Python: Sort list of elements by first element, but if equal sort by original index-CodePudding

I have a list of elements that looks like this:

['3,5', '5,8', '1,8', '6,9', '1,10', '1,2', '4,7', '9,4']

I would like to sort it by first value of each element, but if two elements have the same value, then I would like to sort it by the original index. In this example I want '1,8' to be before '1,10' and '1,10' before '1,2', so the sorted list would look like this:

['1,8', '1,10', '1,2', '3,5', '4,7', '5,8', '6,9', '9,4']

How can I achieve this?

CodePudding user response：

Python's sort functions are stable, which means if the sort key is the same it will have the behavior you want. So you can just make a key that takes the first number ad ignores the rest. One way to pass the substring up to the "," and make it an integer:

l = ['3,5', '5,8', '1,8', '6,9', '1,10', '1,2', '4,7', '9,4']

sorted(l, key=lambda s: int(s[:s.index(',')]))
# ['1,8', '1,10', '1,2', '3,5', '4,7', '5,8', '6,9', '9,4']