Python - Sorting list

Question

I have a lists like the below:

list1 = [['4.00', '7.00'], ['4.00', '4.00'], ['4.00', '1.00']]
list2 = [['4.00', '7.00'], ['1.00', '7.00'], ['6.00', '7.00']]

If x values are equal, I want to sort the elements in ascending order based on y values and if y values are equal, want to sort based on x values like below:

sorted list1 = [['4.00', '1.00'], ['4.00', '4.00'], ['4.00', '7.00']]
sorted list2 = [['1.00', '7.00'], ['4.00', '7.00'], ['6.00', '7.00']]

I tried to use sorted(). But code becomes complicated when i try to compare the values of x to find out if they are the same and then giving key to sorted() based on that. Can I do it in a better way?

Isn't the way you're comparing them the same way lists would normally be compared? — kindall
– kindall, Commented Oct 3, 2017 at 3:48
yeah. I want to know if there is a simpler way by specifying some condition to check if x or y in list are equal in the key given to sorted() function.. — buddingengineer
– buddingengineer, Commented Oct 3, 2017 at 3:52
If you've reached the point of checking y values, and they're still equal, why would going back to x change anything? You're only testing y when x was equal, so going back to x is pointless. — ShadowRanger
– ShadowRanger, Commented Oct 3, 2017 at 3:52
@ShadowRanger.. I just mentioned 2 different conditions.. 1. If x values are equal, sort y. 2. If y values are equal sort x. — buddingengineer
– buddingengineer, Commented Oct 3, 2017 at 3:54
... but that's the way it already works. What are you trying to do that's different? — kindall
– kindall, Commented Oct 3, 2017 at 4:01

ShadowRanger · Accepted Answer · 2017-10-03 03:57:36Z

2

sorted already does what you want. Comparing lists is done lexicographically: The elements are compared from left to right until an unequal element is found, at which point the result of the comparison is the comparison on that first unequal element. For example, with no key function at all, your lists sort as you expect:

>>> list1 = [['4.00', '7.00'], ['4.00', '4.00'], ['4.00', '1.00']]
>>> list2 = [['4.00', '7.00'], ['1.00', '7.00'], ['6.00', '7.00']]
>>> sorted(list1)
[['4.00', '1.00'], ['4.00', '4.00'], ['4.00', '7.00']]
>>> sorted(list2)
[['1.00', '7.00'], ['4.00', '7.00'], ['6.00', '7.00']]

In the list1 case, it correctly compares the first values, determines they're the same, and sorts on the second value; in the list2 case, the first values all differ, so the second value is never even checked.

answered Oct 3, 2017 at 3:57

ShadowRanger

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