How to get unique sorted list in one statement?

Question

The return value of sort() is None so the following code doesn't work:

def sorted_unique_items(a):
   return list(set(a)).sort()

Any idea for better solution?

Answer 1

Use sorted():

sorted(set(a))

and you can omit the list() call entirely.

Answer 2

Its simple Just use the sorted() method :

data = [10,5,46,4]
sorted_data = sorted(data)
print "Sorted Data ::::",sorted_data

Answer 3

Above solution is only work for hashable type like string,integer and tuple but it not work for unhashable type like list.

for example if you have a list data= [[2],[1],[2],[4]]

for unhashable type best solution is :

from itertools import izip, islice

values= [[2],[1],[2],[4]]

def sort_unique_unhashable(values):

    values = sorted(values)
    if not values:
        return []
    consecutive_pairs = izip(values, islice(values, 1, len(values)))
    result = [a for (a, b) in consecutive_pairs if a != b]
    result.append(values[-1])
    return result
print sort_unique_unhashable(values)

Answer 4

There is Two options:

a = [2,34,55,1,22,11,22,55,1]

#First option using sorted and set. 
sorted(set(a))

#second option using list and set.
list(set(a))