找到列表中的对数

Question

Suppose lst = [7,1,5,4,2,3,6] , (7, 2), (5, 4), (6, 3) are some of the pairs and in total there are 6 pairs that adds up to 9

(i) The order of numbers in a pair matters. For example, (7, 2) and (2, 7) are two different pairs. (ii) A number cannot pair with itself. (iii) There is no duplicate element in the list

def find_pairs(lst, key):
    count = 0
    if sum(lst[count:count+1]) == key:
        count += 1
        return count
    else:
        return find_pairs(lst[1:],key)  

This is my code. What's wrong ?? I am getting an error input find_pairs([7,1,5,4,2,3,6], 9) give 6

find_pairs(list(range(1, 100, 2)), 55) #0
find_pairs(list(range(1, 100, 2)), 56) #28

Answer 1

There's a built-in for this in the itertools module :

def find_pairs(lst, key):
    return [(a,b) for a,b in itertools.permutations(lst, 2) if a+b==key]

or, more generically:

def find_tuples(lst, key, num=2):
    return [i for i in itertools.permutations(lst, num) if sum(i)==key]

You can use it like this:

>>> find_tuples(lst, 9)
[(7, 2), (5, 4), (4, 5), (2, 7), (3, 6), (6, 3)]
>>> find_tuples(lst, 9, 3)
[(1, 5, 3), (1, 2, 6), (1, 3, 5), (1, 6, 2), (5, 1, 3), (5, 3, 1), (4, 2, 3), 
 (4, 3, 2), (2, 1, 6), (2, 4, 3), (2, 3, 4), (2, 6, 1), (3, 1, 5), (3, 5, 1), 
 (3, 4, 2), (3, 2, 4), (6, 1, 2), (6, 2, 1)]

Answer 2

Your code is

1. only considering a single value at a time ( lst[i:i+1] is a slice containing a single item, identical to [lst[i]] )

2. (once that is fixed) your code only considers adjacent pairs of values - in your example, (7, 2) would never be found because 7 is not next to 2 in the input list

3. using recursion for absolutely no good reason

Here is a more efficient version ( O(n) instead of O(n**2) ):

def find_pairs_count(lst, pair_sum):
    upto = (pair_sum - 1) // 2
    vals = set(lst)
    return 2 * sum(i <= upto and (pair_sum - i) in vals for i in lst)