Recursive functions for sorting

Question

I made a few recursive functions for learning purposes which do a variety of tasks. Here is my current and functioning code:

def separate(p,l):
 ''' recursive function when is passed a predicate and a list returns a 2-tuple
 whose 0 index is a list of all the values in the argument list for which the 
 predicate returns True,and whose 1 index is a list of all the values in the 
 argument list for which the predicate returns False.'''
 if len(l) == 0:
 return ([],[])
 else:
 (true_list, false_list) = separate(p,l[1:])
 if p(l[0]):
 return ([l[0]] + true_list, false_list)
 else:
 return (true_list, [l[0]] + false_list)
def is_sorted(s):
 ''' recursive function when passed a list returns a bool telling whether or not 
 the values in the list are in non-descending order: lowest to highest allowing 
 repetitions. '''
 if len(s) <= 1:
 return True
 elif s[0] < s[1]:
 return is_sorted(s[1:])
 else:
 return False
def sort(l):
 ''' recursive function when passed a list; it returns a new list (not mutating 
 the passed one) with every value in the original list, but ordered in non-
 descending order. '''
 if len(l) == 0:
 return []
 else:
 (before, after) = separate(lambda i: i < l[0], l[1:])
 return sort(before) + [l[0]] + sort(after)
def compare(a,b):
 ''' a recursive function when is passed two str arguments; it returns one of 
 three str values: '<’, '=’, or '>’ which indicates the relationship between the 
 first and second parameter.'''
 if a == '' and b == '':
 return '='
 if a == '' and b != '':
 return '<'
 if a != '' and b == '':
 return '>'
 if a[0] > b[0]:
 return '>'
 if a[0] < b[0]:
 return '<'
 else:
 return compare(a[1:],b[1:])

Is there a way to write these recursive functions in a cleaner/concise way? Any help would be great.

• 200_success: Please ask a separate question for code_metric(). It is neither recursive, nor does it reuse your other functions. (I believe you have missed the point of the exercise as well.)
• LucyBen: @200_success- WIll do. I know I missed the point of the exercise, but that's what I needed help with.

Answer 1

is_sorted() does not behave as described or as expected. It requires elements to be strictly increasing rather than non-descending.

The implementation and docstring could be shorter as well.

def is_sorted(list):
 """Recursively checks if the elements in the list are in non-descending order."""
 return len(list) <= 1 or (list[0] <= list[1] and is_sorted(list[1:]))

I'd expect that trimming lists at the end would be more efficient (though the code definitely looks weirder):

def is_sorted(list):
 """Recursively checks if the elements in the list are in non-descending order."""
 return len(list) <= 1 or (list[-2] <= list[-1] and is_sorted(list[:-1]))

---

compare() has some redundant checks.

Here, I feel that the else is incongruous. Either rely on the early returns, or use else in conjunction with elif everywhere.

def compare(a,b):
 """Lexicographically compares strings a and b, returning '<', '=', or '>'."""
 if a == '' and b == '':
 return '='
 if a == '':
 return '<'
 if b == '':
 return '>'
 if a[0] > b[0]:
 return '>'
 if a[0] < b[0]:
 return '<'
 return compare(a[1:], b[1:])

Instead of checking for == '', consider checking the length, so that the function can operate on lists as well.

• LucyBen: @200_success- using the is_sorted function when I try to do is_sorted([]) it should return True but it gives me error stating return len(list) <= 1 or (list[0] <= list[1] and is_sorted(list[1:])) TypeError: object of type 'type' has no len()
• LucyBen: @200_success- compare has the same issue. when I do compare('','') it returns < when it should be =. compare('','abc') return > when it should be <. Similarly, the problem persists if any of the strings compared is empty. Works perfectly fine when comparing with non-empty string args.
• 200_success: @LucyBen If you're getting TypeError: object of type 'type' has no len(), then you probably missed the detail that I renamed the parameter from s to list.
• 200_success: How is it possible that compare('', '') returns '<'? That's handled by the first conditional, which I didn't change at all.
• LucyBen: @200_success- nm..I made some minor errors in transcribing the code. It works perfectly. Thank you.

Answer 2

I recommend you be more consistent in not combining if followed by a statement sequence ending in return with an else (I left out the comments):

def separate(p,l):
 if len(l) == 0:
 return ([],[])
 (true_list, false_list) = separate(p,l[1:])
 if p(l[0]):
 return ([l[0]] + true_list, false_list)
 return (true_list, [l[0]] + false_list)

as you do e.g. in compare().

In compare() I would nest the conditions (arbitrarily based on a):

def compare(a,b):
 if a == '':
 if b == '':
 return '='
 return '<'
 if a != '':
 if b == '':
 return '>'
 if a[0] > b[0]:
 return '>'
 if a[0] < b[0]:
 return '<'
 return compare(a[1:],b[1:])

That way it is more clear to me that compare() never ends without a return value.

Your comments should at least use triple double quotes (PEP8) and you should try to conform to PEP 257 (docstring conventions).

• LucyBen: @Alice- when I do compare('','') it returns < when it should be =. compare('','abc') return > when it should be <. Similarly, the problem persists if any of the strings compared is empty. Works perfectly fine when comparing with non-empty string args.
• Alice: @LucyBen it doesn't in my test program. Are you sure you are running the right tests? And the same ones that give the correct results on your own code?
• LucyBen: @Alice- I made some minor errors in transcribing the code. It works perfectly. Thank you. any suggestions for sort function??