python sort list of tuples on custom sort

Question

I am trying to sort a list of tuples, on the first element of the tuple, using a custom pre-defined list as the desired order. So I have the list as

my_list=(['chr1',12],['chrX',32],['chr2',1],['chr1',79],['chr2',6])

and the predefined list on the first element is

custom_list=['chr1','chr2','chrX']

I want the output to be

(['chr1',12],['chr1',79],['chr2',1],['chr2',6],['chrX',32])

(For now, sorting on the second element is not needed.) I can't figure out how to do this . Can anyone help please?

Answer 1

You can use the list.index() function to turn position in the custom_list into a sort key:

sorted(my_list, key=lambda x: (custom_list.index(x[0]), x[1]))

You may want to turn your custom_list into a dictionary however, for faster mapping:

custom_list_indices = {v: i for i, v in enumerate(custom_list)}
sorted(my_list, key=lambda x: (custom_list_indices.get(x[0]), x[1]))

Dictionary lookups take constant time, list.index() time is directly proportional to the length of the list.

Another advantage is that with a dictionary, you can return a default value for entries not found in the dictionary (None in this example); list.index() will raise a ValueError exception instead.

Demo:

>>> my_list=(['chr1',12],['chrX',32],['chr2',1],['chr1',79],['chr2',6])
>>> custom_list=['chr1','chr2','chrX']
>>> sorted(my_list, key=lambda x: (custom_list.index(x[0]), x[1]))
[['chr1', 12], ['chr1', 79], ['chr2', 1], ['chr2', 6], ['chrX', 32]]
>>> custom_list_indices = {v: i for i, v in enumerate(custom_list)}
>>> sorted(my_list, key=lambda x: (custom_list_indices.get(x[0]), x[1]))
[['chr1', 12], ['chr1', 79], ['chr2', 1], ['chr2', 6], ['chrX', 32]]