Explanation of older post: Appending a List and Recursion (Hamming Distance) - Python 3

Question

An older post had this code in it:

def hamming(num, dist):
    if dist == 0:
        return num
    else:
        outputlist = list()
        for item in range(len(num)):
            if len(num[item:]) > dist - 1:
                if num[item] == "0":
                    restoflist = hamming(num[item + 1:], dist - 1)
                    outputlist.append(num[:item] + "1" + str(restoflist))
                else:
                    restoflist = hamming(num[item + 1:], dist - 1)
                    outputlist.append(num[:item] + "0" + str(restoflist))                
        return outputlist

What is the reasoning behind if len(num[item:]) > dist - 1 ?

http://stackoverflow.com/questions/22072112/appending-a-list-and-recursion-hamming-distance-python-3 — user3290553, May 27 '14 at 02:35

score 2 · Accepted Answer · answered May 27 '14 at 02:55

The test makes sure that the sliced end of num is long enough to have a distance of dist-1 from the current string. If not, there's no point in recursing. Another way of writing that test (which would be a bit more efficient) is len(num)-item >= dist.

But really, the check should be dropped and the for loop's bounds changed instead:

for item in range(len(num) - dist + 1):
    if num[item] == "0":
        # ...

score 0 · Answer 2 · answered May 27 '14 at 02:37

0

num[item:] will return a list starting at the index of item going to the end. So the length of that will get the elements until the end of the list.

dist -1 is the equivalent of saying >= dist.

answered May 27 '14 at 02:37

Ian Leaman

135
7

It erms of Hamming distance, why is that important though? – user3290553 May 27 '14 at 02:45

Explanation of older post: Appending a List and Recursion (Hamming Distance) - Python 3

2 Answers2