CODEWITHSUNDEEP
pythonlistpython-2.7sorting
daeda
daeda

Reputation: 389

sort list of objects by return of object method

I've got a list of objects, that have a function which does some calculations on several member variables and returns a value. I want to sort the list according to this return value.

I've found how to sort a list of objects by member variables but I want to keep them private (_variable_name).

Is there an easy way to accomplish this similar to the method of sorting the list directly by the members of the objects?

Upvotes: 2

Views: 3836

Answers (1)

Martijn Pieters
Martijn Pieters

Reputation: 1123400

Just use the key argument to the sorted() function or list.sort() method:

sorted_list = sorted(list_of_objects, key=function_that_calculates)

The function_that_calculates is called for each entry in list_of_objects and its result informs the sort.

If you meant that each object has a method, you can use a lambda or the operator.methodcaller() object to call the method on each element:

sorted_list = sorted(list_of_objects, key=lambda obj: obj.method_name())

or

from operator import methodcaller

sorted_list = sorted(list_of_objects, key=methodcaller('method_name'))

Note that in Python, there is no such thing as a private attribute; your sorting function can still just access it. The leading underscore is just a convention. As such, sorting by a specific attribute can be done with either a lambda again, or using the operator.attrgetter() object:

sorted_list = sorted(list_of_objects, key=lambda obj: obj._variable_name)

or

from operator import attrgetter

sorted_list = sorted(list_of_objects, key=attrgetter('_variable_name'))

Upvotes: 7

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