匹配两个列表字段 python 的最快方法

Question

I have an issue with time in my latest python script. In essence, i have two lists, eg List1: ([a,1],[b,2]) List2: ([a,3],[b,4])

Now in the example above i have provided two entries in each list. However, in reality there is about 150,000.

In my current script I retrieve the first field from the first list [a] and loop through the entire List2 till there is a match. The two list entries are then appended.

The final result would be: ([a,1,3],[b,2,4])

However, given the size of the lists this is taking forever.

Is there a way i can use the field of list1 [a] and in constant time retrieve all entries in list2 that have [a]

I have seen some answers online suggesting sets, but i am unsure as to how to implement one and use it to solve the solution above.

Any help would be appreciated.

Further example:

l1=(['abc123','hi'], ['efg456','bye']) - l1 has around 2000 tuples

l2=(['abc123','letter'],['abc123','john'],['abc123','leaf']) - l2 has around 100,000+ tuples

Output: l3=(['abc123','hi','letter'],['abc123','hi','john'],['abc123','hi','leaf'])

Answer 1

If your a and b values are unique, you can convert the "lists" (what you have is actually a tuple of lists, not a list of lists) into dictionaries and then merge them. For example:

l1 = (['a', 1], ['b', 2], ['c', 5])
l2 = (['a', 3], ['b', 4])

d1 = { k : [v] for [k, v] in l1 }
d2 = { k : [v] for [k, v] in l2 }

for k in d1.keys():
    d1[k] += d2.get(k, [])
    
print(d1)

Output:

{'a': [1, 3], 'b': [2, 4], 'c': [5]}

You can convert that dictionary back to a tuple of lists using a comprehension:

print(tuple([k, *v] for k, v in d1.items()))

Output:

(['a', 1, 3], ['b', 2, 4], ['c', 5])

Answer 2

Not so hard, just use a dict for list1 and a for loop for list2.

dict1 = {key1: [value1] for key1, value1 in list1}  # convert list1 to dict
                                                    # and the values should be converted to dict
for key2, value2 in list2:
    try:
        dict1[key2].append(value2)
    except KeyError:
        continue  # I'm not sure what do you want to do if the keys in list2 didn't exist in list1, so just ignore them
list3 = tuple([key3, *value3] for key3, value3 in dict1.items())
print(list3)