CODEWITHSUNDEEP
pythonsortingdictionary
Niklas Rosencrantz
Niklas Rosencrantz

Reputation: 26637

Sorting a python dictionary by name?

I have a python dictionary that I want to sort according to name:

location_map_india = {
  101: {'name': 'Andaman & Nicobar Islands', 'lat': 11.96, 'long': 92.69, 'radius': 294200},  
  108: {'name': 'Andhra Pradesh', 'lat': 17.04, 'long': 80.09, 'radius': 294200},
...
}

It doesn't come out the way I expect it. I tried:

location_map_india = sorted(location_map_india.iteritems(), key=lambda x: x.name)

The above didn't work. What should I do?

Update

The following was allowed and behaved as expected. Thanks for all the help. Code:

location_map_india = sorted(location_map_india.iteritems(), key=lambda x: x[1]["name"])

Template:

{% for key,value in location_map_india %}<option value="{{key}}" >{{value.name}}</option>{% endfor %}

Upvotes: 2

Views: 426

Answers (4)

wleao
wleao

Reputation: 2346

If you're using python 2.7 or superior, take a look at OrderedDict. Using it may solve your problem.

>>> OrderedDict(sorted(d.items(), key=lambda x: x[1]['name']))

Upvotes: 3

krenel
krenel

Reputation: 776

If you are using a dictionary that must have order in any way, you are not using the correct data structure.

Try list or tuples instead.

Upvotes: 1

Xavier Combelle
Xavier Combelle

Reputation: 11235

you should try 

location_map_india = sorted(location_map_india.items(), key=lambda x: x[1]['name'])

Upvotes: 1

Avaris
Avaris

Reputation: 36725

You are close. Try:

location_map_india = sorted(location_map_india.iteritems(), key=lambda x: x[1]["name"])

but the result would be a list not a dict. dict is orderless.

Upvotes: 6

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