Input of a function contains output of another function - python

A bit of a general question that I cannot find the solution for,
I currently have two functions
def func1(*args, **kwargs):
...
return a,b
and
def func2(x,y):
...
return variables
I would like my code to evaluate
variables = func2(func1())
Which python does not accept as it says func2 requires two arguments when only one is given. My current solution is doing an intermediate dummy redefinition but makes my code extremely cluttered (my "func1" has an output of many parameters).
Is there an elegant solution to this?

def func1():
return 10, 20
def func2(x, y):
return x + y
results = func2(*func1())
print results
--output:--
30
A function can only return one thing, so func1() actually returns the tuple (10, 20). In order to get two things, you need to explode the tuple with *.

Related

Python distinguish between returned tuple and multiple values

I want to write a wrapper function which call one function and pass the results to another function. The arguments and return types of the functions are the same, but I have problem with returning lists and multiple values.
def foo():
return 1,2
def bar():
return (1,2)
def foo2(a,b):
print(a,b)
def bar2(p):
a,b=p
print(a,b)
def wrapper(func,func2):
a=func()
func2(a)
wrapper(bar,bar2)
wrapper(foo,foo2)
I am searching for a syntax which works with both function pairs to use it in my wrapper code.
EDIT: The definitions of at least foo2 and bar2 should stay this way. Assume that they are from an external library.
There is no distinction. return 1,2 returns a tuple. Parentheses do not define a tuple; the comma does. foo and bar are identical.
As I overlooked until JacobIRR's comment, your problem is that you need to pass an actual tuple, not the unpacked values from a tuple, to bar2:
a = foo()
foo2(*a)
a = bar()
bar2(a)
I don't necessarily agree with the design, but following your requirements in the comments (the function definitions can't change), you can write a wrapper that tries to execute each version (packed vs. unpacked) since it sounds like you might not know what the function expects. The wrapper written below, argfixer, does exactly that.
def argfixer(func):
def wrapper(arg):
try:
return func(arg)
except TypeError:
return func(*arg)
return wrapper
def foo():
return 1,2
def bar():
return (1,2)
#argfixer
def foo2(a,b):
print(a,b)
#argfixer
def bar2(p):
a,b=p
print(a,b)
a = foo()
b = bar()
foo2(a)
foo2(b)
bar2(a)
bar2(b)
However, if you aren't able to put the #argfixer on the line before the function definitions, you could alternatively wrap them like this in your own script before calling them:
foo2 = argfixer(foo2)
bar2 = argfixer(bar2)
And as mentioned in previous comments/answers, return 1,2 and return (1,2) are equivalent and both return a single tuple.
This code does not run because of arg differences. It runs if you use def foo2(*args): and def bar2(*p):.
The return 1, 2 and return (1, 2) are equivalent. The comma operator just creates a tuple, whether it is enclosed in parentheses or not.
All programming languages that I know of return a single value, so, since you want to return multiple, those values must be wrapped into a collection type, in this case, a tuple.
The problem is in the way you call the second function. Make it bar2(a) instead of bar2(*a), which breaks the tuple into separate arguments.

Calling the result of a function in another function

The code is very long so I won't type it in.
What I am confused about as a beginner programmer, is function calling. So I had a csv file that the function divided all the content (they were integers) by 95 to get the normalised scores.
I finished the function by returning the result. its called return sudentp_file
Now I want to continue this new variable into another function.
So this new function will get the average of the studentp_file. So I made a new function. Ill add the other function as a template of what im doing.
def normalise(student_file, units_file)
~ Do stuff here ~
return studentp_file
def mean(studentp_file):
mean()
What I get confused about is what to put in the mean(). Do I keep it or remove it? I understand you guys don't know the file I'm working with my a little basic understanding of how functions and function calling works would be appreciated. Thanks.
When you call your function you need to pass in the parameters it needs (based on what you specified in your def statement. So you code might look like this:
def normalise(student_file, units_file)
~ Do stuff here ~
return studentp_file
def mean(studentp_file):
~ other stuff here ~
return mean
# main code starts here
# get student file and units file from somewhere, I'll call them files A and B. Get the resulting studentp file back from the function call and store it in variable C.
C = normalize(A, B)
# now call the mean function using the file we got back from normalize and capture the result in variable my_mean
my_mean = mean(C)
print(my_mean)
i assume that normalise function is executed prior to mean function? if so try out this structure:
def normalise(student_file, units_file):
#do stuff here
return studentp_file
def mean(studentp_file):
#do stuff here
sp_file = normalise(student_file, units_file)
mean(sp_file)
functions in python(2/3) are made for reusability and to keep your code organized in a block. these functions may or may not return a value, based on arguments you pass (if it accepts arguments). think of it as if functions are like real life factories making finished products. raw goods are fed into factories, so that they produce a finished product. functions are also like that. :)
now, notice that i assigned a variable called sp_file with the value of the function call normalise(...). this function call - accepted parameters (student_file, units_file) - which are your 'raw' goods to be fed towards your function normalise.
return - basically returns whatever value towards the point in your code which called your function. in this case return, returns the value of studentp_file back to sp_file. sp_file would then get studentp_file's value and can be then passed to mean() function.
/ogs
Well, it's unclear buy why not just (dummy example):
def f(a,b):
return f2(3)+a+b
def f2(c):
return c+1
Call the f2 in f and do return in f2
If the results from function one will always be called to function two you could do this.
def f_one(x, y):
return (f_two(x, y))
def f_two(x, y):
return x + y
print(f_one(1, 1))
2
Or just a thought... You could set up a variable z that works as a switch, if its 1 it passes the result to function to the next function , or if 2 returns result of function one
def f_one(x, y, z):
result = x + y
if z == 1:
return (f_two(result))
elif z == 2:
return result
def f_two(x):
return x - 1
a = f_one(1, 1, 1)
print(a)
b = f_one(1, 1, 2)
print(b)

strange returning value in a python function

def cons(a, b):
def pair(f):
return f(a, b)
return pair
def car(f):
def left(a, b):
return a
return f(left)
def cdr(f):
def right(a, b):
return b
return f(right)
Found this python code on git.
Just want to know what is f(a,b) in cons definition is, and how does it work?
(Not a function I guess)
cons is a function, that takes two arguments, and returns a function that takes another function, which will consume these two arguments.
For example, consider the following function:
def add(a, b):
return a + b
This is just a function that adds the two inputs, so, for instance, add(2, 5) == 7
As this function takes two arguments, we can use cons to call this function:
func_caller = cons(2, 5) # cons receives two arguments and returns a function, which we call func_caller
result = func_caller(add) # func_caller receives a function, that will process these two arguments
print(result) # result is the actual result of doing add(2, 5), i.e. 7
This technique is useful for wrapping functions and executing stuff, before and after calling the appropriate functions.
For example, we can modify our cons function to actually print the values before and after calling add:
def add(a, b):
print('Adding {} and {}'.format(a, b))
return a + b
def cons(a, b):
print('Received arguments {} and {}'.format(a, b))
def pair(f):
print('Calling {} with {} and {}'.format(f, a, b))
result = f(a, b)
print('Got {}'.format(result))
return result
return pair
With this update, we get the following outputs:
func_caller = cons(2, 5)
# prints "Received arguments 2 and 5" from inside cons
result = func_caller(add)
# prints "Calling add with 2 and 5" from inside pair
# prints "Adding 2 and 5" from inside add
# prints "Got 7" from inside pair
This isn't going to make any sense to you until you know what cons, car, and cdr mean.
In Lisp, lists are stored as a very simple form of linked list. A list is either nil (like None) for an empty list, or it's a pair of a value and another list. The cons function takes a value and a list and returns you another list just by making a pair:
def cons(head, rest):
return (head, rest)
And the car and cdr functions (they stand for "Contents of Address|Data Register", because those are the assembly language instructions used to implement them on a particular 1950s computer, but that isn't very helpful) return the first or second value from a pair:
def car(lst):
return lst[0]
def cdr(lst):
return lst[1]
So, you can make a list:
lst = cons(1, cons(2, cons(3, None)))
… and you can get the second value from it:
print(car(cdr(lst))
… and you can even write functions to get the nth value:
def nth(lst, n):
if n == 0:
return car(lst)
return nth(cdr(lst), n-1)
… or print out the whole list:
def printlist(lst):
if lst:
print(car(lst), end=' ')
printlist(cdr(lst))
If you understand how these work, the next step is to try them on those weird definitions you found.
They still do the same thing. So, the question is: How? And the bigger question is: What's the point?
Well, there's no practical point to using these weird functions; the real point is to show you that everything in computer science can be written with just functions, no built-in data structures like tuples (or even integers; that just takes a different trick).
The key is higher-order functions: functions that take functions as values and/or return other functions. You actually use these all the time: map, sort with a key, decorators, partial… they’re only confusing when they’re really simple:
def car(f):
def left(a, b):
return a
return f(left)
This takes a function, and calls it on a function that returns the first of its two arguments.
And cdr is similar.
It's hard to see how you'd use either of these, until you see cons:
def cons(a, b):
def pair(f):
return f(a, b)
return pair
This takes two things and returns a function that takes another function and applies it to those two things.
So, what do we get from cons(3, None)? We get a function that takes a function, and applies it to the arguments 3 and None:
def pair3(f):
return f(3, None)
And if we call cons(2, cons(3, None))?
def pair23(f):
return f(2, pair3)
And what happens if you call car on that function? Trace through it:
def left(a, b):
return a
return pair23(left)
That pair23(left) does this:
return left(2, pair3)
And left is dead simple:
return 2
So, we got the first element of (2, cons(3, None)).
What if you call cdr?
def right(a, b):
return a
return pair23(right)
That pair23(right) does this:
return right(2, pair3)
… and right is dead simple, so it just returns pair3.
You can work out that if we call car(cdr(pair23)), we're going to get the 3 out of it.
And now you can write lst = cons(1, cons(2, cons(3, None))), write the recursive nth and printlist functions above, and trace through how they work on lst.
I mentioned above that you can even get rid of integers. How do you do that? Read about Church numerals. You define zero and successor functions. Then you can define one as successor(zero) and two as successor(one). You can even recursively define add so that add(x, zero) is x but add(x, successor(y)) is successor(add(x, y)), and go on to define mul, etc.
You also need a special function you can use as a value for nil.
Anyway, once you've done that, using all of the other definitions above, you can do lst = cons(zero(cons(one, cons(two, cons(three, nil)))), and nth(lst, two) will give you back one. (Of course writing printlist will be a bit trickier…)
Obviously, this is all going to be a lot slower than just using tuples and integers and so on. But theoretically, it’s interesting.
Consider this: we could write a tiny dialect of Python that has only three kinds of statements—def, return, and expression statements—and only three kinds of expressions—literals, identifiers, and function calls—and it could do everything normal Python does. (In fact, you could get rid of statements altogether just by having a function-defining expression, which Python already has.) That tiny language would be a pain to use, but it would a lot easier to write a program to reason about programs in that tiny language. And we even know how to translate code using tuples, loops, etc. into code in this tiny subset language, which means we can write a program that reasons about that real Python code.
In fact, with a couple more tricks (curried functions and/or static function types, and lazy evaluation), the compiler/interpreter could do that kind of reasoning on the fly and optimize our code for us. It’s easy to tell programmatically that car(cdr(cons(2, cons(3, None)) is going to return 3 without having to actually evaluate most of those function calls, so we can just skip evaluating them and substitute 3 for the whole expression.
Of course this breaks down if any function can have side effects. You obviously can’t just substitute None for print(3) and get the same results. So instead, you need some clever trick where IO is handled by some magic object that evaluates functions to figure out what it should read and write, and then the whole rest of the program, the part that users write, becomes pure and can be optimized however you want. With a couple more abstractions, we can even make IO something that doesn’t have to be magical to do that.
And then you can build a standard library that gives you back all those things we gave up, written in terms of defining and calling functions, so it’s actually usable—but under the covers it’s all just reducing pure function calls, which is simple enough for a computer to optimize. And then you’ve basically written Haskell.

Python - multiple functions - output of one to the next

I know this is super basic and I have been searching everywhere but I am still very confused by everything I'm seeing and am not sure the best way to do this and am having a hard time wrapping my head around it.
I have a script where I have multiple functions. I would like the first function to pass it's output to the second, then the second pass it's output to the third, etc. Each does it's own step in an overall process to the starting dataset.
For example, very simplified with bad names but this is to just get the basic structure:
#!/usr/bin/python
# script called process.py
import sys
infile = sys.argv[1]
def function_one():
do things
return function_one_output
def function_two():
take output from function_one, and do more things
return function_two_output
def function_three():
take output from function_two, do more things
return/print function_three_output
I want this to run as one script and print the output/write to new file or whatever which I know how to do. Just am unclear on how to pass the intermediate outputs of each function to the next etc.
infile -> function_one -> (intermediate1) -> function_two -> (intermediate2) -> function_three -> final result/outfile
I know I need to use return, but I am unsure how to call this at the end to get my final output
Individually?
function_one(infile)
function_two()
function_three()
or within each other?
function_three(function_two(function_one(infile)))
or within the actual function?
def function_one():
do things
return function_one_output
def function_two():
input_for_this_function = function_one()
# etc etc etc
Thank you friends, I am over complicating this and need a very simple way to understand it.
You could define a data streaming helper function
from functools import reduce
def flow(seed, *funcs):
return reduce(lambda arg, func: func(arg), funcs, seed)
flow(infile, function_one, function_two, function_three)
#for example
flow('HELLO', str.lower, str.capitalize, str.swapcase)
#returns 'hELLO'
edit
I would now suggest that a more "pythonic" way to implement the flow function above is:
def flow(seed, *funcs):
for func in funcs:
seed = func(seed)
return seed;
As ZdaR mentioned, you can run each function and store the result in a variable then pass it to the next function.
def function_one(file):
do things on file
return function_one_output
def function_two(myData):
doThings on myData
return function_two_output
def function_three(moreData):
doMoreThings on moreData
return/print function_three_output
def Main():
firstData = function_one(infile)
secondData = function_two(firstData)
function_three(secondData)
This is assuming your function_three would write to a file or doesn't need to return anything. Another method, if these three functions will always run together, is to call them inside function_three. For example...
def function_three(file):
firstStep = function_one(file)
secondStep = function_two(firstStep)
doThings on secondStep
return/print to file
Then all you have to do is call function_three in your main and pass it the file.
For safety, readability and debugging ease, I would temporarily store the results of each function.
def function_one():
do things
return function_one_output
def function_two(function_one_output):
take function_one_output and do more things
return function_two_output
def function_three(function_two_output):
take function_two_output and do more things
return/print function_three_output
result_one = function_one()
result_two = function_two(result_one)
result_three = function_three(result_two)
The added benefit here is that you can then check that each function is correct. If the end result isn't what you expected, just print the results you're getting or perform some other check to verify them. (also if you're running on the interpreter they will stay in namespace after the script ends for you to interactively test them)
result_one = function_one()
print result_one
result_two = function_two(result_one)
print result_two
result_three = function_three(result_two)
print result_three
Note: I used multiple result variables, but as PM 2Ring notes in a comment you could just reuse the name result over and over. That'd be particularly helpful if the results would be large variables.
It's always better (for readability, testability and maintainability) to keep your function as decoupled as possible, and to write them so the output only depends on the input whenever possible.
So in your case, the best way is to write each function independently, ie:
def function_one(arg):
do_something()
return function_one_result
def function_two(arg):
do_something_else()
return function_two_result
def function_three(arg):
do_yet_something_else()
return function_three_result
Once you're there, you can of course directly chain the calls:
result = function_three(function_two(function_one(arg)))
but you can also use intermediate variables and try/except blocks if needed for logging / debugging / error handling etc:
r1 = function_one(arg)
logger.debug("function_one returned %s", r1)
try:
r2 = function_two(r1)
except SomePossibleExceptio as e:
logger.exception("function_two raised %s for %s", e, r1)
# either return, re-reraise, ask the user what to do etc
return 42 # when in doubt, always return 42 !
else:
r3 = function_three(r2)
print "Yay ! result is %s" % r3
As an extra bonus, you can now reuse these three functions anywhere, each on it's own and in any order.
NB : of course there ARE cases where it just makes sense to call a function from another function... Like, if you end up writing:
result = function_three(function_two(function_one(arg)))
everywhere in your code AND it's not an accidental repetition, it might be time to wrap the whole in a single function:
def call_them_all(arg):
return function_three(function_two(function_one(arg)))
Note that in this case it might be better to decompose the calls, as you'll find out when you'll have to debug it...
I'd do it this way:
def function_one(x):
# do things
output = x ** 1
return output
def function_two(x):
output = x ** 2
return output
def function_three(x):
output = x ** 3
return output
Note that I have modified the functions to accept a single argument, x, and added a basic operation to each.
This has the advantage that each function is independent of the others (loosely coupled) which allows them to be reused in other ways. In the example above, function_two() returns the square of its argument, and function_three() the cube of its argument. Each can be called independently from elsewhere in your code, without being entangled in some hardcoded call chain such as you would have if called one function from another.
You can still call them like this:
>>> x = function_one(3)
>>> x
3
>>> x = function_two(x)
>>> x
9
>>> x = function_three(x)
>>> x
729
which lends itself to error checking, as others have pointed out.
Or like this:
>>> function_three(function_two(function_one(2)))
64
if you are sure that it's safe to do so.
And if you ever wanted to calculate the square or cube of a number, you can call function_two() or function_three() directly (but, of course, you would name the functions appropriately).
With d6tflow you can easily chain together complex data flows and execute them. You can quickly load input and output data for each task. It makes your workflow very clear and intuitive.
import d6tlflow
class Function_one(d6tflow.tasks.TaskCache):
function_one_output = do_things()
self.save(function_one_output) # instead of return
#d6tflow.requires(Function_one)
def Function_two(d6tflow.tasks.TaskCache):
output_from_function_one = self.inputLoad() # load function input
function_two_output = do_more_things()
self.save(function_two_output)
#d6tflow.requires(Function_two)
def Function_three():
output_from_function_two = self.inputLoad()
function_three_output = do_more_things()
self.save(function_three_output)
d6tflow.run(Function_three()) # executes all functions
function_one_output = Function_one().outputLoad() # get function output
function_three_output = Function_three().outputLoad()
It has many more useful features like parameter management, persistence, intelligent workflow management. See https://d6tflow.readthedocs.io/en/latest/
This way function_three(function_two(function_one(infile))) would be the best, you do not need global variables and each function is completely independent of the other.
Edited to add:
I would also say that function3 should not print anything, if you want to print the results returned use:
print function_three(function_two(function_one(infile)))
or something like:
output = function_three(function_two(function_one(infile)))
print output
Use parameters to pass the values:
def function1():
foo = do_stuff()
return function2(foo)
def function2(foo):
bar = do_more_stuff(foo)
return function3(bar)
def function3(bar):
baz = do_even_more_stuff(bar)
return baz
def main():
thing = function1()
print thing

Python lazy evaluator

Is there a Pythonic way to encapsulate a lazy function call, whereby on first use of the function f(), it calls a previously bound function g(Z) and on the successive calls f() returns a cached value?
Please note that memoization might not be a perfect fit.
I have:
f = g(Z)
if x:
return 5
elif y:
return f
elif z:
return h(f)
The code works, but I want to restructure it so that g(Z) is only called if the value is used. I don't want to change the definition of g(...), and Z is a bit big to cache.
EDIT: I assumed that f would have to be a function, but that may not be the case.
I'm a bit confused whether you seek caching or lazy evaluation. For the latter, check out the module lazy.py by Alberto Bertogli.
Try using this decorator:
class Memoize:
def __init__ (self, f):
self.f = f
self.mem = {}
def __call__ (self, *args, **kwargs):
if (args, str(kwargs)) in self.mem:
return self.mem[args, str(kwargs)]
else:
tmp = self.f(*args, **kwargs)
self.mem[args, str(kwargs)] = tmp
return tmp
(extracted from dead link: http://snippets.dzone.com/posts/show/4840 / https://web.archive.org/web/20081026130601/http://snippets.dzone.com/posts/show/4840)
(Found here: Is there a decorator to simply cache function return values? by Alex Martelli)
EDIT: Here's another in form of properties (using __get__) http://code.activestate.com/recipes/363602/
You can employ a cache decorator, let see an example
from functools import wraps
class FuncCache(object):
def __init__(self):
self.cache = {}
def __call__(self, func):
#wraps(func)
def callee(*args, **kwargs):
key = (args, str(kwargs))
# see is there already result in cache
if key in self.cache:
result = self.cache.get(key)
else:
result = func(*args, **kwargs)
self.cache[key] = result
return result
return callee
With the cache decorator, here you can write
my_cache = FuncCache()
#my_cache
def foo(n):
"""Expensive calculation
"""
sum = 0
for i in xrange(n):
sum += i
print 'called foo with result', sum
return sum
print foo(10000)
print foo(10000)
print foo(1234)
As you can see from the output
called foo with result 49995000
49995000
49995000
The foo will be called only once. You don't have to change any line of your function foo. That's the power of decorators.
There are quite a few decorators out there for memoization:
http://wiki.python.org/moin/PythonDecoratorLibrary#Memoize
http://code.activestate.com/recipes/498110-memoize-decorator-with-o1-length-limited-lru-cache/
http://code.activestate.com/recipes/496879-memoize-decorator-function-with-cache-size-limit/
Coming up with a completely general solution is harder than you might think. For instance, you need to watch out for non-hashable function arguments and you need to make sure the cache doesn't grow too large.
If you're really looking for a lazy function call (one where the function is only actually evaluated if and when the value is needed), you could probably use generators for that.
EDIT: So I guess what you want really is lazy evaluation after all. Here's a library that's probably what you're looking for:
http://pypi.python.org/pypi/lazypy/0.5
Just for completness, here is a link for my lazy-evaluator decorator recipe:
https://bitbucket.org/jsbueno/metapython/src/f48d6bd388fd/lazy_decorator.py
Here's a pretty brief lazy-decorator, though it lacks using #functools.wraps (and actually returns an instance of Lazy plus some other potential pitfalls):
class Lazy(object):
def __init__(self, calculate_function):
self._calculate = calculate_function
def __get__(self, obj, _=None):
if obj is None:
return self
value = self._calculate(obj)
setattr(obj, self._calculate.func_name, value)
return value
# Sample use:
class SomeClass(object):
#Lazy
def someprop(self):
print 'Actually calculating value'
return 13
o = SomeClass()
o.someprop
o.someprop
Curious why you don't just use a lambda in this scenario?
f = lambda: g(z)
if x:
return 5
if y:
return f()
if z:
return h(f())
Even after your edit, and the series of comments with detly, I still don't really understand. In your first sentence, you say the first call to f() is supposed to call g(), but subsequently return cached values. But then in your comments, you say "g() doesn't get called no matter what" (emphasis mine). I'm not sure what you're negating: Are you saying g() should never be called (doesn't make much sense; why does g() exist?); or that g() might be called, but might not (well, that still contradicts that g() is called on the first call to f()). You then give a snippet that doesn't involve g() at all, and really doesn't relate to either the first sentence of your question, or to the comment thread with detly.
In case you go editing it again, here is the snippet I am responding to:
I have:
a = f(Z)
if x:
return 5
elif y:
return a
elif z:
return h(a)
The code works, but I want to
restructure it so that f(Z) is only
called if the value is used. I don't
want to change the definition of
f(...), and Z is a bit big to cache.
If that is really your question, then the answer is simply
if x:
return 5
elif y:
return f(Z)
elif z:
return h(f(Z))
That is how to achieve "f(Z) is only called if the value is used".
I don't fully understand "Z is a bit big to cache". If you mean there will be too many different values of Z over the course of program execution that memoization is useless, then maybe you have to resort to precalculating all the values of f(Z) and just looking them up at run time. If you can't do this (because you can't know the values of Z that your program will encounter) then you are back to memoization. If that's still too slow, then your only real option is to use something faster than Python (try Psyco, Cython, ShedSkin, or hand-coded C module).

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