我在弄清楚这个压缩代码时遇到麻烦

Question

def duplicate_count(s):
    return len([c for c in set(s.lower()) if s.lower().count(c)>1])

I'm having difficulty understanding how this code works. I was doing a codewars challenge to return the number of elements with duplicates in a string.

eg. Asasd --> 2

I came up with my own implmentation but I wasn't able to really understand what this code does. If anyone could point me in the direction, it would be much appreciated :)

Answer 1

This is, first of all, a highly inefficient solution to the problem. But let's break it down:

• s.lower() would convert all the characters in a string to lower case:

   In [1]: s = "Hello, WORLD" In [2]: s.lower() Out[2]: 'hello, world' 

• set(s.lower()) would create a set (make sure to read about what sets are) of characters from a string - eliminating all the duplicates:

   In [3]: set(s.lower()) Out[3]: {' ', ',', 'd', 'e', 'h', 'l', 'o', 'r', 'w'} 

• for c in set(s.lower()) iterates over every single character in a set we created above
• for every character in this set, we apply this if condition: if s.lower().count(c)>1 . The count(c) here would count how many times c appears in the string. The >1 helps us to leave characters that are met more than 1 time in a string
• [c for c in set(s.lower()) if s.lower().count(c)>1] is called a list comprehension . It is basically a short way of creating a list. Here, we are creating a list of characters that occur in a string more than one time. Check out this topic about how to verbalize and read the list comprehensions .
• len() then just gets us the length of the list

To summarize, you iterate over the unique characters in a given string and count which of them occur in a string more than one time.

Answer 2

set(s.lower()) # gives unique elements in lower case

and

s.lower().count(c)>1 #checks if an element shows up more than once

All in all the function finds number of not unique elements in a string ,ignoring case.

I believe using collections.Counter is more efficient:

In [7]: from collections import Counter

In [8]: sum(v > 1 for v in Counter("Aabcc".lower()).values())

Out[8]: 2

Answer 3

def duplicate_count(s):

    result = []

    for c in set(s.lower()):
        if s.lower().count(c) > 1:
            result.append(c)

    return len(result)