I came across a pretty clever little function that takes two functions, applies one on top of each other given an argument x:
def compose(f,g):
return lambda *x: f(g(*x))
Now my issue is with *x, as I don't see it really doing anything here. Why couldn't it be simple x (without the asterisk)?
Here are my tests:
>>> def compose(f,g):
... return lambda *x: f(g(*x))
...
>>> this = lambda i: i+1
>>> that = lambda b: b+1
>>> compose(this,that)(2)
4
>>> def compose(f,g):
... return lambda x: f(g(x))
...
>>> compose(this,that)(2)
4
>>> def compose(f,g):
... return lambda *x: f(g(*x))
...
>>> compose(this,that)(2,2)
TypeError: <lambda>() takes exactly 1 argument (2 given)
If g (that in your tests) can also take a variable number of arguments, then lambda *x: f(g(*x)) can be useful.
Otherwise, not so much.
The aim is to allow the composed function to be invoked with any number of arguments, and for all these arguments to be passed to the inner function in the composition.
To complement #Frederic's answer, it would be most flexible if compose used the standard *args, **kwargs construct:
def compose(f,g):
return lambda *args, **kwargs: f(g(*args, **kwargs))
This way, compose works with any g function, regardless of g's signature.
The one in your question is only using the first part, i.e. the *args, and using a different (unconventional) name for it, *x.
The problem with your tests is that your composition functions this and that only accept a single argument, so the point of the * in your compose function is lost (and you receive a "takes exactly 1 argument " error).
The power of the * operator becomes apparent when you want to pass in unpacked tuples, and your lambdas support this:
Try this simple map reduce example:
this = lambda *i: *[x**2 for x in i] # map tuple i
that = lambda *b: sum(b) # reduce tuple b
def compose(f,g):
return lambda *x: f(*g(*x))
compose(that,this)(2,2)
Related
For lambda functions in the following code,
def myfunc(n):
return lambda a : a * n
mydoubler = myfunc(2)
print(mydoubler(11))
I am trying to understand why mydoubler becomes <class 'function'> and how I can call mydoubler(11) without defining it as a function.
A lambda is a function, but with only one expression (line of code).
That expression is executed when the function is called, and the result is returned.
So the equivalent of this lambda:
double = lambda x: x * 2
is this def function:
def double(x):
return x * 2
You can read more here
A lambda is a function, so your code is doing something like
def myfunc(n):
def result(a):
return a * n
return result # returns a function
mydoubler = myfunc(2)
print(f(11))
You're asking how to call mydoubler without defining it as a function, which isn't the clearest question, but you can call it without naming it like so
print( myfunc(2)(11) )
Your myfunc is returning a lambda. Lambda is a small anonymous function. A lambda can take any number of arguments, but can only have one expression.
So after execution of the 3rd line, your mydoubler will become a lambda that's why when you try print(type(mydoubler)) it will return <class 'function'>.
Also in order to call mydoubler with 11, it must be function.
A lambda expression, like a def statement, defines functions. Your code could be equivalently written as
def myfunc(n):
def _(a):
return a * n
return _
mydoubler = myfunc(2)
print(mydoubler(11))
Because the inner function is simple enough to be defined as a single expression, using a lambda expression saves you the trouble of coming up with the otherwise unused name the def statement requires.
The key here is that the inner function closes over the value of n, so that the function returned by myfunc retains a reference to the value of the argument passed to myfunc. That is, mydoubler is hard-coded to multiply its argument by 2, rather than whatever value n may get later. (Indeed, the purpose of the closure is to create a new variable n used by the inner function, one which cannot easily be changed from outside myfunc.)
using decorator you can achive this
from functools import wraps
def decorator_func_with_args(arg1):
def decorator(f):
#wraps(f)
def wrapper(val):
result = f(val)
return result(arg1)
return wrapper
return decorator
#decorator_func_with_args(arg1=2)
def myfunc(n):
return lambda arg:arg*n
result = myfunc(1211)
print(result)
output
2422
Do you mean this?
mydoubler = lambda a : a * 2
mydoubler(11)
Lets say we have functions in python:
def _abs(iterable): #cause its normally for one element only
return [abs(i) for i in iterable]
def A(list, foo):
return foo(list)
list = [2,3,-5]
print( A(list,foo=sum) )
>> 0
while I may pass foo=sum to A, I am looking for an elegant way to pass something like foo=sum(_abs) to perform sum(_abs(list)).
The only way I see it now is to send a list of functions [sum, _abs] and apply them in order. Is there a better way?
Or, to compose more generally (i.e. with an arbitrary number of argument functions):
from functools import partial, reduce
def compose(*funcs):
return partial(reduce, lambda x, f: f(x), reversed(funcs))
(see the docs on partial and reduce - note you don't need to import reduce in Python 2.x)
Then:
>>> compose(sum, _abs)([2, 3, -5])
10
You could make an explicit compose function
>>> def compose(f, g):
... return lambda x: f(g(x))
...
Then use it
>>> A(list, compose(sum, _abs))
10
def compose(funcs, funcargs):
for func in reversed(funcs):
funcargs = func(funcargs)
return funcargs
I am trying to use currying to make a simple functional add in Python. I found this curry decorator here.
def curry(func):
def curried(*args, **kwargs):
if len(args) + len(kwargs) >= func.__code__.co_argcount:
return func(*args, **kwargs)
return (lambda *args2, **kwargs2:
curried(*(args + args2), **dict(kwargs, **kwargs2)))
return curried
#curry
def foo(a, b, c):
return a + b + c
Now this is great because I can do some simple currying:
>>> foo(1)(2, 3)
6
>>> foo(1)(2)(3)
6
But this only works for exactly three variables. How do I write the function foo so that it can accept any number of variables and still be able to curry the result? I've tried the simple solution of using *args but it didn't work.
Edit: I've looked at the answers but still can't figure out how to write a function that can perform as shown below:
>>> foo(1)(2, 3)
6
>>> foo(1)(2)(3)
6
>>> foo(1)(2)
3
>>> foo(1)(2)(3)(4)
10
Arguably, explicit is better than implicit:
from functools import partial
def example(*args):
print("This is an example function that was passed:", args)
one_bound = partial(example, 1)
two_bound = partial(one_bound, 2)
two_bound(3)
#JohnKugelman explained the design problem with what you're trying to do - a call to the curried function would be ambiguous between "add more curried arguments" and "invoke the logic". The reason this isn't a problem in Haskell (where the concept comes from) is that the language evaluates everything lazily, so there isn't a distinction you can meaningfully make between "a function named x that accepts no arguments and simply returns 3" and "a call to the aforementioned function", or even between those and "the integer 3". Python isn't like that. (You could, for example, use a zero-argument call to signify "invoke the logic now"; but that would break special cases aren't special enough, and require an extra pair of parentheses for simple cases where you don't actually want to do any currying.)
functools.partial is an out-of-box solution for partial application of functions in Python. Unfortunately, repeatedly calling partial to add more "curried" arguments isn't quite as efficient (there will be nested partial objects under the hood). However, it's much more flexible; in particular, you can use it with existing functions that don't have any special decoration.
You can implement the same thing as the functools.partial example for yourself like this:
def curry (prior, *additional):
def curried(*args):
return prior(*(args + additional))
return curried
def add(*args):
return sum(args)
x = curry(add, 3,4,5)
y = curry(b, 100)
print y(200)
# 312
It may be easier to think of curry as a function factory rather than a decorator; technically that's all a decorator does but the decorator usage pattern is static where a factory is something you expect to be invoking as part of a chain of operations.
You can see here that I'm starting with add as an argument to curry and not add(1) or something: the factory signature is <callable>, *<args> . That gets around the problem in the comments to the original post.
FACT 1: It is simply impossible to implement an auto currying function for a variadic function.
FACT 2: You might not be searching for curry, if you want the function that will be passed to it * to know* that its gonna be curried, so as to make it behave differently.
In case what you need is a way to curry a variadic function, you should go with something along these lines below (using your own snipped):
def curryN(arity, func):
"""curries a function with a pre-determined number of arguments"""
def curried(*args, **kwargs):
if len(args) + len(kwargs) >= arity:
return func(*args, **kwargs)
return (lambda *args2, **kwargs2:
curried(*(args + args2), **dict(kwargs, **kwargs2)))
return curried
def curry(func):
"""automatically curries a function"""
return curryN(func.__code__.co_argcount, func);
this way you can do:
def summation(*numbers):
return sum(numbers);
sum_two_numbers = curryN(2, summation)
sum_three_numbers = curryN(3, summation)
increment = curryN(2, summation)(1)
decrement = curryN(2, summation)(-1)
I think this is a decent solution:
from copy import copy
import functools
def curry(function):
def inner(*args, **kwargs):
partial = functools.partial(function, *args, **kwargs)
signature = inspect.signature(partial.func)
try:
signature.bind(*partial.args, **partial.keywords)
except TypeError as e:
return curry(copy(partial))
else:
return partial()
return inner
This just allow you to call functools.partial recursively in an automated way:
def f(x, y, z, info=None):
if info:
print(info, end=": ")
return x + y + z
g = curry_function(f)
print(g)
print(g())
print(g(2))
print(g(2,3))
print(g(2)(3))
print(g(2, 3)(4))
print(g(2)(3)(4))
print(g(2)(3, 4))
print(g(2, info="test A")(3, 4))
print(g(2, info="test A")(3, 4, info="test B"))
Outputs:
<function curry.<locals>.inner at 0x7f6019aa6f28>
<function curry.<locals>.inner at 0x7f6019a9a158>
<function curry.<locals>.inner at 0x7f6019a9a158>
<function curry.<locals>.inner at 0x7f6019a9a158>
<function curry.<locals>.inner at 0x7f6019a9a0d0>
9
9
9
test A: 9
test B: 9
I have a bunch of functions in a list:
funcs = [f1, f2, f3, f4, f5]
and all of the functions take in return a single argument, eg.
f1 = lambda x: x*2
I'd like to map all these functions together
result = lambda x: f5(f4(f3(f2(f1(x)))))
or, iterating over funcs
def dispatch(x):
for f in funcs:
x = f(x)
return x
dispatch works fine, but I couldn't figure out a clean way to do this using iterools. Is it possible? Does this sequential function mapping idiom have a name?
There is no point in using itertools here; you are producing one output, and you could not apply this to an infinite iterable. You have to have a finite number of functions in the input iterable for this to work at all.
Use the reduce() function:
from functools import reduce
x = reduce(lambda res, func: func(res), funcs, x)
The functools.reduce() import helps the above work in both Python 2 and 3.
reduce(), together with map(), filter() and, yes, itertools, is an often used tool in functional programming.
Another (less efficient, alas) way of looking at Martijn's answer is to realize that you want to compose the list of functions.
# function composition: compose(f,g)(x) = f(g(x))
def compose(f, g):
return lambda x: f(g(x))
# Identity for function composition
# compose(f, identity)(x) = f(x)
identity = lambda x: x
# result(x) = f1(f2(...fn(x)...))
result = reduce(compose, funcs, identity)
I would like to do something like the following:
def getFunction(params):
f= lambda x:
do stuff with params and x
return f
I get invalid syntax on this. What is the Pythonic/correct way to do it?
This way I can call f(x) without having to call f(x,params) which is a little more messy IMO.
A lambda expression is a very limited way of creating a function, you can't have multiple lines/expressions (per the tutorial, "They are syntactically restricted to a single expression"). However, you can nest standard function definitions:
def getFunction(params):
def to_return(x):
# do stuff with params and x
return to_return
Functions are first-class objects in Python, so once defined you can pass to_return around exactly as you can with a function created using lambda, and either way they get access to the "closure" variables (see e.g. Why aren't python nested functions called closures?).
It looks like what you're actually trying to do is partial function application, for which functools provides a solution. For example, if you have a function multiply():
def multiply(a, b):
return a * b
... then you can create a double() function1 with one of the arguments pre-filled like this:
from functools import partial
double = partial(multiply, 2)
... which works as expected:
>>> double(7)
14
1 Technically a partial object, not a function, but it behaves in the same way.
You can't have a multiline lambda expression in Python, but you can return a lambda or a full function:
def get_function1(x):
f = lambda y: x + y
return f
def get_function2(x):
def f(y):
return x + y
return f