i have an array and fill it like this
arr = [[],[]]
arr[0].append(post.attrib['href']) #strings
arr[1].append(int(klass)) # int
and i need to sort whole array by arr[1] for example input:
[['string3', 'string45', 'string25'], [46, 2, 12]]
and output:
[['string45', 'string25', 'string3'], [2, 12, 46]]
i tried
arr[0].sort(key = x[1].__getitem__)
arr[0].sort(key=lambda x: x[1])
arr = sorted(arr, key=itemgetter(1))
but everything of this isn't working correct
As you are sorting composite elements, create a pair of those elements you intend to sort using (zip or itertools.izip), and use operator.itemgetter, to select the correct key. Finally transpose back to the original format using zip.
from operator import itemgetter
zip(*sorted(zip(*arr), key = itemgetter(1)))
[('string45', 'string25', 'string3'), (2, 12, 46)]
This idea can be extended to multiple keys in a list.
Steps of Execution
>>> arr #Original Array
[['string3', 'string45', 'string25'], [46, 2, 12]]
>>> zip(*arr) #Unpacking the array, transposes it
[('string3', 46), ('string45', 2), ('string25', 12)]
>>> sorted(zip(*arr), key = itemgetter(1)) #Sort the Transpose Array based on second element
[('string45', 2), ('string25', 12), ('string3', 46)]
>>> zip(*sorted(zip(*arr), key = itemgetter(1))) #Transpose the result to generate the original format
[('string45', 'string25', 'string3'), (2, 12, 46)]
* is used for Unpacking Argument Listszip rather than to sorted. The Output of sorted is unpacked and passed to zip. Let me add the steps of execution to the answerHave you considered storing your data as one list of 2-element pairs (e.g. [(46, 'string3'), (2, 'string45'), (12, 'string25')]) instead? Because if you did that, your problem would solve itself in one line of code:
l = [(46, 'string3'), (2, 'string45'), (12, 'string25')]
print sorted(l)
# [(2, 'string45'), (12, 'string25'), (46, 'string3')]
By the way, they're lists, not arrays. Arrays in Python mean something else (namely, NumPy arrays).
I find packing and unpacking arguments a little bit unintuitive at times, so
l = [['string3', 'string45', 'string25'], [46, 2, 12]]
[[y[1] for y in sorted(zip(l[1], x))] for x in l]
[['string45', 'string25', 'string3'], [2, 12, 46]]
