在python中的列表之间映射的最快方法是什么

Question

I have pairs of 4 lists a and b with integer values such as list_1a = [1,2,3,...] and list_1b = [8,11,15,...] . The idea is that the integer values in list_1a are now represented by the integer values in list_1b , and the same for list_2a and list_2b etc.

Now I have a list of 4 columns final_list which contained integer values corresponding to the a lists. I want to map the values in final_list to the values in the b lists. What is the quickest way to do this in python ?

Is there a quicker way than using lists ?

Edit:

To clarify the question, take the following example:

list_1a = [1,2,3]
list_1b = [8,11,15] 

list_2a = [5,6,7,8]
list_2b = [22,26,30,34]

list_3a = [11,12,13,14,18]
list_3b = [18,12,25,28,30]

list_4a = [51,61,72,82]
list_4b = [73,76,72,94]

• Note that some of these lists can contain more than a million entries (So maybe memory can be an issue)
• The lists do not have the same length
• All of the integer values in these lists are unique to their lists, ie list_1a + list_1b will never have a repeating integer value.

final_list should look like final_list_b after the mapping occurs

final_list_a = [[1,6,11,51],[3,6,14,72]]
final_list_b = [[8,26,18,73],[15,26,28,72]]

To put things into perspective, this questions is for a database application where these "lists" contain auto-generated key values

Answer 1

I think what you want is a dictionary , which associates keys with values. Unless i'm confused about what you want to do here.

So if I make 4 short example lists.

list_1a = [1,2,3,4]
list_1b = [8,11,15,18]
list_2a = [5,6,7,8]
list_2b = [22,26,30,34]

and make them into a big list of all "a" values and all "b" values.

a_list = list_1a + list_2a
b_list = list_1b + list_2b

I can then use zip to merge the lists into a dictionary

my_dict = dict(zip(a_list, b_list))

print(my_dict)

See: how to merge 2 list as a key value pair in python

for some other ways to do this last bit.

result:

{1: 8, 2: 11, 3: 15, 4: 18, 5: 22, 6: 26, 7: 30, 8: 34}

Now your "a" list makes up the keys of this dictionary.. while the "b" list make up the values. You can access the values by using the keys. here's some examples.

print(my_dict.keys())
print(my_dict.values())
print(my_dict[5])

gives me:

[1, 2, 3, 4, 5, 6, 7, 8]
[8, 11, 15, 18, 22, 26, 30, 34]
22

Is this what you want?

EDIT: I feel that I should note that while my dictionary has printed in order, dictionaries are actually not ordered like lists. You might want to look into collections.OrderedDict or sorted if this is important to you.

Update:

For what you want to do, maybe consider nested dictionaries. You can make a dictionary whose values are dictionaries, also note that when 1a and 1b don't match in length, zip doesn't care and just excludes 60:

list_1a = [1,2,3,4]
list_1b = [8,11,15,18,60]
list_2a = [5,6,7,8]
list_2b = [22,26,30,34]

a_dict = dict(zip(list_1a, list_2a))
b_dict = dict(zip(list_1b, list_2b))

my_dict = {"a" : a_dict, "b" : b_dict}

print(my_dict)

Result:

{'a': {1: 5, 2: 6, 3: 7, 4: 8}, 'b': {8: 22, 18: 34, 11: 26, 15: 30}}

Now you can access the inner values in a different way:

print(my_dict["a"].keys())
print(my_dict["a"].values())
print(my_dict["a"][4])

Result:

[1, 2, 3, 4]
[5, 6, 7, 8]
8