Python save function with argumnets to variable without to call a function

Question

It is easy to save function into a variable like

pr = print
pr(5)  # 5

But if this possible to save function with arguments without a call, like

some_var = defer print(5)  # No call! 
some_var()  # 5

I tried to use lambda, but it's lead to syntaxys error `l = lambda 5:

Why I need it? For example to no repeat multiple "if" branches:

example:

def foo()
l1 = lambda: 1
l2 = lambda: 2
if 1:
    func = l1
elif 2:
    func = l2
else:
   func = some_outer_func, some_inner_func  
return func  # To use "func" need additional "if" branches for type and length of a returned value

docs.python.org/3.8/library/functools.html#functools.partial — Iain Shelvington
– Iain Shelvington, Commented Aug 5, 2020 at 3:20

Iain Shelvington · Accepted Answer · 2020-08-05 03:24:45Z

1

functools.partial takes a function of many arguments and returns a function with fewer arguments with some of the arguments "saved"

from functools import partial

print_five = partial(print, 5)
print_five()  # 5

It also works with keyword arguments

def foo(a, b=False):
    print(a, b)

bar = partial(foo, b=True)
foo(1)  # 1 False
bar(1)  # 1 True

answered Aug 5, 2020 at 3:24

Iain Shelvington

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2 Comments

salius Over a year ago

Very unfortunately that python does not have special key word for this. Anyway I can't use it with chunks of functions like outer(inner(defered_func)) # need partial for every func

Iain Shelvington Over a year ago

Are you looking to chain function calls?

Kevin Mayo · Accepted Answer · 2020-08-05 03:33:02Z

1

The way with lambda:

pr = lambda: print(5)
pr()

answered Aug 5, 2020 at 3:33

Kevin Mayo

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