Merge Sort in python also calculates inversions

mergesort.py

def mergeSort(A):
    if len(A)>1:
        [x1,x2]=divide(A)
        #print x1,x2
        x1,leftInv=mergeSort(x1)
        x2,rightInv=mergeSort(x2)
        #print leftInv,rightInv
        merged,splitInv=merge(x1,x2)
        #print splitInv
        return merged,(leftInv+rightInv+splitInv)
    else:
        return A,0


def divide(A):
    size=len(A)
    halfSize=size/2
    x1=[]
    x2=[]
    for index in xrange(0,halfSize):
        x1.append(A[index])
    for index in xrange(halfSize,size):
        x2.append(A[index])
    return [x1,x2]


def merge(x1,x2):
    x=[]
    x1Index=0
    x2Index=0
    invCount=0
    size1=len(x1)
    size2=len(x2)
    while ( x1Index < size1 and x2Index < size2):
        if x1[x1Index]<=x2[x2Index]:
            x.append(x1[x1Index])
            x1Index=x1Index+1
        else:
            x.append(x2[x2Index])
            x2Index=x2Index+1
            invCount += (size1-x1Index)

    for i in range(x1Index,size1):
         x.append(x1[i])

    for i in range(x2Index,size2):
         x.append(x2[i])

    return x,invCount
            

a=[6,5,4,3,2,1]
print 'before sorting',a             
x=mergeSort(a)
print 'after sorting',x

5 years later. . .can't you make divide:
def divide(A):
return [A[:len(A)/2] , A[len(A)/2:]]