Question

I have a method, that computes the number of differences in two strings, and outputs where the differences are.

def method(a):
    count=0
    s1="ABC"
    for i in range (len(a)):
        if not a[i]==s1[i]:
            count=count+1
        else:
            count=count+0
    return a,count,difference(a, s1)

On input ex CBB, this method outputs

('CBB', 2, [1, 0, 1])

What I really need is for this method to do the same, but where is not only compares to a single string in s1, but to a list of strings

s1 = ['ACB', 'ABC', 'ABB']

Anyone with a smart method to do this?

Answer 1

the compare function calculates the number of differences (and map of differences that you had been creating with difference()). I rewrote the compare function to take a base string to be compared to, src, so that you don’t get stuck with comparing to "ABC" all the time.

def compare(src, test):
    if len(src) != len(test):
        return # must be the same length
    diffmap = [0]*len(src)
    count = 0
    for i, c in enumerate(src):
        if not c == test[i]:
            count = count+1
            diffmap[i] = 1
    return test, count, diffmap

The compare_to_many function simply goes through a list of strings to compare to, srcs, and creates a list of the comparisons between those base strings and a test string test.

def compare_to_many(srcs, test):
    return map(lambda x: compare(x, test), srcs)

EDIT:

After clarification in the comments, @X-Pender needs the source list to be hardcoded. This can be reflected by the following, single function:

def compare(test):
    def compare_one(src, test):
        diffmap = [0]*len(src)
        count = 0
        for i, c in enumerate(src):
        if not c == test[i]:
            count = count+1
            diffmap[i] = 1
        return test, count, diffmap
    sources = ["ABC", "CDB", "EUA"] # this is your hardcoded list
    return map(lambda x: compare_one(x, test), sources)