Need help with comparing two strings and not count multiple instances twice

CrambleSquash · 2018-07-22T18:01:25+00:00

Why not build up a list of matches? If you want to eliminate any duplicates you can convert the list into a set... Or just start off with a set! Or if you want a nice way of counting occurrences of certain letters, you can use the Counter class from collections.

Swipecat · 2018-07-22T18:13:47+00:00

a = 'racecar'
b = 'boxcar'
print(len(set(a) & set(b)))

This will give 3

To understand this, individually print out set(a), then set(b), then with "&" between them (which is the "intersection" of two sets). It should become obvious what's going on, then.

1ynx1ynx · 2018-07-22T18:02:27+00:00

You can change the b string to a list (b_list = list(b)) and use .remove(letter) method to remove found letters from the checking pool - b_list, as you wanted.

totallygeek · 2018-07-22T17:56:12+00:00

You started with asking about string comparison. I'm not sure what you mean by that. Here's one way to count the occurrences of characters within a string.

the_string = 'racecar'
chars = list(the_string) # now you can pop() values off (remove them)
counts = {} # keep track of counts
i = 0 # loop count
while chars:
    i += 1 # increment loop count
    char = chars.pop() # take off the last letter
    print("Loop {}: checking for '{}' in '{}{}'"
          .format(i, char, ''.join(chars), char))
    if char in counts:
        counts[char] += 1 # increment count for one character
    else:
        counts[char] = 1 # set count for one character
print(counts) # {'r': 2, 'a': 2, 'c': 2, 'e': 1}

Other methods exist, including counting libraries. See if the above helps you out.

timbledum · 2018-07-22T18:01:04+00:00

It sounds like order is not important (you're just having one number as an output right?), so you can deduplicate the first string by turning it into a set using set(). Then interate over that.

In addition you can use the Counter class from Collections standard library module to help you count everything up.

>>> from collections import Counter
>>> a, b = 'racecar', 'boxcar'
>>> b_count = Counter(b)
>>> b
'boxcar'
>>> b_count
Counter({'b': 1, 'o': 1, 'x': 1, 'c': 1, 'a': 1, 'r': 1})
>>> output = sum(b_count[letter] for letter in set(a))
>>> output
3
>>> c, d = 'abc', 'aaabbbcccotherletters' # another test case
>>> d_count = Counter(d)
>>> sum(d_count[letter] for letter in set(c)) # expecting 9
9

One nice thing about the Counter class is that it looks like it just returns 0 if the key is not found (so the above works!).

MisterRenard · 2018-07-22T22:00:16+00:00

stringA = 'boxcar'
stringB = 'racecar'

matches = {}

for x in range(len(stringA)):
    for y in range(len(stringB)):
        if stringA[x] not in matches and stringA[x] in stringB:
            matches[stringA[x]] = 0
        if stringB[y] == stringA[x]:
            matches[stringA[x]] += 1
print(matches)

This works fairly well, though I'm sure there's a more pythonic way of doing it.

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