Sort list of objects by a specific attribute of these objects?

Question

I created class named Student() with attributes name and surname. In that class I created class variable -> all_students = []

When creating an object, it is being added to that class variable = all_students.

How to sort that list of objects by the lenght of the attribute name from longest to shortest?

import operator
class Student:

    all_students = []
    def __init__(self, name, surname):
        self.name = name
        self.surname = surname
        Student.all_students.append(self)

s_1 = Student("Jacobbbb", "anything")

s_2 = Student("Mat", "anything1")

s_3 = Student("Marcooooooo", "sss")


new_list = sorted(Student.all_students, key=operator.attrgetter(len('name')))

for student in new_list:
    print(student.name)

I tried with operator, but I can't do that in this way. Will be gratefull for help.

Answer 1

You're almost there. You just need to change your key a bit to use a lambda:

sorted(Student.all_students, key=lambda x: -len(x.name))

This'll sort your student list in descending order of name length (notice the -len(...).

If you wanted to use attrgetter, you'd need to change your key to key=lambda x: -len(operator.attrgetter('name')(x)). But at that point, it becomes simpler to just use x.name.

---

Here's a demonstration. I've added a __repr__ method so it becomes clearer.

In [320]: def __repr__(self): return 'Student(%s %s)' %(self.name, self.surname)

In [321]: Student.__repr__ = __repr__

In [322]: sorted(Student.all_students, key=lambda x: -len(x.name))
Out[322]: [Student(Marcooooooo sss), Student(Jacobbbb anything), Student(Mat anything1)]

Answer 2

I haven't enough reputation to comment yet:

COLDSPEED's solution is good, but more pythonic still to use a key function (not lambda), and use the reverse arg

def name_length(student):
    return len(student.name)


sorted(Student.all_students, key=name_length, reverse=True)

Answer 3

You can't do it with attrgetter. Use a lambda:

sorted(Student.all_students, key=lambda s: len(s.name))