binary search for list in python

Question

My code for binary search function in a list returns true for a value in the list, but returns None (instead of false) for values not in the list.

Can someone please explain me what I'm doing wrong?

The program is:

def searchlist(x,alist):
        end=int(len(alist)-1)
        mid=int(len(alist)/2)

        while len(alist)>2:
                if  x==alist[mid] or x==alist[0] or x==alist[end] :
                        return("true")
                        break
                elif x>alist[mid]:
                        alist=alist[mid:]
                        mid=int(len(alist)/2)
                        end=int(len(alist)-1)


                elif x<alist[mid]:
                        alist=alist[:mid]
                        mid=int(len(alist)/2)
                        end=int(len(alist)-1)

                else:
                        return("false")

aList=[2,3,5,7,9,12,14,23,34,45,67,89,101]

xnum=int(input("enter a number:"))
searchlist(xnum,aList)
print(searchlist(xnum,aList))

Without diving into the code, alist is going down to 2 values and it's ending the while loop, causing it to return None. I'd suggest print out the current state of the list each loop, and manually run through your code to figure at what point it's not working as expected. Edit: Actually, it looks like your else statement won't ever execute. Put return False after the while loop instead and it should work (I'd suggest use True and False instead of strings). — Peter
– Peter, Commented Nov 19, 2018 at 13:09
If you want to return false if you don't find what you want in your list you should put the return false after the loop, not in the loop — Baptiste Gavalda
– Baptiste Gavalda, Commented Nov 19, 2018 at 13:12
your second elif perfomed. If you delete this block, then return ('false') — Rudolf Morkovskyi
– Rudolf Morkovskyi, Commented Nov 19, 2018 at 13:17

Yakov Dan · Accepted Answer · 2018-11-19 13:14:05Z

1

You get None when your function does not return a value. This happens because the while loop terminates without going into the "else" branch. A better practice would be to return True (not the string, but the Boolean value) when you find the value in the list, and return False after the loop.

answered Nov 19, 2018 at 13:14

Yakov Dan

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Ali Kargar · Accepted Answer · 2018-11-19 13:33:43Z

Your while loop cannot catch the else statement. you don't need that else. try this :

def searchlist(x,alist):
    end=int(len(alist)-1)
    mid=int(len(alist)/2)
    result = False
    while len(alist)>2:
            if  x==alist[mid] or x==alist[0] or x==alist[end] :
                    result = True
            elif x>alist[mid]:
                    alist=alist[mid:]
                    mid=int(len(alist)/2)
                    end=int(len(alist)-1)

            elif x<alist[mid]:
                    alist=alist[:mid]
                    mid=int(len(alist)/2)
                    end=int(len(alist)-1)

    return result

aList=[2,3,5,7,5,67,89,101]

xnum=int(input("enter a number:"))
print(searchlist(xnum,aList))

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