This question already has answers here:
I need help wrapping my head around the return statement with Python and its role in this recursive statement
(2 answers)
Closed 8 years ago.
def fudz(n):
if n <= 2:
return 1
print("nom" * n)
return fudz(n-1) + fudz(n//2)
result = fudz(4)
Can someone give me a step by step of this function?
This is a recursive function, which means it calls an instance of itself on a simplified instance of the problem until it gets to a base case for which it already knows the answer. Remember that any call to a function occurs in its own stack frame, so it's going to have its own value for n.
You can walk through it yourself. First, think about what happens when n==2. Then think about n==3, and increase n until it makes sense to you.
Related
This question already has answers here:
Why does my recursive function return None?
(4 answers)
What is the purpose of the return statement? How is it different from printing?
(15 answers)
Closed 5 months ago.
I'm currently playing with recursive functions and I have a question.
I don't understand why this one work
a, b = map(int, input().split())
def gcd(a,b):
if a%b == 0:
return b
else:
return gcd(b,a%b)
print(gcd(a,b))
but this one doesn't
a, b = map(int, input().split())
def gcd(a,b):
if a%b == 0:
return b
gcd(b,a%b)
print(gcd(a,b))
Though the answer appears in the comments, I thought this is an opportunity to demonstrate how python tools can help you find similar errors like this in the future.
$ pylint --disable=invalid-name,redefined-outer-name,missing-function-docstring,missing-module-docstring main.py
************* Module Downloads.main
main.py:2:0: R1710: Either all return statements in a function should return an expression, or none of them should. (inconsistent-return-statements)
-------------------------------------------------------------------
Your code has been rated at 8.33/10 (previous run: 10.00/10, -1.67)
So here you see pylint points your attention to implicit return values from some path(s) in your function. Note that some of pylint's reports (the ones I disabled for instance) may generate "noise" rather than be helpful). Still, using this tool (and mypy and others) may sure help you
This question already has answers here:
Obtaining closures at runtime [duplicate]
(1 answer)
How to open a closure in python?
(5 answers)
Closed 2 years ago.
I would like to know if there is any method to check whether two functions have the same arguments at runtime in python 3.
Basically, I have this function (func) that takes two arguments and perform some sort of computation. I want to check whether or not a and b have the same arguments' values at runtime
a = func(2, 3)
b = func(2, 3)
a.argsvalue == b.argsvalue
It is not feasible to run the code before and check the results because I am implementing a lazy framework. My main goal is to be able to understand what are the arguments of the function because there is one variable argument that I do not care but there is one static that is created before running the function.
##EDIT
I actually solved this problem using the inspect module (getclosure) for those who are interested. Thank you so much for the comments it helped me to familiarize myself with the terminology. I was actually looking for the closure, which I assigned dynamically.
when you do this - a.argsvalue == b.argsvalue you try to access a member of the value returned from the function.
so, if your "func" would return an object having the args you called it with (which sound like a weird thing to do) you would be able to access it.
anyway, if you need these values, just store them before sending them to the function, and then you can do whatever you want with them.
This question already has answers here:
"Least Astonishment" and the Mutable Default Argument
(33 answers)
Closed 2 years ago.
I'm doing some discrete mathematics stuff for a teacher, he asked me to try and do everything recursively. For some strange reason the list i'm using for a function is the same as it was the last time I called the function.
Here's my code:
def extended_euclidean_algorithm(a:int, b:int, equations=list()):
#This line is just for debugging and checking that my hipothesis was correct
print (len(equations))
if b==0:
return
if a<b:
b,a=a,b
quotient=a//b
remainder=a%b
equations.append(f"{a}={quotient}*{b}+{remainder}")
if extended_euclidean_algorithm(b, remainder, equations):
return equations
for i, equation in enumerate(equations):
equations[i]=equation.split('+')
equations[i][0]=equations[i][0].split('=')
equations[i][0]="-".join(equations[i][0])
equations[i]='='.join(equations[i])
return True
First time I call it, it's Ok. But second time I call it includes the numbers from the last time I called it.
Since you set the list in the arguments, by default the list exists and can be appended to.
You better use:
def extended_euclidean_algorithm(a:int, b:int, equations=None):
equations = equations if equations else []
In this way, equations - if not set - is None rather than an empty list.
Check out the pythontutor visualization here:
This question already has answers here:
How do I clone a list so that it doesn't change unexpectedly after assignment?
(24 answers)
Closed 7 years ago.
I am sorry that the title of this question is vague. I could not find an appropriate statement for my question. I have been struggling with something in python. I have an array which I enter python manually. Then I want to use it inside a function and I want to put the function inside a while loop. My code is something like this:
C=[8592,2092,9284,1136,8267,349,5623,2034,2834,4404]
def h(x):
Emp=C
.
.
.
Emp=Emp-xxxx
return xxx
while xxxx:
xxx=h(x)
well the problem starts from here. I want to keep C unchanged during the whole process but unfortunately values of C change with variability in values of Emp. I have never faced such a problem in Matlab and I cant understand why this is happening in Pyhton. Any help is appreciated.
I want to keep C unchanged during the whole process
Your issue is Emp = C, in Python this creates a reference to C, it doesn't copy it like some statistical computation languages do with that syntax
You can copy C with
import copy
x = copy.copy(C)
while condition:
h(x)
This question already has answers here:
Why aren't python nested functions called closures?
(10 answers)
Closed 8 years ago.
I'm struggling to understand Function closures properly. For example in the code below I am unclear as to how the function knows that in the statement times3(2) that x=2? Also, after reading documentations I still can't fully understand the purpose of closures.
def make_multiplier_of(n):
def multiplier(x):
return x * n
return multiplier
times3 = make_multiplier_of(3)
times3(2) #How does the function know that x=2 here?
Thanks a lot
When you are calling make_multiplier_of(3), the function is returning multiplier such that
def multiplier(x):
return x*3
So times3 = make_multipiler(3) is assigning this particular multiplier function to times3. The same way that when you are doing myLength=len, myLength is the len function and you can call myLength("foo")
times3 is thus this multiplier function. So when you times3(2), you are doing (this particular) multiplier(2)