Sort a list based on dictionary values in python?

Question

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.

georg · Accepted Answer · 2012-10-20 10:11:05Z

21

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:

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

edited Oct 20, 2012 at 10:11

answered Oct 20, 2012 at 9:49

georg

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6 Comments

Mark Amery Over a year ago

Oi, stop jumping in on the questions I'm answering and giving cleverer answers! :) More seriously: there is one difference in behavior between thg's answer and mine. Mine will raise an exception if trial_values contains a value that is not a key of trial_dict, whereas thg's will silently sort that value to the front. Either may be appropriate depending upon context.

tijko Over a year ago

@thg435 I should have double checked the docs, I was clearly misunderstanding .get.

georg Over a year ago

@MarkAmery: You were good, kid, real good. But as long as I'm around, you'll always be second best, see? ))

georg Over a year ago

@MarkAmery: seriously, yes, the lambda gives you more control and in most cases this is what should be used.

Mark Amery Over a year ago

:) Also serious: which of our approaches do you think is stylistically better here? I like the lambda for style reasons even in this case because it means you get to see the expression being sorted on (that is, trial_dict[x]) without doing any further thought; with your answer, I need to visualize the function being called to see that we're sorting on trial_dict.get(x), which requires an extra cognitive step. So I reckon I prefer my approach to yours. Whatcha think?

|

Mark Amery · Accepted Answer · 2014-05-11 16:36:22Z

8

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])

edited May 11, 2014 at 16:36

answered Oct 20, 2012 at 9:46

Mark Amery

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1 Comment

tijko Over a year ago

I should be keeping lambda in mind for a smaller function like this. I'm marking thg435 as answered but, this was real helpful to, I'll have to think to use the lambda for future use, thanks! :)

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