We Keep Coding

Python: How to exit all the recursions in a recursive function [duplicate]

I'm wondering how to break out of a recursive loop to the main function. I am trying to do a simple palindrome exercise. The function should return True for "redivider" but the return True is being passed to is_pal() and the function isn't breaking. Short of adding a second variable to is_pal to track True/False, what is the proper way to break out of this recursive loop?
def first(word):
return word[0]
def last(word):
return word[-1]
def middle(word):
return word[1:-1]
def is_pal(str):
if len(str) == 1:
return True
if first(str) == last(str) and len(str) > 1:
is_pal(middle(str))
print is_pal("redivider")

One way to break out of a recursive function in Python is to throw an exception and catch that at the top level. Some people will say that this is not the right way to think about recursion, but it gets the job done. Furthermore, if the task is to identify "problem" elements in an array/array of arrays/ndarray etc., a break technique is convenient, because it stops the algorithm from continuing after the global solution has been identified.
def solve_problem(lst):
def solve_rec(l):
'''has impl. that may throw an exception '''
try:
solve_rec(lst)
return True
except:
return False

def is_pal(str):
if len(str) <= 1:
return True
if first(str) == last(str):
return is_pal(middle(str))
else:
return False
That way, if they don't match, False is returned; if it makes it all the way to the end, True is returned. I also eliminated a redundant conditional and checked for the edge-case of an even-length palindrome.

You don't "break" out of recursive functions. Trying to do so says you're thinking about them the wrong way. Currently your recursive call is ignoring the output, which means that the recursion is pointless; whatever is_pal(middle(str)) returns has no effect on the return value of your function.
A recursive algorithm solves the input problem by decomposing the problem into a smaller problem, getting the solution to the smaller problem recursively, and then using the smaller solution to construct a correct solution to the larger problem. You don't "break" out of the inner calls, you return a solution back up one level. You don't know (or need to know) whether you're in an inner call or a top level call. In either case, your function should do the same thing: return True if the argument is a palindrome, and False if it isn't.
The algorithm you're trying to implement is basically this:
If the string is of length 1, it's a palindrome (return True)
Otherwise, if the first character is the same as the last character, then the input is a palindrome if the middle characters are a palindrome.
So what this means is that once you've established the first and last characters are the same, the answer to "is my input a palindrome" is exactly the same as the answer to "are the middle characters a palindrome". You need to return that answer to fulfil your contract. So the recursive call should be return is_pal(middle(str)) rather than just is_pal(middle(str)). If this was a top level call, then that's the answer; if this wasn't a top-level call, then the outer call is going to need this answer to work out the answer to the outer problem (in this case, by simply returning it).
Btw, your algorithm also has some other problems.
You never return False, so the answer can never be False (in this case you happen to accidentally return None by falling off the end of the function if the first and last character don't match, and None will probably do as a stand in for False in most cases, but it's still not really correct).
If the string's length is zero rather than 1 (which will happen if an empty string is passed in or if a palindrome of even length is passed in once all the pairs of equal first and last characters are stripped off), then you don't return the correct answer, and in fact you try to get the first and last character of the empty string, which will cause an exception.

You can exit the program after printing the results using the exit() function.
That may not be a good practice, but it might be what you're looking for.

You're missing a return. Also, don't use str as a variable name. Last thing, the first and last functions could be named slightly better.
def first_letter(word):
return word[0]
def last_letter(word):
return word[-1]
def middle(word):
return word[1:-1]
def is_pal(word):
if len(word) == 1:
return True
if first_letter(word) == last_letter(word) and len(word) > 1:
return is_pal(middle(word))
print is_pal("redivider")

You need to return False in case they don't match and add a return statement. Also you will probably want to check against len(str)==0 and len(str)==1:
def is_pal(str):
if len(str) in [0, 1]:
return True
if first(str) == last(str) and len(str) > 1:
return is_pal(middle(str))
else :
return False

YOU CAN BREAK RECURSION BY RETURNING 1 in 'if' statement before you write your 'function()' 2nd time.
I mean that's what we do to find factorial !! RIGHT? :)

What is wrong with my python hyperoperation recursive function?

I tried to use the definition from Wikipedia for Hyperoperations and translate it into Python code.
My goal was to make it readable, making it fast was a task for later.
Here's the definition I used and below it is my code.
from Wikipedia: https://en.wikipedia.org/wiki/Hyperoperation#Definition
And now here's my code:
def hyperation(n, a, b):
if n == 0:
return b+1
if n == 1 and b == 0:
return a
if n == 2 and b == 0:
return 0
if n >= 3 and b == 0:
return 1
return hyperation(n-1, a, hyperation(n, a, b-1))
When I tried it with hyperation(n=1, a=3, b=3), what should basically be the same as
3 + 3
I always hit the recursion limit, doesn't matter how high I set it.
When trying this on paper by hand, it works perfectly fine. And while debugging I couldn't figure out what happens.
Thanks for every help!!!

In your last line the return keyword is missing.

First, your code was not formatted right, cause python is a indentation sensitive language.
Second, The last line of the error log indicates that the error is the addition of NoneType and int. You didn't return anything in the final else block, so that will return NoneType. Just add return at the beginning of the last sentence.
Third, you can delete every elif after return and just use if.

Why is my condition expression giving a syntax error?

def singleNumber(nums):
for num in set(nums):
return num if nums.count(num) != 2 #error occurs here
print(singleNumber([1,1,4,5,5]))
This follows the usual python condition expression format. I don't understand why it's giving me a syntax error here.
The purpose of this function is to find number that doesn't occur twice.

If you're going to actually execute a return, you must return something no matter what.
So you could try:
return num if nums.count(num) != 2 else None
However, that's not going to work in this case since it will return on the first check rather checking all the elements for what you want.
In other words, let's say the first element checked is the first 1 in [1,1,4,5,5]. It will work out that there are two copies of that value in the array and then return None, skipping the rest of the elements.
I'd probably rewrite it as:
if nums.count(num) != 2: return num
which basically does what you need in that, if you don't explicitly return something (i.e., all of the values occur twice), the caller gets None implicitly when the function exits.
If you're the type that dislikes implicit things, you can explicitly add return None at the end of the function, though it's not really necessary.

Because you forgot to write what the result should be if the condition is false.
return num if nums.count(num) != 2 else ...what?
And if you didn't want that then you should write a normal if statement instead.

When you use a ternary conditional operator both if and else part are mandatory.
return num if nums.count(num) != 2 else <missing???>

when to use if vs elif in python

If I have a function with multiple conditional statements where every branch gets executed returns from the function. Should I use multiple if statements, or if/elif/else? For example, say I have a function:
def example(x):
if x > 0:
return 'positive'
if x < 0:
return 'negative'
return 'zero'
Is it better to write:
def example(x):
if x > 0:
return 'positive'
elif x < 0:
return 'negative'
else:
return 'zero'
Both have the same outcome, but is one more efficient or considered more idiomatic than the other?
Edit:
A couple of people have said that in the first example both if statements are always evaluated, which doesn't seem to be the case to me
for example if I run the code:
l = [1,2,3]
def test(a):
if a > 0:
return a
if a > 2:
l.append(4)
test(5)
l will still equal [1,2,3]

I'll expand out my comment to an answer.
In the case that all cases return, these are indeed equivalent. What becomes important in choosing between them is then what is more readable.
Your latter example uses the elif structure to explicitly state that the cases are mutually exclusive, rather than relying on the fact they are implicitly from the returns. This makes that information more obvious, and therefore the code easier to read, and less prone to errors.
Say, for example, someone decides there is another case:
def example(x):
if x > 0:
return 'positive'
if x == -15:
print("special case!")
if x < 0:
return 'negative'
return 'zero'
Suddenly, there is a potential bug if the user intended that case to be mutually exclusive (obviously, this doesn't make much sense given the example, but potentially could in a more realistic case). This ambiguity is removed if elifs are used and the behaviour is made visible to the person adding code at the level they are likely to be looking at when they add it.
If I were to come across your first code example, I would probably assume that the choice to use ifs rather than elifs implied the cases were not mutually exclusive, and so things like changing the value of x might be used to change which ifs execute (obviously in this case the intention is obvious and mutually exclusive, but again, we are talking about less obvious cases - and consistency is good, so even in a simple example when it is obvious, it's best to stick to one way).

Check this out to understand the difference:
>>> a = 2
>>> if a > 1: a = a+1
...
>>> if a > 2: a = a+1
...
>>> a
4
versus
>>> a = 2
>>> if a > 1: a = a+1
... elif a > 2: a = a+1
...
>>> a
3
The first case is equivalent to two distinct if's with empty else statements (or imagine else: pass); in the second case elif is part of the first if statement.

In some cases, elif is required for correct semantics. This is the case when the conditions are not mutually exclusive:
if x == 0: result = 0
elif y == 0: result = None
else: result = x / y
In some cases it is efficient because the interpreter doesn't need to check all conditions, which is the case in your example. If x is negative then why do you check the positive case? An elif in this case also makes code more readable as it clearly shows only a single branch will be executed.

In general (e.g. your example), you would always use an if..elif ladder to explicitly show the conditions are mutually-exclusive. It prevents ambiguity, bugs etc.
The only reason I can think of that you might ever not use elif and use if instead would be if the actions from the body of the preceding if statement (or previous elif statements) might have changed the condition so as to potentially make it no longer mutually exclusive. So it's no longer really a ladder, just separate concatenated if(..elif..else) blocks. (Leave an empty line between the separate blocks, for good style, and to prevent someone accidentally thinking it should have been elif and 'fixing' it)
Here's a contrived example, just to prove the point:
if total_cost>=10:
if give_shopper_a_random_discount():
print 'You have won a discount'
total_cost -= discount
candidate_prime = True
if total_cost<10:
print 'Spend more than $10 to enter our draw for a random discount'
You can see it's possible to hit both conditions, if the first if-block applies the discount, so then we also execute the second, which prints a message which would be confusing since our original total had been >=10.
An elif here would prevent that scenario.
But there could be other scenarios where we want the second block to run, even for that scenario.
if total_cost<10:
<some other action we should always take regardless of original undiscounted total_cost>

In regards to the edit portion of your question when you said:
"A couple of people have said that in the first example both if statements are always evaluated, which doesn't seem to be the case to me"
And then you provided this example:
l = [1,2,3]
def test(a):
if a > 0:
return a
if a > 2:
l.append(4)
test(5)
Yes indeed the list l will still equal [1,2,3] in this case, ONLY because you're RETURNING the result of running the block, because the return statement leads to exiting the function, which would result in the same thing if you used elif with the return statement.
Now try to use the print statement instead of the return one, you'll see that the 2nd if statement will execute just fine, and that 4 will indeed be appended to the list l using append.
Well.. now what if the first ifstatement changes the value of whatever is being evaluated in the 2nd if statement?
And yes that's another situation. For instance, say you have a variable x and you used if statement to evaluate a block of code that actually changed the x value.
Now, if you use another if statement that evaluates the same variable x will be wrong since you're considering x value to be the same as its initial one, while in fact it was changed after the first if was executed. Therefore your code will be wrong.
It happens pretty often, and sometimes you even want it explicitly to be changed. If that's how you want your code to behave, then yes you should use multiple if's which does the job well. Otherwise stick to elif.
In my example, the 1st if block is executed and changed the value of x, which lead to have the 2nd if evaluates a different x (since its value was changed).
That's where elif comes in handy to prevent such thing from happening, which is the primary benefit of using it.
The other secondary good benefit of using elif instead of multiple if's is to avoid confusion and better code readability.

Consider this For someone looking for a easy way:
>>> a = ['fb.com', 'tw.com', 'cat.com']
>>> for i in a:
... if 'fb' in i:
... pass
... if 'tw' in i:
... pass
... else:
... print(i)
output:
fb.com
cat.com
And
>>> a = ['fb.com', 'tw.com', 'cat.com']
>>> for i in a:
... if 'fb' in i:
... pass
... elif 'tw' in i:
... pass
... else:
... print(i)
Output:
cat.com
'If' checks for the first condition then searches for the elif or else, whereas using elif, after if, it goes on checking for all the elif condition and lastly going to else.

elif is a bit more efficient, and it's quite logical: with ifs the program has to evaluate each logical expression every time. In elifs though, it's not always so. However, in your example, this improvement would be very, very small, probably unnoticeable, as evaluating x > 0 is one of the cheapest operations.
When working with elifs it's also a good idea to think about the best order. Consider this example:
if (x-3)**3+(x+1)**2-6*x+4 > 0:
#do something 1
elif x < 0:
#do something 2
Here the program will have to evaluate the ugly expression every time! However, if we change the order:
if x < 0:
#do something 2
elif (x-3)**3+(x+1)**2-6*x+4 > 0:
#do something 1
Now the program will first check if x < 0 (cheap and simple) and only if it isn't, will it evaluate the more complicated expression (btw, this code doesn't make much sense, it's just a random example)
Also, what perreal said.

Flow of recursion in python roman to arabic

So I just solved the last quiz on recursion on codeacademy. But while I was trying to understand the flow of execution of this code on pythontutor.org, I just couldnt follow the flow of this code.
def rval(letr):
if(letr=='I'):
return 1
elif(letr=='V'):
return 5
elif(letr=='X'):
return 10
elif(letr=='L'):
return 50
elif(letr=='C'):
return 100
elif(letr=='D'):
return 500
elif(letr=='M'):
return 1000
else:
return "error"
def arabic(n):
if len(n)==0:
return 0
elif len(n)==1:
return rval(n)
elif len(n)==2:
if rval(n[0])>rval(n[1]):
return rval(n[0])+rval(n[1])
else:
return rval(n[1])-rval(n[0])
else:
return arabic(n[len(n)-2:])+arabic(n[:len(n)-2])
arabic('DXCVI')
======PROBLEM FACED======
My question is this --
Suppose I run arabic('DXCVI'), then how does this line arabic(n[len(n)-2:])+arabic(n[:len(n)-2]) get executed
Do both arabic(n[len(n)-2:]) & arabic(n[:len(n)-2]) start getting executed simultaneously or does the second term wait until the first is done/returns a value ?

The left part arabic(n[len(n)-2:]) is always called first. An easy way to find this out is to put a print statement in your arabic(n) function.
Try something like this:
def arabic(n):
print n
if len(n)==0:
return 0
elif len(n)==1:
return rval(n)
elif len(n)==2:
# more code
Which would output this:
DXCVI
VI
DXC
XC
D
Proving that the left side, arabic(n[len(n)-2:]), gets called before +arabic(n[:len(n)-2]).

First arabic(n[len(n)-2:]) (the left part) is executed, then arabic(n[:len(n)-2]) (the right part) is executed.
Remember that the first part may also have recursive calls, so the second one must wait until it returns a value to get executed.