递归中的全局变量。 Python

Question

OK, i'm using Python 2.7.3 and here is my code:

def lenRecur(s): 

    count = 0

    def isChar(c):
        c = c.lower()
        ans=''
        for s in c:
            if s in 'abcdefghijklmnopqrstuvwxyz':
                ans += s
        return ans

    def leng(s):
        global count
        if len(s)==0:
            return count
        else:
            count += 1
            return leng(s[1:])

    return leng(isChar(s))

I'm trying to modify the variable count inside the leng function. Here are the things that I've tried:

1. If I put the variable count outside the lenRecur function it works fine the first time, but if I try again without restarting python shell, the count (obviously) doesn't restart, so it keeps adding.
2. If I change the count += 1 line for count = 1 it also works, but the output is (obviously) one.

So, my goal here is to get the length of the string using recursion, but I don't know how to keep track of the number of letters. I've searched for information about global variables, but I am still stuck. I don't know if i haven't understood it yet, or if I have a problem in my code.

Thanks in advance!

Answer 1

count in lenRecur is not a global. It is a scoped variable.

You'll need to use Python 3 before you can make that work in this way; you are looking for the nonlocal statement added to Python 3.

In Python 2, you can work around this limitation by using a mutable (such as a list) for count instead:

def lenRecur(s): 

    count = [0]

    # ...

    def leng(s):
        if len(s)==0:
            return count[0]
        else:
            count[0] += 1
            return lenIter(s[1:])

Now you are no longer altering the count name itself; it remains unchanged, it keeps referring to the same list. All you are doing is altering the first element contained in the count list.

An alternative 'spelling' would be to make count a function attribute:

def lenRecur(s): 

    # ...

    def leng(s):
        if len(s)==0:
            return leng.count
        else:
            leng.count += 1
            return lenIter(s[1:])

    leng.count = 0

Now count is no longer local to lenRecur() ; it has become an attribute on the unchanging lenRecur() function instead.

For your specific problem, you are actually overthinking things. Just have the recursion do the summing:

def lenRecur(s):
    def characters_only(s):
        return ''.join([c for c in s if c.isalpha()])

    def len_recursive(s):
        if not s:
            return 0
        return 1 + len_recursive(s[1:])

    return len_recursive(characters_only(s))

Demo:

>>> def lenRecur(s):
...     def characters_only(s):
...         return ''.join([c for c in s if c.isalpha()])
...     def len_recursive(s):
...         if not s:
...             return 0
...         return 1 + len_recursive(s[1:])
...     return len_recursive(characters_only(s))
... 
>>> lenRecur('The Quick Brown Fox')
16

Answer 2

I think You can pass count as second argument

def anything(s):
    def leng(s, count):
        if not s:
            return count
        return leng(s[1:], count + 1)

    return leng(isChar(s), 0)

this should work better than muting objects from outer scope such as using mutable objects ( list or dict ) or monkey-patching function itself for example.

Answer 3

You need to make the variable count a function variable like

def lenRecur(s):
    lenRecur.count = 0

However, I see a few problems with the code.

1) If you are trying to find the number of alphabets in a string through recursion, this one will do:

def lenRecur(s):
    def leng(s, count = 0):
            if not s:
                    return count
            else:
                    count += int(s[0].isalpha())
                    return leng(s[1:], count)
    return leng(s)

But still I would prefer having a single function to do the task, like there will be no leng method at all.

2) If your goal is just to find the number of alphabets in a string, I would prefer list comprehension

def alphalen(s):
    return sum([1 for ch in s if ch.isalpha()])

If this is anything other than learning purpose, I suggest you to avoid recursion. Because, the solution cannot be used for larger strings(lets say, finding the alphabet count from contents of a file). You might hit the RunTimeError of Maximum Recursion Depth Exceeded.

Even though you can work around this through setting the recursion depth through setrecursionlimit function, I suggest you to go for other easy ways. More info on setting the recursionlimithere .

Answer 4

Define it outside all function definitions, if you want to use it as a global variable:

count = 0
def lenRecur(s): 

or define it as a function attribute:

def lenRecur(s): 
    lenRecur.count = 0
    def isChar(c):

This has been fixed in py3.x where you can use the nonlocal statement:

def leng(s):
    nonlocal count
    if len(s)==0:

Answer 5

You don't need count. The below function should work.

    def leng(s):
        if not s:
            return 0
        return 1 + leng(s[1:])

Answer 6

Global variable in recursion is very tricky as the depth reaches to its last state and starts to return back to the first recursive call the values of local variables change so we use global variables. the issue with global variables is that when u run the func multiple times the global variable doesn't reset.