SortedDict in Django

Question

I have a dictionary with a list of times I want to display in a template:

from django.utils.datastructures import SortedDict

time_filter = SortedDict({
    0 : "Eternity",
    15 : "15 Minutes",
    30 : "30 Minutes",
    45 : "45 Minutes",
    60 : "1 Hour",
    90 : "1.5 Hours",
    120 : "2 Hours",
    150 : "2.5 Hours",
    180 : "3 Hours",
    210 : "3.5 Hours",
    240 : "4 Hours",
    270 : "4.5 Hours",
    300 : "5 Hours"
})

I want to create a drop down in the template:

<select id="time_filter">
    {% for key, value in time_filter.items %}
        <option value="{{ key }}">{{ value }}</option>
    {% endfor %}
</select>

But the elements in the drop down aren't coming through in the order defined in the dictionary. What am I missing?

Answer 1

Look here.

You are doing the "That does not work" thing, i.e. giving an unsorted dictionary as the input to the sorted dictionary.

You want

SortedDict([
    (0, 'Eternity'),
    (15, '15 minutes'),
    # ...
    (300, '300 minutes'),
])

Answer 2

Consider using one of Python's many dictionary implementations that maintains the keys in sorted order. For example, the sortedcontainers module is pure-Python and fast-as-C implementations. It supports fast get/set/iter operations and keeps the keys sorted. There's also a performance comparison that benchmarks the implementation against several other popular choices.

Answer 3

You instantiate the SortedDict with a "normal" dict as an argument - your ordering is lost. You have to instantiate the SortedDict with an iterable that preserves the ordering, e.g.:

SortedDict((
   (0, "Eternity"),
   (15, "15 Minutes"),
   # ...
))

Answer 4

This answer may not exactly answer the question, you can use "dictsort" and "dictsortreversed" from django template tags to sort normal dict. So there is no need to use SortedDict.