comparing two strings in python with duplicates

Question

I am trying to compare the two strings: 'apple' and 'pear' and return letters that do not belong to the other string.

For example, 'apple' does not contain 'r' in 'pear'

'pear' does not contain 'l' and 'p' in apple (pear contains p but does not contains two p's).

So I want to have a function that returns 'r', 'l', and 'p'.

I tried set, but it ignores the duplicates (p, in this example).

def solution(A, B):
    N = len(A)
    M = len(B)
    letters_not_in_B = list(set([c for c in A if c not in B]))
    letters_not_in_A = list(set([c for c in B if c not in A]))
    answer = len(letters_not_in_B) + len(letters_not_in_A)
    return answer

If you want duplicates, than you shouldn't use sets since they only have unique values, just use lists — Maaaaa
– Maaaaa, Commented Jan 5, 2018 at 19:17
You say you want the function to return 'r', 'l', and 'p' but your attempt, after having letters, instead returns a number. Make up your mind? — Stefan Pochmann
– Stefan Pochmann, Commented Jan 5, 2018 at 19:28

Ajax1234 · Accepted Answer · 2018-01-05 19:17:12Z

6

You can compare the character counts for each separate string resulting from the concatenation of the parameters a and b:

def get_results(a, b):
  return list(set([i for i in a+b if a.count(i) != b.count(i)]))

print(get_results('apple', 'pear'))

Output:

['p', 'r', 'l']

answered Jan 5, 2018 at 19:17

Ajax1234

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3 Comments

John Gordon Over a year ago

This returns identical results for get_results('apppppple', 'pear'), which is incorrect.

Ajax1234 Over a year ago

@JohnGordon the OP wants data formatted with no duplicates where there occurs a letter that does not have equal counts in each string.

John Gordon Over a year ago

I interpret "return letters that do not belong to the other string" as meaning "If the first string has one p and the second string has five ps, I want four ps returned." Shame on OP for not being precise, I suppose.

Beefster · Accepted Answer · 2018-01-05 19:22:54Z

5

Use a Counter

from collections import Counter
Counter('apple') - Counter('pear') # --> Counter({'p': 1, 'l': 1})
Counter('pear') - Counter('apple') # --> Counter({'r': 1})

answered Jan 5, 2018 at 19:22

Beefster

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Comments

John Gordon · Accepted Answer · 2018-01-05 19:20:11Z

-1

def solution(a, b):

    # create mutable list copies of a and b
    list_a = list(a)
    list_b = list(b)

    for ch in a:
        if ch in list_b:
            list_b.remove(ch)

    for ch in b:
        if ch in list_a:
            list_a.remove(ch)

    return list_a + list_b

answered Jan 5, 2018 at 19:20

John Gordon

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