CODEWITHSUNDEEP
pythonlistsortingdictionary
tijko
tijko

Reputation: 8322

Sort a list based on dictionary values in python?

Say I have a dictionary and then I have a list that contains the dictionary's keys. Is there a way to sort the list based off of the dictionaries values?

I have been trying this:

trial_dict = {'*':4, '-':2, '+':3, '/':5}
trial_list = ['-','-','+','/','+','-','*']

I went to use:

sorted(trial_list, key=trial_dict.values())

And got:

TypeError: 'list' object is not callable

Then I went to go create a function that could be called with trial_dict.get():

def sort_help(x):
    if isinstance(x, dict):
        for i in x:
            return x[i]

sorted(trial_list, key=trial_dict.get(sort_help(trial_dict)))

I don't think the sort_help function is having any affect on the sort though. I'm not sure if using trial_dict.get() is the correct way to go about this either.

Upvotes: 21

Views: 25716

Answers (2)

Mark Amery
Mark Amery

Reputation: 155176

The key argument in the sorted builtin function (or the sort method of lists) has to be a function that maps members of the list you're sorting to the values you want to sort by. So you want this:

sorted(trial_list, key=lambda x: trial_dict[x])

Upvotes: 8

georg
georg

Reputation: 215049

Yes dict.get is the correct (or at least, the simplest) way:

sorted(trial_list, key=trial_dict.get)

As Mark Amery commented, the equivalent explicit lambda:

sorted(trial_list, key=lambda x: trial_dict[x])

might be better, for at least two reasons:

  1. the sort expression is visible and immediately editable
  2. it doesn't suppress errors (when the list contains something that is not in the dict).

Upvotes: 21

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