I like using lambdas more then I should so more than once I wanted to solve nested if statement with it and failed using standard syntax.
But then I tried something like this:
lambda x: 1 if x > 100 else 2 if x > 50 else 3
Surprisingly it works, although proper function is supposed to have elif:
def foo(x):
if x > 100:
return 1
elif x > 50:
return 2
else:
return 3
Any ideas why?
You don't have to use elif. You can use if inside another else which is the same as what lambda function is doing:
def foo(x):
if x > 100:
return 1
else:
if x > 50:
return 2
else:
return 3
lambda x: 3 - bisect.bisect_left([50, 100], x) :-)
I'd recommend
lambda x: 1 if x > 100 else (2 if x > 50 else 3)
for readability because explicit is better than implicit.
Indeed, due to evaluation from left to right, the ternary operator is right associative as the following code demonstrates
def f(a, b):
return 1 if a else 2 if b else 3
def left(a, b):
return (1 if a else 2) if b else 3
def right(a, b):
return 1 if a else (2 if b else 3)
tpl = "{:^8}{:^8}{:^6}{:^6}{:^6}"
print(tpl.format('a', 'b', 'f', 'left', 'right'))
for (a, b) in [(False, False),(False, True),(True, False),(True, True)]:
print(tpl.format(repr(a), repr(b), f(a,b), left(a,b), right(a,b)))
""" my output ->
a b f left right
False False 3 3 3
False True 2 2 2
True False 1 3 1
True True 1 1 1
"""
The ternary expression always gives the same result as the expression where everything after the first else is parenthesized.
When the conditions are (x>100) and (x>50), the case True, False can never happen, so the three expressions give the same result. Nevertheless, explicit is better than implicit!
You are actually using something like elif by writing
else 2 if x > 50
in
lambda x: 1 if x > 100 else 2 if x > 50 else 3
Its just a different way of writing something like elif in lambda
lambda x: 1 if x > 100 else 2 if x > 50 else 3
Means :
# 1 if x > 100
if x > 100:
return 1
# else 2 if x > 50 else 3
else:
# 2 if x > 50 else 3
if x > 50:
return 2
else:
return 3
Which does the same thing that your second snippet with elif.
You cannot use elif in a lambda expression. As far as I know, it should raise a SyntaxError.
You have concatenated two conditional expressions.
One
2 if x > 50 else 3
provides the value for the else clause of the other.
1 if x > 100 else ...
You have the conditional expression
x if C else y
where
x = 1
C = 100
and
y = 2 if x > 50 else 3
From the docs, y can be an expression so y can be a conditional expression.
lambda x: 1 if x > 100 else (2 if x > 50 else 3)for readability because explicit is better than implicit.elif?