Sorting list based on order of substrings in another list

Question

I have two lists of strings.

list_one = ["c11", "a78", "67b"]
list_two = ["a", "b", "c"]

What is the shortest way of sorting list_one using strings from list_two to get the following output?

["a78", "67b", "c11"]

Edit 1: There is a similar question Sorting list based on values from another list?, but in that question he already has the list of required indexes for resulting string, while here I have just the list of substrings.

Edit 2: Since the example of list above might be not fully representative, I add another case.

list_one is ["1.cde.png", "1.abc.png", "1.bcd.png"] list_two is ["abc", "bcd", "cde"]. The output is supposed to be [ "1.abc.png", "1.bcd.png", "1.cde.png"]

If, for example, list_one is shorter than list_two, it should still work:

list_one is ["1.cde.png", "1.abc.png"] list_two is ["abc", "bcd", "cde"] The output is supposed to be [ "1.abc.png", "1.cde.png"]

Possible duplicate of Sorting list based on values from another list? — Tom Wojcik
– Tom Wojcik, Commented Jul 19, 2018 at 9:07
Your expected output doesn't make sense. Why is 67b second? — blhsing
– blhsing, Commented Jul 19, 2018 at 9:09
what is the relation between list_one and list_two? It seems list_one's objects are included list_two's values. Is it the criteria? — Ali Yesilli
– Ali Yesilli, Commented Jul 19, 2018 at 9:11
@TomWojcik in stackoverflow.com/questions/6618515/… he already has the list of required order for resulting string, while here I have just the list of substrings — Aliaksei Laurynovich
– Aliaksei Laurynovich, Commented Jul 19, 2018 at 9:11
@blhsing because in the list_two, substring "b" is the second — Aliaksei Laurynovich
– Aliaksei Laurynovich, Commented Jul 19, 2018 at 9:12

blhsing · Accepted Answer · 2018-07-19 09:55:03Z

4

key = {next((s for s in list_one if v in s), None): i for i, v in enumerate(list_two)}
print(sorted(list_one, key=key.get))

This outputs:

['a78', '67b', 'c11']

edited Jul 19, 2018 at 9:55

answered Jul 19, 2018 at 9:20

blhsing

109k9 gold badges88 silver badges132 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

Aliaksei Laurynovich Over a year ago

Your code doesn't work if list_one is smaller than list_two. It gives an error "IndexError: list index out of range"

blhsing Over a year ago

Indeed. Fixed then.

tobias_k Over a year ago

Why key=key.__getitem__ instead of key=key.get?

Aliaksei Laurynovich Over a year ago

I'm not sure what exactly you have changed, but it still doesn't work for the case when list_one is shorter than list_two. For example, I tested your code for the case when list_one = ["c11", "a78"], list_two = ["a", "b", "c"]. The output is supposed to be ["a78", "c11"]. But it gives me an error "IndexError: list index out of range".

blhsing Over a year ago

@tobias_k Good point. __getitem__ was entirely unnecessary. Revised as suggested. Thanks.

|

khelili miliana · Accepted Answer · 2018-07-19 09:38:23Z

3

Try this

list_one = ["c11", "a78", "67b"]
list_two = ["a", "b", "c"]

[x for y in list_two for x in list_one if y in x]

Output :

["a78", "67b", "c11"]

edited Jul 19, 2018 at 9:38

answered Jul 19, 2018 at 9:16

khelili miliana

3,8222 gold badges17 silver badges30 bronze badges

1 Comment

cawecoy Over a year ago

if list_one has itens without "a", "b" or "c", and you want to keep them at the end of the output list, add this: OUTPUT_LIST + list(set(list_one) - set(OUTPUT_LIST))

tobias_k · Accepted Answer · 2018-07-19 09:56:53Z

Assuming that each item in list_one contains exactly one of the characters from list_two, and that you know the class of those characters, e.g. letters, you can extract those using a regex and build a dictionary mapping the characters to the element. Then, just look up the correct element for each character.

>>> list_one = ["c11", "a78", "67b"]
>>> list_two = ["a", "b", "c"]
>>> d = {re.search("[a-z]", s).group(): s for s in list_one}
>>> list(map(d.get, list_two))
['a78', '67b', 'c11']
>>> [d[c] for c in list_two]
['a78', '67b', 'c11']

Other than the other approaches posted so far, which all seem to be O(n²), this is only O(n).

Of course, the approach can be generalized to e.g. more than one character, or characters in specific positions of the first string, but it will always require some pattern and knowledge about that pattern. E.g., for your more recent example:

>>> list_one = ["1.cde.png", "1.abc.png", "1.bcd.png"]
>>> list_two = ["abc", "cde"]
>>> d = {re.search("\.(\w+)\.", s).group(1): s for s in list_one}
>>> d = {s.split(".")[1]: s for s in list_one}  # alternatively without re
>>> [d[c] for c in list_two if c in d]
['1.abc.png', '1.cde.png']

Sunitha · Accepted Answer · 2018-07-19 09:51:10Z

-1

>>> sorted(list_one, key=lambda x: [i for i,e in enumerate(list_two) if e in x][0])
['a78', '67b', 'c11']

answered Jul 19, 2018 at 9:51

Sunitha

12.1k2 gold badges23 silver badges23 bronze badges

Collectives™ on Stack Overflow

Sorting list based on order of substrings in another list

4 Answers 4

6 Comments

1 Comment

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

6 Comments

1 Comment

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related