Question
Does Python have a ternary conditional operator?
Is there a ternary conditional operator in Python?
Question
Is there a ternary conditional operator in Python?
Solution
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:
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).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:
Solution
Solution
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.
Solution
<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
Solution
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.
Solution
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
Solution
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.
Solution
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.
Solution
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
Solution
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
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.
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.
Solution
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
Addendum:
A special case of boolean indexing is when you need a single character. E.g.:
sign = '+-'[n < 0]
And finally, the plural s
:
print(f"total: {n} item{'s'[n==1:]}")
Solution
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.
Solution
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'
Solution
a if condition else b
Just memorize this pyramid if you have trouble remembering:
condition
if else
a b
Solution
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
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.
Solution
The ternary conditional operator simply allows testing a condition in a single line replacing the multiline if-else making the code compact.
[on_true] if [expression] else [on_false]
# 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
# 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
# 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
Solution
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.
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];
Solution
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.
Solution
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'
Solution
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")
Solution
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)
Solution
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.
Solution
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 simplify 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.
Solution
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"
Solution
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")
Solution
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")
Solution
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"
Solution
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
Solution
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.
Solution
I have data coming from a device as a string and leaving it as a rebuild string. Conditional expression need to be limited by (). This allow to have multiple conditions to build the string in one line. If not it seems like whatever after the "else" will be accounted.
d0 = "-679 58 1029"
d1 = d0.split(" ")
strg = (d1[0][:-2] if len(d1[0])>= 3 else "0") + " " +d1[0][-2:]+ " "+ (d1[1][:-2] if len(d1[1])>= 3 else "0") + " " + d1[1][-2:] + " " +d1[2]
print(strg)