Lets say I have three functions in one module as defined below:
def add_nums(a, b):
return a + b
def sum_and_square_one(add_nums, a, b):
result = add_nums(a,b)
return result*result
def sum_and_square_two(a, b):
result = add_nums(a,b)
return result*result
Both functions sum_and_square_one and sum_and_square_two do the same task. But the former takes add_nums as an argument while the latter calls add_nums inside. My question is which one is the better way. Passing a function as an argument or calling inside a function?
If you always want to call the same function then there is no point in passing it as an argument every time. (this is the case in your example)
If you, however, want to dynamically call different (maybe similar) functions, then passing the appropiate function as an argument would make sense.
Depends on your use case. Do you need the function to change the way it works depending on that argument?
Or, rewriting your meaningless example into a bit less meaningless example (but still quite):
def add_nums(a, b):
return a + b
def do_something_then_square(a, b, what_to_do):
result = what_to_do(a, b)
return result * result
def sum_then_square(a, b):
result = add_nums(a, b)
return result * result
# Then you do:
sum_then_square(2, 5) # 49
do_something_then_square(2, 5, what_to_do=add_nums) # 49
do_something_then_square(2, 5, what_to_do=lambda a, b: a + b) # 49
do_something_then_square(2, 5, what_to_do=lambda a, b: a * b) # 100
do_something_then_square(2, 5, what_to_do=min) # 4
do_something_then_square(2, 5, what_to_do=complex) # (-21+20j)
Only question is: do you need that added flexibility? Otherwise it's just useless additional typing. That question must be answered on a case by case basis; it needs a complete example to give a useful answer.
Related
I have problem in understanding this code in Python
x = layers.Flatten()(last_output)
Since Flatten is a function, how does the function get the data from last_output written outside the function call parenthesis. Don't remember seeing this kind of code in Java.
Thanks and Regards
Flatten() is the class instantiation (which is probably clear to you) and the second calls the instance with that parameter. For this to work the class must have a __call__ function defined.
Example:
class Sum:
def __call__(self, a, b, c):
return a + b + c
s = Sum()
print(s(3, 4, 5))
print(Sum()(3,4,5))
Also same behavior can be obtained with a function that returns another function with arguments:
def Sum2():
def Sum3(a, b, c):
return a + b + c
return Sum3
s2 = Sum2()
print(s2(3, 4, 5))
print(Sum2()(3, 4, 5))
Consider this
def outer():
def print_thrice(string):
for _ in range(3):
print (string)
return print_thrice
If you call outer, it will return the print_thrice function which you can then call. So you'd use it like this
x = outer()
x("hello")
Or more compactly, outer()("hello") which is what's going on here.
Following is an example of what I would like to do.
def f(a, b, c):
if a == 'method1':
c = 0
return b + c
In this function, parameter c is unneeded if the condition a='method1' is satisfied.
Still, I can call the function with f(a='method1', b=1, c=2), which will have the same effect as f(a='method1', b=1, c=0).
A solution to this is to set the parameter c default to 0
def f(a, b, c=0):
if a == 'method1':
c = 0
return b + c
Now I can call f(a='method1',b=1), which is exactly what I want. The problem is, I can still change the parameter c in the call f(a='method1',b=1,c=1), which I do not want the user to be able to.
Can I enforce this condition in the function signature, and not in the body (i.e. I would not like to use if in the body). Or if there is another better solution, please tell.
Something like
def f(a, b, c = 0 if a == 'method1', else c is required):
return b + c
Thanks in advance.
a, b and c are all assigned dynamically at runtime. There is no way you can make up for this in the signature. It needs to be detected at runtime and you might as well do that in an if as anywhere else. You can specialize at the function name level, though, and python will take care of detecting the number of parameters.
def f(b,c):
return b + c
def f_method1(b):
return f(b, 0)
def f_method2(half_a_c):
return f(0, half_a_c*2)
Hmm... this almost seems like something that you should be able to do with functools.singledispatch and typing.Literal, but I can't quite get them to work together at least in python 3.7 (with Literal coming from the typing_extensions module). I think that in general singledispatch is probably the only tool that will really get what you've asked for as the different registered functions can have entirely different signatures. However, to do this our methods need to be different classes.
from functools import singledispatch
class Method1():
pass
class OtherMethods():
pass
#singledispatch
def f(method, b, c):
return b + c
#f.register(Method1)
def _(method, b):
return b
f(Method1(), 12) # returns 12
f(Method1(), 12, 7) # raises an error
f(OtherMethods(), 12) # raises an error
f(OtherMethods(), 12, 7) # returns 19
Now this is not exactly what you asked for but the arguments are enforced in the signature.
If someone who knows more about the implementation of singledispatch and Literal comes by maybe they can explain the interaction between the two.
One easier thing to do would be to simply define c to default to some invalid value.
def f(a, b, c=None):
if a == 'method1':
c = 0
return b + c
This solves the problem that if the user forgets to set c for a method besides method1 they'll receive a (possibly cryptic) error message. However it doesn't fix the fact that if they set c when using method1 that value will be silently ignored and possibly cause confusion.
Consider the Python function line defined as follows:
def line(m, b):
def inner_function(x):
return m * x + b
return inner_function
This function has the property that for any floats m and b, the object line(m, b) is a Python function, and when line(m, b) is called on a float x, it returns a float line(m, b)(x). The float line(m, b)(x) can be interpreted as the value of the line with slope m and y-intercept b at the point x. This is one method for writing a Python function that "depends on parameters" m and b.
Is there a special name for this method of writing a Python function that depends on some parameters?
Is there a more Pythonic and/or computationally efficient way to write a function that does the same thing as line above?
This is called a closure, and it's a perfectly reasonable way to write one, as well as one of the most efficient means of doing so (in the CPython reference interpreter anyway).
The only other common pattern I know of is the equivalent of C++'s functors, where a class has the state as attributes, and the additional parameters are passed to __call__, e.g. to match your case:
class Line:
def __init__(self, m, b):
self.m = m
self.b = b
def __call__(self, x):
return self.m * x + self.b
It's used identically, either creating/storing an instance and reusing it, or as in your example, creating it, using it once, and throwing it away (Line(m, b)(x)). Functors are slower than closures though (as attribute access is more expensive than reading from nested scope, at least in the CPython reference interpreter), and as you can see, they're more verbose as well, so I'd generally recommend the closure unless your needs require the greater flexibility/power of class instances.
I support #ShaddowRanger's answer. But using partial is another nice approach.
import functools
def f(m, b, x):
return m * x + b
line = functools.partial(f, 2, 3)
line(5)
=> 13
One thing which is worth pointing out is that lambda objects, and OP's inner_function aren't pickleable, whereas line here, as well as #ShaddowRanger's Line objects are, which makes them a bit more useful.
This is a little shorter:
def line(m,b):
return lambda x: m*x+b;
Is it possible to define a closure for a function which is already defined?
For example I'd like to have a "raw" function and a function which already has some predefined values set by a surrounding closure.
Here is some code showing what I can do with a closure to add predefined variables to a function definition:
def outer(a, b, c):
def fun(d):
print(a + b + c - d)
return fun
foo = outer(4, 5, 6)
foo(10)
Now I want to have a definition of fun outside of a wrapping closure function, to be able to call fun either with variables from a closure or by passing variables directly. I know that I need to redefine a function to make it usable in a closure, thus I tried using lambda for it:
def fun(a, b, c, d): # raw function
print(a + b + c - d)
def clsr(func): # make a "closure" decorator
def wrap(*args):
return lambda *args: func(*args)
return wrap
foo = clsr(fun)(5, 6, 7) # make a closure with values already defined
foo(10) # raises TypeError: fun() missing 3 required positional arguments: 'a', 'b', and 'c'
fun(5, 6, 7, 10) # prints 8
What I also tried is using wraps from functools, but I was not able to make it work.
But is this even possible? And if yes: Is there any module which already implements decorators for this?
You can just define the wrap on the fly:
def fun(a, b, c, d): # raw function
print(a + b + c - d)
def closed(d): fun(5,6,7,d)
closed(10)
You can use this with lambda, but #juanpa points out you should not if there is no reason to. The above code will result in 8. This method by the way is not Python specific, most languages would support this.
But if you need a closure in a sense that it relies on the wrapper variables, than no, and there is good reason not to. This will create essentially a non-working function, that relies on wrapping. In this case using a class maybe better:
class fun:
def __init__(self,*args): #Can use specific things, not just *args.
self.args = args #Or meaningful names
def __call__(self,a, b, c, d): # raw function
print(a + b + c - d,self.args)
def closed(d):
fun("some",3,"more",['args'])(5,6,7,d)
closed(10)
or using *args/**kwargs directly and passing extra variables through that. Otherwise I am not familiar with a "inner function" construct that only works after wrapping.
I've been having trouble understanding what h(a)(b) means. I'd never seen one of those before yesterday, and I couldn't declare a function this way:
def f (a)(b):
return a(b)
When I tried to do def f (a, b):, it didn't work either. What do these functions do? How can I declare them? And, finally, what's the difference between f(a, b)and f(a)(b)?
Functions with multiple parameter brackets don't exist, as you saw when you tried to define one. There are, however, functions which return (other) functions:
def func(a):
def func2(b):
return a + b
return func2
Now when you call func() it returns the inner func2 function:
>>> func2 = func(1) # You don't have to call it func2 here
>>> func2(2)
3
But if you don't need the inner function later on, then there's no need to save it into a variable and you can just call them one after the other:
>>> func(1)(2) # func(1) returns func2 which is then called with (2)
3
This is a very common idiom when defining decorators that take arguments.
Notice that calling func() always creates a new inner function, even though they're all named func2 inside of the definition of our func:
>>> f1 = func(1)
>>> f2 = func(1)
>>> f1(1), f2(1)
(2, 2)
>>> f1 is f2
False
And, finally, what's the difference between f(a, b)and f(a)(b)?
It should be clear now that you know what f(a)(b) does, but to summarize:
f(a, b) calls f with two parameters a and b
f(a)(b) calls f with one parameter a, which then returns another function, which is then called with one parameter b
f(a)(b) just means that the expression f(a) returns a value that is itself callable. It's a short form of
g = f(a)
g(b)
You might be more comfortable adding a pair of redundant parentheses to emphasize that this is not a single syntactic construct.
(f(a))(b) # f(a) is evaluated first, then the result is applied to b
It is exactly analogous to the same doubling of square brackets for indexing nested dictionaries.
d1[x][y]
is equivalent to
d2 = d1[x]
d2[y]
Lets say we have an expression like
f(a)(b)
then, f(a) returns a function itself which gets invoked with argument b. Consider the following example
def f(a):
def g(b):
return a * b
return g
Then f(5)(4) evaluates to 5 * 4, since f(5) returns a function which is basically
def g(b):
return 5 * b
One could now do stuff like this
mult_by_5 = f(5)
[mult_by_5(x) for x in range(10)]
Let's be fancy, what about more nested functions?:
def f(a):
def g(b):
def h(c):
return a * b *c
return h
return g
f(2)(3)(4) # 24