I'm new to python automation and wrote a script to get some port handles from Ixia and store into a list. I;m trynig to sort that port-handle where I see a problem.
I tried using the sort method but doesn;t work
>>> a
['1/1/11', '1/1/6']
>>> a.sort()
>>> a
['1/1/11', '1/1/6']
>>> d = a.sort()
>>> print(d)
None
>>>
Am i missing anything here .. kindly clarify
I want the output in the following format
1/1/6 1/1/11
You are trying to sort a list of strings. Strings are naturally sorted in lexicographical_order, i.e. "10" < "11" < "2" < "5" < ..., so Python executes correctly what you want it to do. This being said, you need to transform your data into something that will be sorted as you want.
>>> a = ['1/1/11', '1/1/6']
>>> a
['1/1/11', '1/1/6']
>>> def to_tuple(string_representation):
... return tuple(int(i) for i in string_representation.split('/'))
...
>>> b = [to_tuple(element) for element in a]
>>> b.sort()
>>> b
[(1, 1, 6), (1, 1, 11)]
>>> a.sort(key=to_tuple)
>>> a
['1/1/6', '1/1/11']
Here we use the fact that tuple is sorted by default exactly how we want it to be sorted in your case (actually, it is also a lexicographical order, but now 11 is one element of a sequence and not two).
List b contains a transformed list, where each element is a tuple. And now sort will work as you want.
The second option, will be using a custom key operator. It is a function that returns a key to compare different elements of your list. In this case, key will be a corresponding tuple of integers and will be compared as you want.
The second approach (with the key operator) will create an additional overhead during sorting as it will be called O(NlogN) times.
You also tried using the result of sort function as a value, but it changes the given list in-place. If you want a sorted copy of your list, use sorted.
