Python implementation of “median of medians” algorithm

Question

I\'ve written this implementation of the median of medians algorithm in python, but it doesn\'t seem to output the right result, and it also does not seem of linear complexi

Answer 1

1) Your code indentation is wrong, try this:

def select(L):
    if len(L) < 10:
        L.sort()
        return L[int(len(L)/2)]
    S = []
    lIndex = 0
    while lIndex+5 < len(L)-1:
        S.append(L[lIndex:lIndex+5])
        lIndex += 5
    S.append(L[lIndex:])
    Meds = []
    for subList in S:
        print(subList)
        Meds.append(select(subList))
    L2 = select(Meds)
    L1 = L3 = []
    for i in L:
        if i < L2:
            L1.append(i)
        if i > L2:
            L3.append(i)
    if len(L) < len(L1):
        return select(L1)
    elif len(L) > len(L1) + 1:
        return select(L3)
    else:
        return L2

2) The method you use does not return the median, it just return a number which is not so far from the median. To get the median, you need to count how many number are greater than your pseudo-median, if a majority is greater, repeat the algorithm with the numbers greater than the pseudo-median, else repeat with the other numbers.

def select(L, j):
    if len(L) < 10:
        L.sort()
        return L[j]
    S = []
    lIndex = 0
    while lIndex+5 < len(L)-1:
        S.append(L[lIndex:lIndex+5])
        lIndex += 5
    S.append(L[lIndex:])
    Meds = []
    for subList in S:
        Meds.append(select(subList, int((len(subList)-1)/2)))
    med = select(Meds, int((len(Meds)-1)/2))
    L1 = []
    L2 = []
    L3 = []
    for i in L:
        if i < med:
            L1.append(i)
        elif i > med:
            L3.append(i)
        else:
            L2.append(i)
    if j < len(L1):
        return select(L1, j)
    elif j < len(L2) + len(L1):
        return L2[0]
    else:
        return select(L3, j-len(L1)-len(L2))

Warning: L = M = [] is not L = [] and M = []