More easy to understand (What is Lambda actually doing):
ls2=[[0,1,'f'],[4,2,'t'],[9,4,'afsd']]
def thirdItem(ls):
#return the third item of the list
return ls[2]
#Sort according to what the thirdItem function return
ls2.sort(key=thirdItem)
array.sort(key = lambda x:x[1])
You can easily sort using this snippet, where 1 is the index of the element.
I think lambda function can solve your problem.
old_list = [[0,1,'f'], [4,2,'t'],[9,4,'afsd']]
#let's assume we want to sort lists by last value ( old_list[2] )
new_list = sorted(old_list, key=lambda x: x[2])
#Resulst of new_list will be:
[[9, 4, 'afsd'], [0, 1, 'f'], [4, 2, 't']]
Sorting a Multidimensional Array execute here
arr=[[2,1],[1,2],[3,5],[4,5],[3,1],[5,2],[3,8],[1,9],[1,3]]
arr.sort(key=lambda x:x[0])
la=set([i[0] for i in Points])
for i in la:
tempres=list()
for j in arr:
if j[0]==i:
tempres.append(j[1])
for j in sorted(tempres,reverse=True):
print(i,j)
in place
>>> l = [[0, 1, 'f'], [4, 2, 't'], [9, 4, 'afsd']]
>>> l.sort(key=lambda x: x[2])
not in place using sorted:
>>> sorted(l, key=lambda x: x[2])
Like this:
import operator
l = [...]
sorted_list = sorted(l, key=operator.itemgetter(desired_item_index))
**old_list = [[0,1,'f'], [4,2,'t'],[9,4,'afsd']]
#let's assume we want to sort lists by last value ( old_list[2] )
new_list = sorted(old_list, key=lambda x: x[2])**
correct me if i'm wrong but isnt the 'x[2]' calling the 3rd item in the list, not the 3rd item in the nested list? should it be x[2][2]?
Itemgetter lets you to sort by multiple criteria / columns:
sorted_list = sorted(list_to_sort, key=itemgetter(2,0,1))
Make sure that you do not have any null or NaN values in the list you want to sort. If there are NaN values, then your sort will be off, impacting the sorting of the non-null values.
Check out Python: sort function breaks in the presence of nan
multiple criteria can also be implemented through lambda function
sorted_list = sorted(list_to_sort, key=lambda x: (x[1], x[0]))
Source: Stackoverflow.com