How do I pass a value into a python function without calling?

Question

I was unable to find a reasonable way to create a variable which calls a function requiring parameters.

Here is a simplified version of my code. I would like print_hello to print hello when it is called, and not when it is defined.

print_hello = print('hello')

When I define print_hello, it calls print('hello'). When I call print_hello, it gives me an error. How do I fix this?

Answer 1

If you just want a function that does precisely what you describe, Sheldore's answer is the simplest way to go (and more Pythonic than using a named lambda).

An alternative approach is to make a partial application of the function with functools.partial, which allows you to pass additional arguments at call time:

from functools import partial

print_hello = partial(print, "hello")

print_hello()  # Prints "hello" to stdout

print_hello(file=sys.stderr)  # Prints "hello" to stderr

print_hello("world")  # Prints "hello world" to stdout

Answer 2

Just define print_hello as a lambda function

>>> print_hello = lambda: print('hello')
>>> print_hello()
hello

To delay execution, you'll have to wrap the call to print in another function. A lambda is less code than defining another function.

Note: that pep08 recommends using a def function rather than a lambda when assigning to a variable. See here. So @Sheldores answer is probably the way to go.