alphabetically sorted if statement not working

Question

The if statement below has a problem in it somewhere and I can not figure it out. Any conventions or method misuses that might be causing it to not function right? checkList is a user inputed sentence and lis is a large list of words.

def realCheck(checkList):  
        string = "".join(checkList)  
    print string  
    wordList = string.split()  
    if match(wordList, lis).sort(key=str.lower) == wordList.sort(key=str.lower):  
        return True  
    else:  
        return False  

Answer 1

1. If checkList is a string, then there is no need for "".join(checkList). It just gives you back the same string:

   In [94]: checkList="This is a sentence"    
   In [95]: "".join(checkList)
   Out[95]: 'This is a sentence'
   

2. The first line, string = "".join(checkList) has the wrong indentation. Move it back to be flush with the other lines in the definition.

3. Don't name a variable string. It overrides the standard Python module of the same name.

4. Presumably match(wordList, lis) returns a list. The sort method sorts the list, and returns None. Since None == None is True,

   if match(wordList, lis).sort(key=str.lower) ==  wordList.sort(key=str.lower):
   

   is always true.

   More likely, what you want is

   sorted(astr.lower() for astr in match(wordList, lis))==sorted(astr.lower() for astr in wordList)
   

   Unlike the sort method, the sorted function returns the sorted list.

   As Alex Martelli points out,

   sorted(match(wordList, lis),key=str.lower)==sorted(wordList,key=str.lower)
   

   always has the same truth value as

   sorted(match(wordList, lis))==sorted(wordList)
   

   So using str.lower as the key for sorting (rather than as a transformation before comparing with ==) is probably not what you want.

5. The statement

   if condition:
       return True
   else:
       return False
   

   can be simplified to

   return condition