如何使用 python 中的列表有效地对列表列表进行排序

Question

I have a list of lists as follows.

mylist = [['dog', ['dogs details']], ['cat', ['cats details']], ['rat', ['rats details']], ['rabbit', ['rabbits details']], ['mouse', ['mice details']], ['goat', ['goats details']]]

I also have a names list as follows.

mynameslist = ["fish", "rabbit", "cow", "cat", "dog", "owl", "rat", "mouse", "parrot", "goat", "emjenne", "emj"]

I want to order mylist considering the order of mynameslist . ie my output should be as follows.

mylist_new = [['rabbit', ['rabbits details']], ['cat', ['cats details']], ['dog', ['dogs details']], ['rat', ['rats details']], ['mouse', ['mice details']], ['goat', ['goats details']]]

I tried to do this as follows.

mylist_new = []
for item in mynameslist:
   for val in mylist:
       if val[0] == item:
           mylist_new.append(val)
           break

However, I have really long mylist and mynameslist . Since my code is O(n^2) it takes a lot of time to create mylist_new . I am wondering if there is a more pythonic way of doing this with less time.

I am happy to provide more details if needed.

Answer 1

Create a map from animal name to index:

>>> namemap = {e:i for i,e in enumerate(mynameslist)}

And then use that as a key for sorted :

>>> sorted(mylist, key=lambda x: namemap[x[0]])
[['rabbit', ['rabbits details']], ['cat', ['cats details']], ['dog', ['dogs details']], ['rat', ['rats details']], ['mouse', ['mice details']], ['goat', ['goats details']]]
>>>

This will be O(N*logN) same as sorted now, since namemap[x[0]] will be constant time.

Alternatively, depending on if your data is well behaved, you can do this in O(N), make a map out of your original list:

>>> animal_map = dict(mylist)

Then iterate over the names list using the map to reconstruct:

>>> [[name, animal_map[name]] for name in mynameslist if name in animal_map]
[['rabbit', ['rabbits details']], ['cat', ['cats details']], ['dog', ['dogs details']], ['rat', ['rats details']], ['mouse', ['mice details']], ['goat', ['goats details']]]
>>>

This will not work if there are duplicates in mylist

Answer 2

You can use sorted or sort (to sort inplace) with key parameter like this:

mylist = [['dog', ['dogs details']], ['cat', ['cats details']], ['rat', ['rats details']], ['rabbit', ['rabbits details']], ['mouse', ['mice details']], ['goat', ['goats details']]]
mynameslist = ["fish", "rabbit", "cow", "cat", "dog", "owl", "rat", "mouse", "parrot", "goat", "emjenne", "emj"]
sorted(mylist, key=lambda x: mynameslist.index(x[0]))
#mylist.sort(key=lambda x: mynameslist.index(x[0]))

Output:

[['rabbit', ['rabbits details']],
 ['cat', ['cats details']],
 ['dog', ['dogs details']],
 ['rat', ['rats details']],
 ['mouse', ['mice details']],
 ['goat', ['goats details']]]

Edit: as @juanpa.arrivillaga said, this is fast to implement to slow on complexity (O(N^2)) so if the list is big you should consider using his answer instead