.
To restate that. This condition:
not (is_on_floor() && Input.is_action_just_pressed("Jump") && nojump == false)`
Is equivalent to the following condition (by D'Morgan's law):
not is_on_floor() or not Input.is_action_just_pressed("Jump") or not nojump == false
By the way, using and and or is idiomatic in GDScript.
Wait, I can simplify that a bit:
not is_on_floor() or not Input.is_action_just_pressed("Jump") or nojump`
Thus, the execution flow will enter if either:
not is_on_floor() is true. I mean, if is_on_floor() is false.
not Input.is_action_just_pressed("Jump") is true. I mean, if Input.is_action_just_pressed("Jump") is false.
nojump is true.
And I reiterate that is either of them. Only one is sufficient. And I believe most of the time you haven't just pressed "Jump", also being on the air is enough. And I suspect nojump is false most of the time.
So, we can conclude that the execution flow will enter most of the time.
Write what you mean. You say in comments:
#falling without jumping first
Let us do that. What does falling mean? Does it mean on the air?
var falling := not is_on_floor()
Or do you also mean going down?
var falling := not is_on_floor() and velocity.dot(Vector2.DOWN) > 0.0
Alright, your pick. What does jumping first mean? Does it mean the player got in the air as a result of a jump? Ok... So you would do this when the player jumped:
jumped = true
Which I believe is this:
if canjump > 0 && Input.is_action_just_pressed("Jump"):
jumped = true
And reset it on the ground:
if is_on_floor():
jumped = false
Now we can say "falling without jumping first":
if falling and not jumped:
prints("something")
Almost as if it were the comment, but unlike the comment it can be executed.

need a conditional statement for desired output

I need to write a conditional tatement in IF class by using three variable to get some desired output, i have tried many logics with that three variable to get the desired output which i can use in if class but i couldn't succeed, hope anyone can help me with this, Thank you
here us the conditional table i need
A B C resultFalse False False FalseTrue False False TrueTrue True False FalseTrue True True TrueTrue True False False
The result is only true if
1) A is true and B & C are false
2) A, B, and C are true
In other words:
if((A & !B & !C) | (A & B & C)){
print('true')
}
The third and fifth row in your table are equal by the way.

Why i cant check if a variable have a value using == true?

if(10) it is true, but if(10 == true) is false. Can someone tell me why the first case convert the number to bool but the second case didnt?
if (10) is equivalent to if (10 != 0), whereas if (10 == true) is if (10 == 1) (since true is promoted to the value 1 of type int).
In layman's terms: Two things that both happen to satisfy some property aren't automatically the same thing.
(E.g. doughnuts and frisbees are both round, but that doesn't mean a doughnut is equal to a frisbee. Integers and booleans can both be evaluated in a boolean context, but that doesn't mean that every integer that evaluates as true is equal to every true boolean.)
if( ... )
{
// if statement
}
To execute if statement in C++, ... should have a true value.
When you wrote if( 10 ){//something useful}
I think 10 treats as int but not bool variable. The following logic should be applied then
if( 10 ) -> if( bool( 10 ) ) -> if( true )
When you write if( 10 == true ){//something useful}, according to C++ standard, there should be the following logic behind the scene
if( 10 == true ) -> if( 10 == int( true ) ) -> if( 10 == 1 ) -> if( false )
You may write something like
if( 10 != false )
or
if( !!10 == true )
also
if( ( bool ) 10 == true ) // alternatively, if( bool ( 10 ) == true )
In old C (before C99), there is no false or true, but there are 0 or non-0 values.
In modern C (from C99), there is false or true (<stdbool.h>), but they are syntactic sugar for 0 and 1, respectively.
if( 10 ) // evaluates directly since 10 is non-zero value
if( 10 == true ) -> if( 10 == 1 ) -> if( 0 )
Because they are entirely different things.
In C, anything that is NOT false is automatically true, and C has a very strict definition of false.
You can think of if(10) as if(10 != false)
and likewise if (10 == true) as if((10 == true) != false)
10 is clearly not true, in the sense that 10 is a number and true is a boolean.
When you're in an if statement, the compiler has to evaluate the condition until it reaches either a true or a false. If it does not reach a true or a false, it has to convert it to a true or a false. As a general rule, 0 evaluates to false, and everything else evaluates to true.
So if(-1) is true, as is if(234) and so on.
The expression 10 == true already evaluates to false, so no further conversion is needed. if(10) is neither true or false, so the compiler has to convert it, using our rule above, and it becomes true.

Python consecutive if statements [duplicate]

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Is there a ternary conditional operator in Python?
Yes, it was added in version 2.5. The expression syntax is:
a if condition else b
First condition is evaluated, then exactly one of either a or b is evaluated and returned based on the Boolean value of condition. If condition evaluates to True, then a is evaluated and returned but b is ignored, or else when b is evaluated and returned but a is ignored.
This allows short-circuiting because when condition is true only a is evaluated and b is not evaluated at all, but when condition is false only b is evaluated and a is not evaluated at all.
For example:
>>> 'true' if True else 'false'
'true'
>>> 'true' if False else 'false'
'false'
Note that conditionals are an expression, not a statement. This means you can't use statements such as pass, or assignments with = (or "augmented" assignments like +=), within a conditional expression:
>>> pass if False else pass
File "<stdin>", line 1
pass if False else pass
^
SyntaxError: invalid syntax
>>> # Python parses this as `x = (1 if False else y) = 2`
>>> # The `(1 if False else x)` part is actually valid, but
>>> # it can't be on the left-hand side of `=`.
>>> x = 1 if False else y = 2
File "<stdin>", line 1
SyntaxError: cannot assign to conditional expression
>>> # If we parenthesize it instead...
>>> (x = 1) if False else (y = 2)
File "<stdin>", line 1
(x = 1) if False else (y = 2)
^
SyntaxError: invalid syntax
(In 3.8 and above, the := "walrus" operator allows simple assignment of values as an expression, which is then compatible with this syntax. But please don't write code like that; it will quickly become very difficult to understand.)
Similarly, because it is an expression, the else part is mandatory:
# Invalid syntax: we didn't specify what the value should be if the
# condition isn't met. It doesn't matter if we can verify that
# ahead of time.
a if True
You can, however, use conditional expressions to assign a variable like so:
x = a if True else b
Or for example to return a value:
# Of course we should just use the standard library `max`;
# this is just for demonstration purposes.
def my_max(a, b):
return a if a > b else b
Think of the conditional expression as switching between two values. We can use it when we are in a 'one value or another' situation, where we will do the same thing with the result, regardless of whether the condition is met. We use the expression to compute the value, and then do something with it. If you need to do something different depending on the condition, then use a normal if statement instead.
Keep in mind that it's frowned upon by some Pythonistas for several reasons:
The order of the arguments is different from those of the classic condition ? a : b ternary operator from many other languages (such as C, C++, Go, Perl, Ruby, Java, JavaScript, etc.), which may lead to bugs when people unfamiliar with Python's "surprising" behaviour use it (they may reverse the argument order).
Some find it "unwieldy", since it goes contrary to the normal flow of thought (thinking of the condition first and then the effects).
Stylistic reasons. (Although the 'inline if' can be really useful, and make your script more concise, it really does complicate your code)
If you're having trouble remembering the order, then remember that when read aloud, you (almost) say what you mean. For example, x = 4 if b > 8 else 9 is read aloud as x will be 4 if b is greater than 8 otherwise 9.
Official documentation:
Conditional expressions
Is there an equivalent of C’s ”?:” ternary operator?
You can index into a tuple:
(falseValue, trueValue)[test]
test needs to return True or False.
It might be safer to always implement it as:
(falseValue, trueValue)[test == True]
or you can use the built-in bool() to assure a Boolean value:
(falseValue, trueValue)[bool(<expression>)]
For versions prior to 2.5, there's the trick:
[expression] and [on_true] or [on_false]
It can give wrong results when on_true has a false Boolean value.1
Although it does have the benefit of evaluating expressions left to right, which is clearer in my opinion.
1. Is there an equivalent of C’s ”?:” ternary operator?
<expression 1> if <condition> else <expression 2>
a = 1
b = 2
1 if a > b else -1
# Output is -1
1 if a > b else -1 if a < b else 0
# Output is -1
From the documentation:
Conditional expressions (sometimes called a “ternary operator”) have the lowest priority of all Python operations.
The expression x if C else y first evaluates the condition, C (not x); if C is true, x is evaluated and its value is returned; otherwise, y is evaluated and its value is returned.
See PEP 308 for more details about conditional expressions.
New since version 2.5.
An operator for a conditional expression in Python was added in 2006 as part of Python Enhancement Proposal 308. Its form differs from common ?: operator and it looks like this:
<expression1> if <condition> else <expression2>
which is equivalent to:
if <condition>: <expression1> else: <expression2>
Here is an example:
result = x if a > b else y
Another syntax which can be used (compatible with versions before 2.5):
result = (lambda:y, lambda:x)[a > b]()
where operands are lazily evaluated.
Another way is by indexing a tuple (which isn't consistent with the conditional operator of most other languages):
result = (y, x)[a > b]
or explicitly constructed dictionary:
result = {True: x, False: y}[a > b]
Another (less reliable), but simpler method is to use and and or operators:
result = (a > b) and x or y
however this won't work if x would be False.
A possible workaround is to make x and y lists or tuples as in the following:
result = ((a > b) and [x] or [y])[0]
or:
result = ((a > b) and (x,) or (y,))[0]
If you're working with dictionaries, instead of using a ternary conditional, you can take advantage of get(key, default), for example:
shell = os.environ.get('SHELL', "/bin/sh")
Source: ?: in Python at Wikipedia
Unfortunately, the
(falseValue, trueValue)[test]
solution doesn't have short-circuit behaviour; thus both falseValue and trueValue are evaluated regardless of the condition. This could be suboptimal or even buggy (i.e. both trueValue and falseValue could be methods and have side effects).
One solution to this would be
(lambda: falseValue, lambda: trueValue)[test]()
(execution delayed until the winner is known ;)), but it introduces inconsistency between callable and non-callable objects. In addition, it doesn't solve the case when using properties.
And so the story goes - choosing between three mentioned solutions is a trade-off between having the short-circuit feature, using at least Python 2.5 (IMHO, not a problem anymore) and not being prone to "trueValue-evaluates-to-false" errors.
Ternary operator in different programming languages
Here I just try to show some important differences in the ternary operator between a couple of programming languages.
Ternary operator in JavaScript
var a = true ? 1 : 0;
# 1
var b = false ? 1 : 0;
# 0
Ternary operator in Ruby
a = true ? 1 : 0
# 1
b = false ? 1 : 0
# 0
Ternary operator in Scala
val a = true ? 1 | 0
# 1
val b = false ? 1 | 0
# 0
Ternary operator in R programming
a <- if (TRUE) 1 else 0
# 1
b <- if (FALSE) 1 else 0
# 0
Ternary operator in Python
a = 1 if True else 0
# 1
b = 1 if False else 0
# 0
For Python 2.5 and newer there is a specific syntax:
[on_true] if [cond] else [on_false]
In older Pythons, a ternary operator is not implemented but it's possible to simulate it.
cond and on_true or on_false
Though there is a potential problem, which is if cond evaluates to True and on_true evaluates to False then on_false is returned instead of on_true. If you want this behaviour the method is OK, otherwise use this:
{True: on_true, False: on_false}[cond is True] # is True, not == True
which can be wrapped by:
def q(cond, on_true, on_false)
return {True: on_true, False: on_false}[cond is True]
and used this way:
q(cond, on_true, on_false)
It is compatible with all Python versions.
You might often find
cond and on_true or on_false
but this leads to a problem when on_true == 0
>>> x = 0
>>> print x == 0 and 0 or 1
1
>>> x = 1
>>> print x == 0 and 0 or 1
1
Where you would expect this result for a normal ternary operator:
>>> x = 0
>>> print 0 if x == 0 else 1
0
>>> x = 1
>>> print 0 if x == 0 else 1
1
Does Python have a ternary conditional operator?
Yes. From the grammar file:
test: or_test ['if' or_test 'else' test] | lambdef
The part of interest is:
or_test ['if' or_test 'else' test]
So, a ternary conditional operation is of the form:
expression1 if expression2 else expression3
expression3 will be lazily evaluated (that is, evaluated only if expression2 is false in a boolean context). And because of the recursive definition, you can chain them indefinitely (though it may considered bad style.)
expression1 if expression2 else expression3 if expression4 else expression5 # and so on
A note on usage:
Note that every if must be followed with an else. People learning list comprehensions and generator expressions may find this to be a difficult lesson to learn - the following will not work, as Python expects a third expression for an else:
[expression1 if expression2 for element in iterable]
# ^-- need an else here
which raises a SyntaxError: invalid syntax.
So the above is either an incomplete piece of logic (perhaps the user expects a no-op in the false condition) or what may be intended is to use expression2 as a filter - notes that the following is legal Python:
[expression1 for element in iterable if expression2]
expression2 works as a filter for the list comprehension, and is not a ternary conditional operator.
Alternative syntax for a more narrow case:
You may find it somewhat painful to write the following:
expression1 if expression1 else expression2
expression1 will have to be evaluated twice with the above usage. It can limit redundancy if it is simply a local variable. However, a common and performant Pythonic idiom for this use-case is to use or's shortcutting behavior:
expression1 or expression2
which is equivalent in semantics. Note that some style-guides may limit this usage on the grounds of clarity - it does pack a lot of meaning into very little syntax.
One of the alternatives to Python's conditional expression
"yes" if boolean else "no"
is the following:
{True: "yes", False: "no"}[boolean]
which has the following nice extension:
{True: "yes", False: "no", None: "maybe"}[boolean_or_none]
The shortest alternative remains
("no", "yes")[boolean]
which works because issubclass(bool, int).
Careful, though: the alternative to
yes() if boolean else no()
is not
(no(), yes())[boolean] # bad: BOTH no() and yes() are called
but
(no, yes)[boolean]()
This works fine as long as no and yes are to be called with exactly the same parameters. If they are not, like in
yes("ok") if boolean else no() # (1)
or in
yes("ok") if boolean else no("sorry") # (2)
then a similar alternative either does not exist (1) or is hardly viable (2). (In rare cases, depending on the context, something like
msg = ("sorry", "ok")[boolean]
(no, yes)[boolean](msg)
could make sense.)
Thanks to Radek Rojík for his comment
As already answered, yes, there is a ternary operator in Python:
<expression 1> if <condition> else <expression 2>
In many cases <expression 1> is also used as Boolean evaluated <condition>. Then you can use short-circuit evaluation.
a = 0
b = 1
# Instead of this:
x = a if a else b
# Evaluates as 'a if bool(a) else b'
# You could use short-circuit evaluation:
x = a or b
One big pro of short-circuit evaluation is the possibility of chaining more than two expressions:
x = a or b or c or d or e
When working with functions it is more different in detail:
# Evaluating functions:
def foo(x):
print('foo executed')
return x
def bar(y):
print('bar executed')
return y
def blubb(z):
print('blubb executed')
return z
# Ternary Operator expression 1 equals to False
print(foo(0) if foo(0) else bar(1))
''' foo and bar are executed once
foo executed
bar executed
1
'''
# Ternary Operator expression 1 equals to True
print(foo(2) if foo(2) else bar(3))
''' foo is executed twice!
foo executed
foo executed
2
'''
# Short-circuit evaluation second equals to True
print(foo(0) or bar(1) or blubb(2))
''' blubb is not executed
foo executed
bar executed
1
'''
# Short-circuit evaluation third equals to True
print(foo(0) or bar(0) or blubb(2))
'''
foo executed
bar executed
blubb executed
2
'''
# Short-circuit evaluation all equal to False
print(foo(0) or bar(0) or blubb(0))
''' Result is 0 (from blubb(0)) because no value equals to True
foo executed
bar executed
blubb executed
0
'''
PS: Of course, a short-circuit evaluation is not a ternary operator, but often the ternary is used in cases where the short circuit would be enough. It has a better readability and can be chained.
Simulating the Python ternary operator.
For example
a, b, x, y = 1, 2, 'a greather than b', 'b greater than a'
result = (lambda:y, lambda:x)[a > b]()
Output:
'b greater than a'
a if condition else b
Just memorize this pyramid if you have trouble remembering:
condition
if else
a b
The ternary conditional operator simply allows testing a condition in a single line replacing the multiline if-else making the code compact.
Syntax:
[on_true] if [expression] else [on_false]
1- Simple Method to use ternary operator:
# Program to demonstrate conditional operator
a, b = 10, 20
# Copy value of a in min if a < b else copy b
min = a if a < b else b
print(min) # Output: 10
2- Direct Method of using tuples, Dictionary, and lambda:
# Python program to demonstrate ternary operator
a, b = 10, 20
# Use tuple for selecting an item
print( (b, a) [a < b] )
# Use Dictionary for selecting an item
print({True: a, False: b} [a < b])
# lambda is more efficient than above two methods
# because in lambda we are assure that
# only one expression will be evaluated unlike in
# tuple and Dictionary
print((lambda: b, lambda: a)[a < b]()) # in output you should see three 10
3- Ternary operator can be written as nested if-else:
# Python program to demonstrate nested ternary operator
a, b = 10, 20
print ("Both a and b are equal" if a == b else "a is greater than b"
if a > b else "b is greater than a")
Above approach can be written as:
# Python program to demonstrate nested ternary operator
a, b = 10, 20
if a != b:
if a > b:
print("a is greater than b")
else:
print("b is greater than a")
else:
print("Both a and b are equal")
# Output: b is greater than a
Vinko Vrsalovic's answer is good enough. There is only one more thing:
Note that conditionals are an expression, not a statement. This means you can't use assignment statements or pass or other statements within a conditional expression
Walrus operator in Python 3.8
After the walrus operator was introduced in Python 3.8, something changed.
(a := 3) if True else (b := 5)
gives a = 3 and b is not defined,
(a := 3) if False else (b := 5)
gives a is not defined and b = 5, and
c = (a := 3) if False else (b := 5)
gives c = 5, a is not defined and b = 5.
Even if this may be ugly, assignments can be done inside conditional expressions after Python 3.8. Anyway, it is still better to use normal if statement instead in this case.
More a tip than an answer (I don't need to repeat the obvious for the hundredth time), but I sometimes use it as a one-liner shortcut in such constructs:
if conditionX:
print('yes')
else:
print('nah')
, becomes:
print('yes') if conditionX else print('nah')
Some (many :) may frown upon it as unpythonic (even, Ruby-ish :), but I personally find it more natural - i.e., how you'd express it normally, plus a bit more visually appealing in large blocks of code.
You can do this:
[condition] and [expression_1] or [expression_2];
Example:
print(number%2 and "odd" or "even")
This would print "odd" if the number is odd or "even" if the number is even.
The result: If condition is true, exp_1 is executed, else exp_2 is executed.
Note: 0, None, False, emptylist, and emptyString evaluates as False.
And any data other than 0 evaluates to True.
Here's how it works:
If the condition [condition] becomes "True", then expression_1 will be evaluated, but not expression_2.
If we "and" something with 0 (zero), the result will always to be false. So in the below statement,
0 and exp
The expression exp won't be evaluated at all since "and" with 0 will always evaluate to zero and there is no need to evaluate the expression. This is how the compiler itself works, in all languages.
In
1 or exp
the expression exp won't be evaluated at all since "or" with 1 will always be 1. So it won't bother to evaluate the expression exp since the result will be 1 anyway (compiler optimization methods).
But in case of
True and exp1 or exp2
The second expression exp2 won't be evaluated since True and exp1 would be True when exp1 isn't false.
Similarly in
False and exp1 or exp2
The expression exp1 won't be evaluated since False is equivalent to writing 0 and doing "and" with 0 would be 0 itself, but after exp1 since "or" is used, it will evaluate the expression exp2 after "or".
Note:- This kind of branching using "or" and "and" can only be used when the expression_1 doesn't have a Truth value of False (or 0 or None or emptylist [ ] or emptystring ' '.) since if expression_1 becomes False, then the expression_2 will be evaluated because of the presence "or" between exp_1 and exp_2.
In case you still want to make it work for all the cases regardless of what exp_1 and exp_2 truth values are, do this:
[condition] and ([expression_1] or 1) or [expression_2];
Many programming languages derived from C usually have the following syntax of the ternary conditional operator:
<condition> ? <expression1> : <expression2>
At first, the Python's benevolent dictator for life (I mean Guido van Rossum, of course) rejected it (as non-Pythonic style), since it's quite hard to understand for people not used to C language. Also, the colon sign : already has many uses in Python. After PEP 308 was approved, Python finally received its own shortcut conditional expression (what we use now):
<expression1> if <condition> else <expression2>
So, firstly it evaluates the condition. If it returns True, expression1 will be evaluated to give the result, otherwise expression2 will be evaluated. Due to lazy evaluation mechanics – only one expression will be executed.
Here are some examples (conditions will be evaluated from left to right):
pressure = 10
print('High' if pressure < 20 else 'Critical')
# Result is 'High'
Ternary operators can be chained in series:
pressure = 5
print('Normal' if pressure < 10 else 'High' if pressure < 20 else 'Critical')
# Result is 'Normal'
The following one is the same as previous one:
pressure = 5
if pressure < 20:
if pressure < 10:
print('Normal')
else:
print('High')
else:
print('Critical')
# Result is 'Normal'
Yes, Python have a ternary operator, here is the syntax and an example code to demonstrate the same :)
#[On true] if [expression] else[On false]
# if the expression evaluates to true then it will pass On true otherwise On false
a = input("Enter the First Number ")
b = input("Enter the Second Number ")
print("A is Bigger") if a>b else print("B is Bigger")
Other answers correctly talk about the Python ternary operator. I would like to complement by mentioning a scenario for which the ternary operator is often used, but for which there is a better idiom. This is the scenario of using a default value.
Suppose we want to use option_value with a default value if it is not set:
run_algorithm(option_value if option_value is not None else 10)
or, if option_value is never set to a falsy value (0, "", etc.), simply
run_algorithm(option_value if option_value else 10)
However, in this case an ever better solution is simply to write
run_algorithm(option_value or 10)
The syntax for the ternary operator in Python is:
[on_true] if [expression] else [on_false]
Using that syntax, here is how we would rewrite the code above using Python’s ternary operator:
game_type = 'home'
shirt = 'white' if game_type == 'home' else 'green'
It's still pretty clear, but much shorter. Note that the expression could be any type of expression, including a function call, that returns a value that evaluates to True or False.
Python has a ternary form for assignments; however there may be even a shorter form that people should be aware of.
It's very common to need to assign to a variable one value or another depending on a condition.
>>> li1 = None
>>> li2 = [1, 2, 3]
>>>
>>> if li1:
... a = li1
... else:
... a = li2
...
>>> a
[1, 2, 3]
^ This is the long form for doing such assignments.
Below is the ternary form. But this isn't the most succinct way - see the last example.
>>> a = li1 if li1 else li2
>>>
>>> a
[1, 2, 3]
>>>
With Python, you can simply use or for alternative assignments.
>>> a = li1 or li2
>>>
>>> a
[1, 2, 3]
>>>
The above works since li1 is None and the interpreter treats that as False in logic expressions. The interpreter then moves on and evaluates the second expression, which is not None and it's not an empty list - so it gets assigned to a.
This also works with empty lists. For instance, if you want to assign a whichever list has items.
>>> li1 = []
>>> li2 = [1, 2, 3]
>>>
>>> a = li1 or li2
>>>
>>> a
[1, 2, 3]
>>>
Knowing this, you can simply such assignments whenever you encounter them. This also works with strings and other iterables. You could assign a whichever string isn't empty.
>>> s1 = ''
>>> s2 = 'hello world'
>>>
>>> a = s1 or s2
>>>
>>> a
'hello world'
>>>
I always liked the C ternary syntax, but Python takes it a step further!
I understand that some may say this isn't a good stylistic choice, because it relies on mechanics that aren't immediately apparent to all developers. I personally disagree with that viewpoint. Python is a syntax-rich language with lots of idiomatic tricks that aren't immediately apparent to the dabbler. But the more you learn and understand the mechanics of the underlying system, the more you appreciate it.
Pythonic way of doing the things:
"true" if var else "false"
But there always exists a different way of doing a ternary condition too:
"true" and var or "false"
There are multiple ways. The simplest one is to use the condition inside the "print" method.
You can use
print("Twenty" if number == 20 else "Not twenty")
Which is equivalent to:
if number == 20:
print("Twenty")
else:
print("Not twenty")
In this way, more than two statements are also possible to print. For example:
if number == 20:
print("Twenty")
elif number < 20:
print("Lesser")
elif 30 > number > 20:
print("Between")
else:
print("Greater")
can be written as:
print("Twenty" if number == 20 else "Lesser" if number < 20 else "Between" if 30 > number > 20 else "Greater")
The if else-if version can be written as:
sample_set="train" if "Train" in full_path else ("test" if "Test" in full_path else "validation")
Yes, it has, but it's different from C-syntax-like programming languages (which is condition ? value_if_true : value_if_false
In Python, it goes like this: value_if_true if condition else value_if_false
Example: even_or_odd = "even" if x % 2 == 0 else "odd"
A neat way to chain multiple operators:
f = lambda x,y: 'greater' if x > y else 'less' if y > x else 'equal'
array = [(0,0),(0,1),(1,0),(1,1)]
for a in array:
x, y = a[0], a[1]
print(f(x,y))
# Output is:
# equal,
# less,
# greater,
# equal
I find the default Python syntax val = a if cond else b cumbersome, so sometimes I do this:
iif = lambda (cond, a, b): a if cond else b
# So I can then use it like:
val = iif(cond, a, b)
Of course, it has the downside of always evaluating both sides (a and b), but the syntax is way clearer to me.

C++ operators precedence

I am using this statement
if ((pm && pn) || (pm == false && pn == false))
it is supposed to return true only if both pm and pn are true or if both are false. But this is also returning true if only only first one (pm) is true.
So now it is acting like this:
0 0 = 1
0 1 = 0
1 0 = 1
1 1 = 1
but I need it to work like this:
0 0 = 1
0 1 = 0
1 0 = 0
1 1 = 1
can you tell me where am I making mistake?
What you want is simply:
if (pm == pn)
You are checking if pm is true twice. You also need to check if both are the same, not whether they are both true. So,
if ((pm == pn)
^^ ^^
pm && pm
should be
pm && pn
^
The whole expression can be simplified to
pm == pn
if the variables already have bool type.
Why not try xor?
if (!(pm ^ pn)) { /*...*/ }
Or simply equal?
if (pm == pn) { /*...*/ }
if ((pm && pm) || (pm == false && pn == false))
it is supposed to return true only if both pm and pn are true or if both are false. But this is also returning true if only only first one (pm) is true.
Because you made a typo. You meant pm && pn.
Instead just write if (pm == pn), which is equivalent along as the only semantic values are indeed true and false for both variables.
Plus, consider making your variable names clearer and more distinct.
Note that operator precedence has nothing to do with this.
Since the question's title asks about precedence, note that || has lower precedence than &&. So the two sets of inner parentheses are redundant, and the original expression is just a longer way of saying
if (pm && pm || pm == false && pn == false)
Now, fixing the obvious typo:
if (pm && pn || pm == false && pn == false)
Removing the unneeded explicit comparisons:
if (pm && pn || !pm && !pn)
And, finally, a less obvious transformation, which others have suggested:
if (pm == pn)