This post discusses how to custom sort one list in python based on the order of elements of another list.
| >>> a=[2, 3, 4, 5, 6, 7, 8] | |
| >>> b=[4, 2, 3, 8, 1, 5, 7, 6] | |
| >>> sorted(a, key = lambda x: b.index(x)) | |
| [4, 2, 3, 8, 5, 7, 6] | |
| >>> a=[2, 3, 4, 5, 6, 7, 8, 9] | |
| >>> sorted(a, key = lambda x: b.index(x)) | |
| Traceback (most recent call last): | |
| File "<stdin>", line 1, in <module> | |
| File "<stdin>", line 1, in <lambda> | |
| ValueError: 9 is not in list | |
| # better way : test 1 | |
| >>> sorted(a, key = lambda x: b.index(x) if x in b else len(a)) | |
| [4, 2, 3, 8, 5, 7, 6, 9] | |
| # better way : test 2 | |
| >>> a=[2, 3, 4, 5, 6, 7, 8, 9, 45] | |
| >>> sorted(a, key = lambda x: b.index(x) if x in b else len(a)) | |
| [4, 2, 3, 8, 5, 7, 6, 9, 45] | |
| # better way : test 3 | |
| >>> a=[2, 3, 4, 5, 6, 108, 7, 8, 9, 45, –1] | |
| >>> sorted(a, key = lambda x: b.index(x) if x in b else len(a)) | |
| [4, 2, 3, 8, 5, 7, 6, 108, 9, 45, –1] | |
| >>> |
