python List index out of Error option in merge sort

Question

I am trying with a simple merge function and leave out the sorting orders in function. The Problem is here I am getting List Index out of Range Error in 20 line, elif (lowList[0] >= highList[0]).

I am checking, if any of the lists are empty in the if statements before this line, so this should not get executed if they are empty. But, Python checks for that, I think. Is there any workaround to overcome this.

def merge(my_l, low, high):

    middle = int((low + high) / 2)
    print(middle)
    lowList = my_l[low : middle]
    highList = my_l[middle + 1 : high]

    my_indx = low
    fullList = []

    while (lowList is not [] or  highList is not []):

        if lowList is []:
            fullList.append(highList.pop(0))

        elif highList is []:
            fullList.append(lowList.pop(0))

        elif (lowList[0] >= highList[0]):
            fullList.append(lowList.pop(0))
        else:
            fullList.append(highList.pop(0))

    return fullList

san = [2, 3, 4, 5, 6, 7]

san = merge(san, 0, len(san) - 1)
print (len(san - 1))

for i in san:
    print(i)

Thanks for the help.

Regards, S

score 1 · Accepted Answer · answered Jan 16 '14 at 04:46

The is keyword checks for identity, not equality.

Change your to == in the elif's and '!=' in the while loop, and this error should not appear.

To see this in action, consider this code:

a = []
b = []

print id(a)
print id(b)
print a is b
print a == b

which outputs

33444128
33415256
False
True

python List index out of Error option in merge sort

1 Answers1