空字典Python中的KeyError

Question

I am doing a simple task: find intersection of two arrays in Python.

I wrote the code:

def intersect(nums1, nums2):
    """
    :type nums1: List[int]
    :type nums2: List[int]
    :rtype: List[int]
    """

    hash_2 = {}

    for x in nums2:
        if x in hash_2:
            hash_2[x] = 1
        else:
            hash_2[x] =  hash_2[x] + 1

    intersection = []

    for x in nums1:
        if x in hash_2:
            if hash_2[x] >0:
                intersection.append(x)
                hash_2[x] =  hash_2[x] - 1

    return intersection    

print(intersect([],[1]))

I get :

    line 14, in intersect
    hash_2[x] =  hash_2[x] + 1
KeyError: 1

I tried debugging but it is not helping. Why is the python program sending 1 to else condition when the dictionary itself is empty?

Is this some issue in Python's dictionary that you should not search in empty dictionaries?

Answer 1

The value in x , 1 , isn't in hash_2 (since hash_2 is empty), so the else branch is taken. And since x isn't in hash_2 , you get a KeyError trying to access hash_2[x] .

It looks like you want your test to be if x not in hash_2 .

Answer 2

I don't get the point of hash_2 and what it is, But if you want to calculate the intersection of two sequences here is a solution:

def intersection(seq1, seq2):
    result = list()
    for item in seq1:
        if item in seq2:
            result.append(item)
    return result

def main():
    a = [10, 20, 30, 40, 'salam', 'bye', ['!', '*']]
    b = [10, 11, 23, 30, 'salam', 'not', ['!', '*']]
    print(intersection(a, b))

main()