Is there a way to do the following in just one line with lambdas in python?
squared = lambda x : x**2
cubed = lambda x : x**3
squareCubed = lambda x : squared(cubed(x))
without using any helper functions or reduce.
print(squareCubed(2)) # > 64
3 Answers
Sure, you can do something like:
squareCube = (lambda f, g: lambda x: f(g(x)))(lambda x: x**2, lambda x: x**3)
But why do this?
In any case, you shouldn't be assigning the result of lambda to a variable if you are following the official PEP8 style guide. What is wrong with:
def square(x):
return x**2
def cube(x):
return x**3
def compose(f, g):
return lambda x: f(g(x))
square_cube = compose(square, cube)
Is much more readable.
answered Aug 11, 2022 at 16:41
juanpa.arrivillaga
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2 Comments
cvb0rg
Thanks this makes sense. I'm a bit confused as to how in
var = () () the first parentheses know to look in the 2nd () for the definitions of f and g?juanpa.arrivillaga
The first parentheses just groups
(lambda f, g: lambda x: f(g(x)))(lambda x: x**2, lambda x: x**3), this results in a function (of course - it is a lambda expression), you then call that function, passing it the corresponding arguments. It's exactly like the more readable version I showed below itIf you want to hard code it:
squareCube = lambda x: (x**3)**2
But generically:
compose = lambda f, g: lambda x: f(g(x))
squareCube = compose(lambda x: x**2, lambda x: x**3)
Or, in one line:
squareCube = (lambda f, g: lambda x: f(g(x)))(lambda x: x**2, lambda x: x**3)
Comments
squareCubed = lambda x : (x**3)**2
print(squareCubed(2)) # gives 64
1 Comment
Trenton McKinney
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cubedwithlambda x: x**3andsquaredwithlambda x: x**2??