Trier la liste de la liste
>>> lis = [[1,4,7],[3,6,9],[2,59,8]]
>>> sorted(lis, key=lambda x: x[0])
[[1, 4, 7], [2, 59, 8], [3, 6, 9]]
Bloody Bug
>>> lis = [[1,4,7],[3,6,9],[2,59,8]]
>>> sorted(lis, key=lambda x: x[0])
[[1, 4, 7], [2, 59, 8], [3, 6, 9]]#1 Changes list
list.sort(reverse=True)
#2 Returns sorted list
sorted(list, reverse=True)# defning A as a list
A.sort(reverse = True)from operator import itemgetter
A = [[10, 8], [90, 2], [45, 6]]
print("Sorted List A based on index 0: % s" % (sorted(A, key=itemgetter(0))))
B = [[50, 'Yes'], [20, 'No'], [100, 'Maybe']]
print("Sorted List B based on index 1: % s" % (sorted(B, key=itemgetter(1))))l=[1,3,2,5]
l= sorted(l)
print(l)
#output=[1, 2, 3, 5]
#or reverse the order:
l=[1,3,2,5]
l= sorted(l,reverse=True)
print(l)
#output=[5, 3, 2, 1]# You can use a lambda function, but you'll have to specify the index of the sorting key.
A = [[100, 'Yes'], [40, 'Maybe'], [60, 'No']]
print("Sorted List A based on index 0: % s" % (sorted(A, key=lambda x:x[0])))
B = [[2, 'Dog'], [0, 'Bird'], [7, 'Cat']]
print("Sorted List A based on index 1: % s" % (sorted(B, key=lambda x:x[1])))
# Also, you can use .sort() if you want to sort just by the first index
A = [[55, 90], [45, 89], [90, 70]]
A.sort()
print("New sorted list A is % s" % (A))
A.sort(reverse=True)
print("New reverse sorted list A is % s" % (A))
# You can even change the key sort, if you want to sort by length for example:
A = [[5, 90, 'Hi', 66], [80, 99], [56, 32, 80]]
A.sort(key=len) # <-
print("New sorted list A is % s" % (A))