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returning value without breaking a loop

I intend to make a while loop inside a defined function. In addition, I want to return a value on every iteration. Yet it doesn't allow me to iterate over the loop.
Here is the plan:
def func(x):
n=3
while(n>0):
x = x+1
return x
print(func(6))
I know the reason to such issue-return function breaks the loop.
Yet, I insist to use a defined function. Therefore, is there a way to somehow iterate over returning a value, given that such script is inside a defined function?

When you want to return a value and continue the function in the next call at the point where you returned, use yield instead of return.
Technically this produces a so called generator, which gives you the return values value by value. With next() you can iterate over the values. You can also convert it into a list or some other data structure.
Your original function would like this:
def foo(n):
for i in range(n):
yield i
And to use it:
gen = foo(100)
print(next(gen))
or
gen = foo(100)
l = list(gen)
print(l)
Keep in mind that the generator calculates the results 'on demand', so it does not allocate too much memory to store results. When converting this into a list, all results are caclculated and stored in the memory, which causes problems for large n.

Depending on your use case, you may simply use print(x) inside the loop and then return the final value.
If you actually need to return intermediate values to a caller function, you can use yield.

You can create a generator for that, so you could yield values from your generator.
Example:
def func(x):
n=3
while(n>0):
x = x+1
yield x
func_call = func(6) # create generator
print(next(func_call)) # 7
print(next(func_call)) # 8

Iteration for the last value of iteration in Python

How can I define a function in python in such a way that it takes the previous value of my iteration where I define the initial value.
My function is defined as following:
def Deulab(c, yh1, a, b):
Deulab = c- (EULab(c, yh1, a, b)-1)*0.3
return (Deulab,yh1, a,b)
Output is
Deulab(1.01, 1, 4, 2)
0.9964391705626454
Now I want to iterate keeping yh1, a ,b fixed and start with c0=1 and iterate recursively for c.

The most pythonic way of doing this is to define an interating generator:
def iterates(f,x):
while True:
yield x
x = f(x)
#test:
def f(x):
return 3.2*x*(1-x)
orbit = iterates(f,0.1)
for _ in range(10):
print(next(orbit))
Output:
0.1
0.2880000000000001
0.6561792000000002
0.7219457839595519
0.6423682207442558
0.7351401271107676
0.6230691859914625
0.7515327214700762
0.5975401280955426
0.7695549549155365
You can use the generator until some stop criterion is met. For example, in fixed-point iteration you might iterate until two successive iterates are within some tolerance of each other. The generator itself will go on forever, so when you use it you need to make sure that your code doesn't go into an infinite loop (e.g. don't simply assume convergence).

It sound like you are after recursion.
Here is a basic example
def f(x):
x += 1
if x < 10:
x = f(x)
return x
print (f(4))
In this example a function calls itself until a criteria is met.

CodeCupboard has supplied an example which should fit your needs.
This is a bit of a more persistent version of that, which would allow you to go back to where you were with multiple separate function calls
class classA:
#Declare initial values for class variables here
fooResult = 0 #Say, taking 0 as an initial value, not unreasonable!
def myFoo1(x):
y = 2*x + fooResult #A simple example function
classA.fooResult = y #This line is updating that class variable, so next time you come in, you'll be using it as part of calc'ing y
return y #and this will return the calculation back up to wherever you called it from
#Example call
rtn = classA.myFoo1(5)
#rtn1 will be 10, as this is the first call to the function, so the class variable had initial state of 0
#Example call2
rtn2 = classA.myFoo1(3)
#rtn2 will be 16, as the class variable had a state of 10 when you called classA.myFoo1()
So if you were working with a dataset where you didn't know what the second call would be (i.e. the 3 in call2 above was unknown), then you can revisit the function without having to worry about handling the data retention in your top level code. Useful for a niche case.
Of course, you could use it as per:
list1 = [1,2,3,4,5]
for i in list1:
rtn = classA.myFoo1(i)
Which would give you a final rtn value of 30 when you exit the for loop.

Recursive vs Iterative Functions Python

I am currently learning Python and would like some clarification on the difference between iterative and recursive functions. I understand that recursive functions call themselves but I am not exactly sure how to define an iterative function.
For instance, I wrote this code
random_list = ['6', 'hello', '10', 'find', '7']
def sum_digits(string):
return sum(int(x) for x in string if x.isdigit())
print "Digits:", sum_digits(random_list)
I thought that this was an iterative function but after doing some research I am not sure. I need to know specifically because the next exercise asks me to write a version of the function that is recursive/iterative (depending on what my first function is).

Recursive function call itself while does not reach the out poin whereas iterative function update calculating value through the iteration over the range.

so the question is "write an iterative and recursive version of sum". great.
don't use the built-in sum method, and write your own. i'll give you the iterative, you should figure out the recursive:
def my_iterative_sum(a):
res = 0
for n in a:
res += a
return res
this is iterative because it iterates over all the values and sums them up.
edit: clearly your post is iterative. are you calling function f from within function f? no.
maybe reading up on what recursion is will help with this. https://www.google.com/search?q=recursion

To those who might still want to see the difference between recursive and iterative function.
iterative
def iterative_sum(n):
result = 1
for i in range(2,n+1):
result *= i
return result
print(iterative_sum(5))
iteration is when a loop repeatedly executes until the controlling condition becomes false
recursive
def recursive_sum(n):
if n == 1:
return 1
else:
return n * recursive_sum(n-1)
print(recursive_sum(5))
recursive function is when a function calls itself
This link explains it much better
https://techdifferences.com/difference-between-recursion-and-iteration-2.html

In case you came here looking for recursive functions in Python, here is an example of a recursive sum
from typing import Optional
>>> def r_sum(v: int, sum: Optional[int] = None) -> int:
... if sum is None:
... sum = 0
... if v == 0:
... return sum
... else:
... sum = sum + v
... prev = v - 1
... return r_sum(prev, sum)
...
>>> r_sum(3)
6
>>> r_sum(4)
10
>>> r_sum(5)
15

calculating current value based on previous value

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.

Lazy evaluation in Python

What is lazy evaluation in Python?
One website said :
In Python 3.x the range() function returns a special range object which computes elements of the list on demand (lazy or deferred evaluation):
>>> r = range(10)
>>> print(r)
range(0, 10)
>>> print(r[3])
3
What is meant by this?

The object returned by range() (or xrange() in Python2.x) is known as a lazy iterable.
Instead of storing the entire range, [0,1,2,..,9], in memory, the generator stores a definition for (i=0; i<10; i+=1) and computes the next value only when needed (AKA lazy-evaluation).
Essentially, a generator allows you to return a list like structure, but here are some differences:
A list stores all elements when it is created. A generator generates the next element when it is needed.
A list can be iterated over as much as you need, a generator can only be iterated over exactly once.
A list can get elements by index, a generator cannot -- it only generates values once, from start to end.
A generator can be created in two ways:
(1) Very similar to a list comprehension:
# this is a list, create all 5000000 x/2 values immediately, uses []
lis = [x/2 for x in range(5000000)]
# this is a generator, creates each x/2 value only when it is needed, uses ()
gen = (x/2 for x in range(5000000))
(2) As a function, using yield to return the next value:
# this is also a generator, it will run until a yield occurs, and return that result.
# on the next call it picks up where it left off and continues until a yield occurs...
def divby2(n):
num = 0
while num < n:
yield num/2
num += 1
# same as (x/2 for x in range(5000000))
print divby2(5000000)
Note: Even though range(5000000) is a generator in Python3.x, [x/2 for x in range(5000000)] is still a list. range(...) does it's job and generates x one at a time, but the entire list of x/2 values will be computed when this list is create.

In a nutshell, lazy evaluation means that the object is evaluated when it is needed, not when it is created.
In Python 2, range will return a list - this means that if you give it a large number, it will calculate the range and return at the time of creation:
>>> i = range(100)
>>> type(i)
<type 'list'>
In Python 3, however you get a special range object:
>>> i = range(100)
>>> type(i)
<class 'range'>
Only when you consume it, will it actually be evaluated - in other words, it will only return the numbers in the range when you actually need them.

A github repo named python patterns and wikipedia tell us what lazy evaluation is.
Delays the eval of an expr until its value is needed and avoids repeated evals.
range in python3 is not a complete lazy evaluation, because it doesn't avoid repeated eval.
A more classic example for lazy evaluation is cached_property:
import functools
class cached_property(object):
def __init__(self, function):
self.function = function
functools.update_wrapper(self, function)
def __get__(self, obj, type_):
if obj is None:
return self
val = self.function(obj)
obj.__dict__[self.function.__name__] = val
return val
The cached_property(a.k.a lazy_property) is a decorator which convert a func into a lazy evaluation property. The first time property accessed, the func is called to get result and then the value is used the next time you access the property.
eg:
class LogHandler:
def __init__(self, file_path):
self.file_path = file_path
#cached_property
def load_log_file(self):
with open(self.file_path) as f:
# the file is to big that I have to cost 2s to read all file
return f.read()
log_handler = LogHandler('./sys.log')
# only the first time call will cost 2s.
print(log_handler.load_log_file)
# return value is cached to the log_handler obj.
print(log_handler.load_log_file)
To use a proper word, a python generator object like range are more like designed through call_by_need pattern, rather than lazy evaluation