What does the return statement do? How should it be used in Python?
How does return differ from print?
See also
Often, people try to use print in a loop inside a function in order to see multiple values, and want to be able to use the results from outside. They need to be returned, but return exits the function the first time. See How can I use `return` to get back multiple values from a loop? Can I put them in a list?.
Often, beginners will write a function that ultimately prints something rather than returning it, and then also try to print the result, resulting in an unexpected None. See Why is "None" printed after my function's output?.
Occasionally in 3.x, people try to assign the result of print to a name, or use it in another expression, like input(print('prompt:')). In 3.x, print is a function, so this is not a syntax error, but it returns None rather than what was displayed. See Why does the print function return None?.
Occasionally, people write code that tries to print the result from a recursive call, rather than returning it properly. Just as if the function were merely called, this does not work to propagate the value back through the recursion. See Why does my recursive function return None?.
Consider How do I get a result (output) from a function? How can I use the result later? for questions that are simply about how to use return, without considering print.
The print() function writes, i.e., "prints", a string in the console. The return statement causes your function to exit and hand back a value to its caller. The point of functions in general is to take in inputs and return something. The return statement is used when a function is ready to return a value to its caller.
For example, here's a function utilizing both print() and return:
def foo():
print("hello from inside of foo")
return 1
Now you can run code that calls foo, like so:
if __name__ == '__main__':
print("going to call foo")
x = foo()
print("called foo")
print("foo returned " + str(x))
If you run this as a script (e.g. a .py file) as opposed to in the Python interpreter, you will get the following output:
going to call foo
hello from inside foo
called foo
foo returned 1
I hope this makes it clearer. The interpreter writes return values to the console so I can see why somebody could be confused.
Here's another example from the interpreter that demonstrates that:
>>> def foo():
... print("hello within foo")
... return 1
...
>>> foo()
hello within foo
1
>>> def bar():
... return 10 * foo()
...
>>> bar()
hello within foo
10
You can see that when foo() is called from bar(), 1 isn't written to the console. Instead it is used to calculate the value returned from bar().
print() is a function that causes a side effect (it writes a string in the console), but execution resumes with the next statement. return causes the function to stop executing and hand a value back to whatever called it.
Think of the print statement as causing a side-effect, it makes your function write some text out to the user, but it can't be used by another function.
I'll attempt to explain this better with some examples, and a couple definitions from Wikipedia.
Here is the definition of a function from Wikipedia
A function, in mathematics, associates one quantity, the argument of the function, also known as the input, with another quantity, the value of the function, also known as the output..
Think about that for a second. What does it mean when you say the function has a value?
What it means is that you can actually substitute the value of a function with a normal value! (Assuming the two values are the same type of value)
Why would you want that you ask?
What about other functions that may accept the same type of value as an input?
def square(n):
return n * n
def add_one(n):
return n + 1
print square(12)
# square(12) is the same as writing 144
print add_one(square(12))
print add_one(144)
#These both have the same output
There is a fancy mathematical term for functions that only depend on their inputs to produce their outputs: Referential Transparency. Again, a definition from Wikipedia.
Referential transparency and referential opaqueness are properties of parts of computer programs. An expression is said to be referentially transparent if it can be replaced with its value without changing the behavior of a program
It might be a bit hard to grasp what this means if you're just new to programming, but I think you will get it after some experimentation.
In general though, you can do things like print in a function, and you can also have a return statement at the end.
Just remember that when you use return you are basically saying "A call to this function is the same as writing the value that gets returned"
Python will actually insert a return value for you if you decline to put in your own, it's called "None", and it's a special type that simply means nothing, or null.
I think the dictionary is your best reference here
Return and Print
In short:
return gives something back or replies to the caller of the function while print produces text
In python, we start defining a function with def, and generally - but not necessarily - end the function with return.
Suppose we want a function that adds 2 to the input value x. In mathematics, we might write something like f(x) = x + 2, describing that relationship: the value of the function, evaluated at x, is equal to x + 2.
In Python, it looks like this instead:
def f(x):
return x + 2
That is: we define a function named f, which will be given an x value. When the code runs we figure out x + 2, and return that value. Instead of describing a relationship, we lay out steps that must be taken to calculate the result.
After defining the function, it can be called with whatever argument you like. It doesn't have to be named x in the calling code, and it doesn't even have to be a variable:
print f(2)
>>> 4
We could write the code for the function in some other ways. For example:
def f(x):
y = x + 2
return y
or even
def f(x):
x = x + 2
return x
Again, we are following steps in order - x = x + 2 changes what x refers to (now it means the result from the sum), and that is what gets returned by return x (because that's the value *at the time that the return happens).
return means "output this value from this function".
print means "send this value to (generally) stdout"
In the Python REPL, a function's return value will be output to the screen by default (this isn't the same as printing it). This output only happens at the REPL, not when running code from a .py file. It is the same as the output from any other expression at the REPL.
This is an example of print:
>>> n = "foo\nbar" #just assigning a variable. No output
>>> n #the value is output, but it is in a "raw form"
'foo\nbar'
>>> print(n) #the \n is now a newline
foo
bar
>>>
This is an example of return:
>>> def getN():
... return "foo\nbar"
...
>>> getN() #When this isn't assigned to something, it is just output
'foo\nbar'
>>> n = getN() # assigning a variable to the return value. No output
>>> n #the value is output, but it is in a "raw form"
'foo\nbar'
>>> print(n) #the \n is now a newline
foo
bar
>>>
This answer goes over some of the cases that have not been discussed above.
The return statement allows you to terminate the execution of a function before you reach the end. This causes the flow of execution to immediately return to the caller.
In line number 4:
def ret(n):
if n > 9:
temp = "two digits"
return temp #Line 4
else:
temp = "one digit"
return temp #Line 8
print("return statement")
ret(10)
After the conditional statement gets executed the ret() function gets terminated due to return temp (line 4).
Thus the print("return statement") does not get executed.
Output:
two digits
This code that appears after the conditional statements, or the place the flow of control cannot reach, is the dead code.
Returning Values
In lines number 4 and 8, the return statement is being used to return the value of a temporary variable after the condition has been executed.
To bring out the difference between print and return:
def ret(n):
if n > 9:
print("two digits")
return "two digits"
else :
print("one digit")
return "one digit"
ret(25)
Output:
two digits
'two digits'
Note that return can also be used for control flow. By putting one or more return statements in the middle of a function, we can say: "stop executing this function. We've either got what we wanted or something's gone wrong!"
For example, imagine trying to implement str.find(sub) if we only had str.index(sub) available (index raises a ValueError if the substring isn't found, whereas find returns -1).
We could use a try/except block:
def find(s: str, sub: str) -> int:
try:
return s.index(sub)
except ValueError:
return -1
This is fine, and it works, but it's not very expressive. It's not immediately clear what would cause str.index to raise a ValueError: a reader of this code must understand the workings of str.index in order to understand the logic of find.
Rather than add a doc-string, saying "...unless sub isn't found, in which case return -1", we could make the code document itself, like this:
def find(s: str, sub: str) -> int:
if sub not in s:
return -1
return s.index(sub)
This makes the logic very clear.
The other nice thing about this is that once we get to return s.index(sub) we don't need to wrap it in a try/except because we already know that the substring is present!
See the Code Style section of the Python Guide for more advice on this way of using return.
To put it as simply as possible:
return makes the value (a variable, often) available for use by the caller (for example, to be stored by a function that the function using return is within). Without return, your value or variable wouldn't be available for the caller to store/re-use.
print, by contrast, prints to the screen - but does not make the value or variable available for use by the caller.
Difference between "return" and "print" can also be found in the following example:
RETURN:
def bigger(a, b):
if a > b:
return a
elif a <b:
return b
else:
return a
The above code will give correct results for all inputs.
PRINT:
def bigger(a, b):
if a > b:
print a
elif a <b:
print b
else:
print a
NOTE: This will fail for many test cases.
ERROR:
----
FAILURE: Test case input: 3, 8.
Expected result: 8
FAILURE: Test case input: 4, 3.
Expected result: 4
FAILURE: Test case input: 3, 3.
Expected result: 3
You passed 0 out of 3 test cases
Here is my understanding. (hope it will help someone and it's correct).
def count_number_of(x):
count = 0
for item in x:
if item == "what_you_look_for":
count = count + 1
return count
So this simple piece of code counts number of occurrences of something. The placement of return is significant. It tells your program where do you need the value. So when you print, you send output to the screen. When you return you tell the value to go somewhere. In this case you can see that count = 0 is indented with return - we want the value (count + 1) to replace 0.
If you try to follow logic of the code when you indent the return command further the output will always be 1, because we would never tell the initial count to change.
I hope I got it right.
Oh, and return is always inside a function.
return should be used for recursive functions/methods or you want to use the returned value for later applications in your algorithm.
print should be used when you want to display a meaningful and desired output to the user and you don't want to clutter the screen with intermediate results that the user is not interested in, although they are helpful for debugging your code.
The following code shows how to use return and print properly:
def fact(x):
if x < 2:
return 1
return x * fact(x - 1)
print(fact(5))
This explanation is true for all of the programming languages not just python.
return is part of a function definition, while print outputs text to the standard output (usually the console).
A function is a procedure accepting parameters and returning a value. return is for the latter, while the former is done with def.
Example:
def timestwo(x):
return x*2
Best thing about return function is you can return a value from function but you can do same with print so whats the difference ?
Basically return not about just returning it gives output in object form so that we can save that return value from function to any variable but we can't do with print because its same like stdout/cout in C Programming.
Follow below code for better understanding
CODE
def add(a, b):
print "ADDING %d + %d" % (a, b)
return a + b
def subtract(a, b):
print "SUBTRACTING %d - %d" % (a, b)
return a - b
def multiply(a, b):
print "MULTIPLYING %d * %d" % (a, b)
return a * b
def divide(a, b):
print "DIVIDING %d / %d" % (a, b)
return a / b
print "Let's do some math with just functions!"
age = add(30, 5)
height = subtract(78, 4)
weight = multiply(90, 2)
iq = divide(100, 2)
print "Age: %d, Height: %d, Weight: %d, IQ: %d" % (age, height, weight, iq)
# A puzzle for the extra credit, type it in anyway.
print "Here is a puzzle."
what = add(age, subtract(height, multiply(weight, divide(iq, 2))))
print "That becomes: ", what, "Can you do it by hand?"
We are now doing our own math functions for add, subtract, multiply, and divide. The important thing to notice is the last line where we say return a + b (in add). What this does is the following:
Our function is called with two arguments: a and b.
We print out what our function is doing, in this case "ADDING."
Then we tell Python to do something kind of backward: we return the addition of a + b. You might say this as, "I add a and b then return them."
Python adds the two numbers. Then when the function ends, any line that runs it will be able to assign this a + b result to a variable.
The simple truth is that print and return have nothing to do with each other. print is used to display things in the terminal (for command-line programs).1 return is used to get a result back when you call a function, so that you can use it in the next step of the program's logic.
Many beginners are confused when they try out code at Python's interpreter prompt2, like
>>> def example():
... return 1
...
>>> example()
1
The value was displayed; doesn't this mean that return displays things? No. If you try the same code in a .py file, you can see for yourself that running the script doesn't cause the 1 to display.
This shouldn't actually be confusing, because it works the same way as any other expression:
>>> 1 + 1
2
This displays at the interactive prompt, but not if we make a script that just says 1 + 1 and try running it.
Again: if you need something to display as part of your script, print it. If you need to use it in the next step of the calculation, return it.
The secret is that the interactive prompt is causing the result to be displayed, not the code. It's a separate step that the prompt does for you, so that you can see how the code works a step at a time, for testing purposes.
Now, let's see what happens with print:
>>> def example():
... return 'test'
...
>>> print(example())
test
The result will display, whether we have this in an interactive prompt or in a script. print is explicitly used to display the value - and as we can see, it displays differently. The interactive prompt uses what is called the repr of the value that was returned from example, while print uses the str of the value.
In practical terms: print shows us what the value looks like, in text form (for a string, that just means the contents of the string as-is). The interactive prompt shows us what the value is - typically, by writing something that looks like the source code we would use to create it.3
But wait - print is a function, right? (In 3.x, anyway). So it returned a value, right? Isn't the interpreter prompt supposed to display that in its separate step? What happened?
There is one more trick: print returns the special value None, which the interpreter prompt will ignore. We can test this by using some expressions that evaluate to None:
>>> None
>>> [None][0]
>>> def example():
... pass # see footnote 4
...
>>> example()
>>>
In each case, there is no separate line at all for output, not even a blank line - the interpreter prompt just goes back to the prompt.
1 It can also be used to write into files, although this is a less common idea and normally it will be clearer to use the .write method.
2 This is sometimes called the REPL, which stands for "read-eval-print loop".
3 This isn't always practical, or even possible - especially once we start defining our own classes. The firm rule is that repr will lean on the .__repr__ method of the object to do the dirty work; similarly, str leans on .__str__.
4 Functions in Python implicitly return None if they don't explicitly return a value.
Return statement -- will return some values according your function.
def example(n):
if n == 5:
return true
else:
return false
if you call above function and you pass number 5 then it will return true else it will return false.
Printing function -- it will print content that you have given to the print function or with in print function bracket.
def example(n):
if n == 5:
print("number is equal")
else:
print("number is not equal")
I am reading Hackers and Painters and am confused by a problem mentioned by the author to illustrate the power of different programming languages.
The problem is:
We want to write a function that generates accumulators—a function that takes a number n, and returns a function that takes another number i and returns n incremented by i. (That’s incremented by, not plus. An accumulator has to accumulate.)
The author mentions several solutions with different programming languages. For example, Common Lisp:
(defun foo (n)
(lambda (i) (incf n i)))
and JavaScript:
function foo(n) { return function (i) { return n += i } }
However, when it comes to Python, the following codes do not work:
def foo(n):
s = n
def bar(i):
s += i
return s
return bar
f = foo(0)
f(1) # UnboundLocalError: local variable 's' referenced before assignment
A simple modification will make it work:
def foo(n):
s = [n]
def bar(i):
s[0] += i
return s[0]
return bar
I am new to Python. Why doesn the first solution not work while the second one does? The author mentions lexical variables but I still don't get it.
s += i is just sugar for s = s + i.*
This means you assign a new value to the variable s (instead of mutating it in place). When you assign to a variable, Python assumes it is local to the function. However, before assigning it needs to evaluate s + i, but s is local and still unassigned -> Error.
In the second case s[0] += i you never assign to s directly, but only ever access an item from s. So Python can clearly see that it is not a local variable and goes looking for it in the outer scope.
Finally, a nicer alternative (in Python 3) is to explicitly tell it that s is not a local variable:
def foo(n):
s = n
def bar(i):
nonlocal s
s += i
return s
return bar
(There is actually no need for s - you could simply use n instead inside bar.)
*The situation is slightly more complex, but the important issue is that computation and assignment are performed in two separate steps.
An infinite generator is one implementation. You can call __next__ on a generator instance to extract successive results iteratively.
def incrementer(n, i):
while True:
n += i
yield n
g = incrementer(2, 5)
print(g.__next__()) # 7
print(g.__next__()) # 12
print(g.__next__()) # 17
If you need a flexible incrementer, one possibility is an object-oriented approach:
class Inc(object):
def __init__(self, n=0):
self.n = n
def incrementer(self, i):
self.n += i
return self.n
g = Inc(2)
g.incrementer(5) # 7
g.incrementer(3) # 10
g.incrementer(7) # 17
In Python if we use a variable and pass it to a function then it will be Call by Value whatever changes you make to the variable it will not be reflected to the original variable.
But when you use a list instead of a variable then the changes that you make to the list in the functions are reflected in the original List outside the function so this is called call by reference.
And this is the reason for the second option does work and the first option doesn't.
The below assignment is taken from here:
Q5. Define the repeated function from Homework 2 by calling reduce with compose1 as the first argument. Add only a single expression to the starter implementation below:
def square(x):
return x*x
def compose1(f, g):
"""Return a function of x that computes f(g(x))."""
return lambda x: f(g(x))
from functools import reduce
def repeated(f, n):
"""Return the function that computes the nth application of f, for n>=1.
f -- a function that takes one argument
n -- a positive integer
>>> repeated(square, 2)(5)
625
>>> repeated(square, 4)(5)
152587890625
"""
assert type(n) == int and n > 0, "Bad n"
return reduce(compose1, "*** YOUR CODE HERE ***" )
To complete this assignment, I would like to understand, to what does g bind to? f binds to the square function.
First, what should repeated(f, 4) return?
A function that, when called on some arbitrary arg, will return f(f(f(f(arg)))).
So, if you want to build that with compose1, you'll need to return either compose1(compose1(compose1(f, f), f), f) or compose1(f, compose1(f, compose1(f, f))).
Now, look at what reduce does, and figure out what it's going to pass to compose1 each time. Clearly your iterable argument has to either start or end with f itself. But what else do you want there to make sure you get one of the two acceptable results?
And meanwhile, inside each call to compose1 except the last, one of the two arguments has to be the repeated function's f, while the other will be the result of another call to compose1. (The last time, of course, they'll both be f.) Figure out which of those is f and which is g, and how you get reduce to pass the right values for each, and you've solved the problem.
Im trying to learn how to utilize the lambda function and i am having some difficulties.
example from HomeWork.
i have the following functions
def square(x):
return x**2
def incr(x):
return x+1
i need to create a function called 'repeated' that will take one of the above functions + an integer and will return some value that depends on the times the functions have been called IE:
repeated (incr,4)(2)
6
repeated (square,2)(5)
625
using a 'compose' function.
def compose(f,g):
return lambda x:f(g(x))
thank in advance for an explanation.
If you should use a lambda, you can use recursion like this:
def repeated(func, times):
return lambda x: repeated(func, times - 1)(func(x)) if times > 0 else x
what repeated return is a function waiting for an argument, x,and then it activates itself times times (descriptive...) on x
When it should no longer activate itself (in other words, times <= 0, it stops and returns the x the last call received.
You can use compose in the same manner as follows:
If you want to use compose you can do this:
def repeated(func, times):
return compose(func, repeated(func, times-1)) if times > 0 else lambda x
What this does is just call compose recursively times - 1 more times, using repeated, until no more composing is needed.
Edit:
Python's standard implementation and recursion don't get along too well, both of these solution are correct, but using a loop is more pythonic. Another answer is available that does just that.
repeated needs to return a callable too; compose() can produce that callable for you, but you need to apply it multiple times:
def repeated(f, count):
callable = f
for _ in range(count - 1):
callable = compose(f, callable)
return callable
So if count is 1, the original f() is returned. For count = 2, f(f()) is returned, etc.
i would like to perform a calculation using python, where the current value (i) of the equation is based on the previous value of the equation (i-1), which is really easy to do in a spreadsheet but i would rather learn to code it
i have noticed that there is loads of information on finding the previous value from a list, but i don't have a list i need to create it! my equation is shown below.
h=(2*b)-h[i-1]
can anyone give me tell me a method to do this ?
i tried this sort of thing, but that will not work as when i try to do the equation i'm calling a value i haven't created yet, if i set h=0 then i get an error that i am out of index range
i = 1
for i in range(1, len(b)):
h=[]
h=(2*b)-h[i-1]
x+=1
h = [b[0]]
for val in b[1:]:
h.append(2 * val - h[-1]) # As you add to h, you keep up with its tail
for large b list (brr, one-letter identifier), to avoid creating large slice
from itertools import islice # For big list it will keep code less wasteful
for val in islice(b, 1, None):
....
As pointed out by #pad, you simply need to handle the base case of receiving the first sample.
However, your equation makes no use of i other than to retrieve the previous result. It's looking more like a running filter than something which needs to maintain a list of past values (with an array which might never stop growing).
If that is the case, and you only ever want the most recent value,then you might want to go with a generator instead.
def gen():
def eqn(b):
eqn.h = 2*b - eqn.h
return eqn.h
eqn.h = 0
return eqn
And then use thus
>>> f = gen()
>>> f(2)
4
>>> f(3)
2
>>> f(2)
0
>>>
The same effect could be acheived with a true generator using yield and send.
First of, do you need all the intermediate values? That is, do you want a list h from 0 to i? Or do you just want h[i]?
If you just need the i-th value you could us recursion:
def get_h(i):
if i>0:
return (2*b) - get_h(i-1)
else:
return h_0
But be aware that this will not work for large i, as it will exceed the maximum recursion depth. (Thanks for pointing this out kdopen) In that case a simple for-loop or a generator is better.
Even better is to use a (mathematically) closed form of the equation (for your example that is possible, it might not be in other cases):
def get_h(i):
if i%2 == 0:
return h_0
else:
return (2*b)-h_0
In both cases h_0 is the initial value that you start out with.
h = []
for i in range(len(b)):
if i>0:
h.append(2*b - h[i-1])
else:
# handle i=0 case here
You are successively applying a function (equation) to the result of a previous application of that function - the process needs a seed to start it. Your result looks like this [seed, f(seed), f(f(seed)), f(f(f(seed)), ...]. This concept is function composition. You can create a generalized function that will do this for any sequence of functions, in Python functions are first class objects and can be passed around just like any other object. If you need to preserve the intermediate results use a generator.
def composition(functions, x):
""" yields f(x), f(f(x)), f(f(f(x)) ....
for each f in functions
functions is an iterable of callables taking one argument
"""
for f in functions:
x = f(x)
yield x
Your specs require a seed and a constant,
seed = 0
b = 10
The equation/function,
def f(x, b = b):
return 2*b - x
f is applied b times.
functions = [f]*b
Usage
print list(composition(functions, seed))
If the intermediate results are not needed composition can be redefined as
def composition(functions, x):
""" Returns f(x), g(f(x)), h(g(f(x)) ....
for each function in functions
functions is an iterable of callables taking one argument
"""
for f in functions:
x = f(x)
return x
print composition(functions, seed)
Or more generally, with no limitations on call signature:
def compose(funcs):
'''Return a callable composed of successive application of functions
funcs is an iterable producing callables
for [f, g, h] returns f(g(h(*args, **kwargs)))
'''
def outer(f, g):
def inner(*args, **kwargs):
return f(g(*args, **kwargs))
return inner
return reduce(outer, funcs)
def plus2(x):
return x + 2
def times2(x):
return x * 2
def mod16(x):
return x % 16
funcs = (mod16, plus2, times2)
eq = compose(funcs) # mod16(plus2(times2(x)))
print eq(15)
While the process definition appears to be recursive, I resisted the temptation so I could stay out of maximum recursion depth hades.
I got curious, searched SO for function composition and, of course, there are numerous relavent Q&A's.