Python-比较两个字典列表并从一个列表中返回字典

Question

I have two lists, a = [dict1, dict2, dict3] and b = [dict1, dict2, dict4, dict5, dict6]

I want to create two more lists, with the dicts from both lists not found in the other. So they would be,

c = [dict3] and d = [dict4, dict5]

I have tried the following but it returns way to many dicts

for i, j in [(i,j) for i in range(len(a)) for j in range(len(b))]:
    if cmp(b[j],a[i]) == 1 or -1:
        new_prods = {}
        new_prods = a[i]
        c.append(new_prods)  

for i, j in [(i,j) for i in range(len(a)) for j in range(len(b))]:
    if cmp(b[j],a[i]) == 0:
        old_prods = {}
        old_prods = b[j]
        d.append(old_prods)  

Thanks in advance

Answer 1

A simple O(n^2) solution:

a = [dict1, dict2, dict3] 
b = [dict1, dict2, dict4, dict5, dict6]

c = dicts_from_a_not_in_b = [x for x in a if x not in b]
d = dicts_from_b_not_in_a = [x for x in b if x not in a]

Answer 2

In my opinion using sets would make perfect sense in this case.

c = set(a).difference(b)    
d = set(b).difference(a)

You can convert the sets to list by simply wrapping them by list(c) & list(d)

Answer 3

Here is an alternate way to do it reusing a function:

a = [dict1, dict2, dict3]
b = [dict1, dict2, dict4, dict5, dict6]

def find_uncommon_elements(list1, list2):
    list3 = []
    for item in list1:
        if item not in list2:
            list3.append(item)
    return list3

c = find_uncommon_elements(a,b)
d = find_uncommon_elements(b,a)

Writing a function helps in the event that you have to reuse this again later with a different set of dictionary lists, without having rewrite the entire function.

This will return:

c = [dict3]
d = [dict4, dict5, dict6]