Python按第一个元素排序2D数组（不是唯一的）

Question

I have a very big 2D array where the second elements are not unique. Something like this:

list = [ ['text43','value43'], 
         ['text23','value23'], 
         ['text12','value12'],
         ['text43','different_val_43'],
         ['text12','another_value12'], 
         ['text04','value04'], 
         ['text43','anohter_value43'] ]

I would like to sort it by the first element but not in alphabetical order, just in the order of appearance of first element. Desired output:

list = [ ['text43','value43'],
         ['text43','different_val_43'],
         ['text43','anohter_value43'],
         ['text23','value23'],
         ['text12','value12'],
         ['text12','another_value12'],
         ['text04','value04'] ]

Answer 1

You can use a custom sorting function that would return the index at which the first element of a sublist is first found, eg:

lst = [['text43','value43'],
       ['text23','value23'],
       ['text12','value12'],
       ['text43','different_val_43'],
       ['text12','another_value12'],
       ['text04','value04'],
       ['text43','anohter_value43']]

d = {}
for i, item in enumerate(lst):
    if item[0] not in d:
        d[item[0]] = i

lst.sort(key=lambda item: d[item[0]])
print(lst)

Output:

[['text43', 'value43'],
 ['text43', 'different_val_43'],
 ['text43', 'anohter_value43'],
 ['text23', 'value23'],
 ['text12', 'value12'],
 ['text12', 'another_value12'],
 ['text04', 'value04']]

Answer 2

Look if this helps. Using sorted()

lst = [ ['text43','value43'], 
         ['text23','value23'], 
         ['text12','value12'],
         ['text43','different_val_43'],
         ['text12','another_value12'], 
         ['text04','value04'], 
         ['text43','anohter_value43'] ]

sorted(lst, reverse=True)

Output:

[['text43', 'value43'],
 ['text43', 'different_val_43'],
 ['text43', 'anohter_value43'],
 ['text23', 'value23'],
 ['text12', 'value12'],
 ['text12', 'another_value12'],
 ['text04', 'value04']]