Basic code:
var1 = ['b', 'a', 'c', 'd']
var2 = ['c', 'a']
print(set(var1).difference(set(var2)))
Output:
{'b', 'd'}
Question
Is it possible to sort this output into alphabetical order? If so, how can I?
This is what I have tried:
print(set(var1).difference(set(var2)).sort())
But error shows up:
print(set(var1).difference(set(var2)).sort())
AttributeError: 'set' object has no attribute 'sort'
Sets have no order, so sorting them makes no sense. But if you pass a set to sorted it will be turned into a list and sorted:
print(sorted(set(var1).difference(set(var2))))
None.
sort() doesn't return the sorted list. Use sorted instead, as the edited answer indicates.Here is the code that will solve the problem:
var1 = ['b', 'a', 'c', 'd']
var2 = ['c', 'a']
print(sorted(set((set(var1).difference(set(var2))))))
Output:
['b', 'd']
You might be wondering that the output is a list and not a set. That's because the whole point of using a set, both in mathematics as a tool and in programming languages as a data structure is that it's not ordered. Meaning the sets {p, q} and {q, p} are the same set!
set.difference is already a set, so you don't have to call set() on it.You can get the list of elements in your set sorted alphabetically by comparing the ord of the different characters.
test_list = ["a", "b", "u", "x", "e", "f", "k", "z"]
test_set = set(test_list)
sorted_list = sorted(test_set, key=ord) # == ['a', 'b', 'e', 'f', 'k', 'u', 'x', 'z']
key=ord doesn't do anything particularly useful. Strings are already sorted lexicographically by default.
key could be something else, so maybe someone deeps further in that topic if they are interested.key exists is hardly a reason to use it unnecessarily.
sorted(set(var1).difference(var2))to turn the set into a sorted list, but you cannot sort the set itself (sets are, by definition, unordered).set(...).differencecan take an arbitrary iterable as its argument; you don't need to wrapvar2in a set first.