sorting a list of dictionary values by date in python

Question

I have a list and I am appending a dictionary to it as I loop through my data…and I would like to sort by one of the dictionary keys.

ex:

data = "data from database"
list = []
for x in data:
     dict = {'title':title, 'date': x.created_on}
     list.append(dict)

I want to sort the list in reverse order by value of ‘date’

Asked By: Joe

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Source

Answer 1

Sort the data (or a copy of the data) directly and build the list of dicts afterwards. Sort using the function sorted with an appropiate key function (operator.attrgetter probably)

Answered By: Tetha

Answer 2

You can do it this way:

list.sort(key=lambda item:item['date'], reverse=True)

Answered By: hvintus

Answer 3

from operator import itemgetter

your_list.sort(key=itemgetter('date'), reverse=True)

Related notes

don’t use list, dict as variable names, they are builtin names in Python. It makes your code hard to read.

you might need to replace dictionary by tuple or collections.namedtuple or custom struct-like class depending on the context

from collections import namedtuple
from operator    import itemgetter

Row = namedtuple('Row', 'title date')
rows = [Row(row.title, row.created_on) for row in data]
rows.sort(key=itemgetter(1), reverse=True)

Example:

>>> lst = [Row('a', 1), Row('b', 2)]
>>> lst.sort(key=itemgetter(1), reverse=True)
>>> lst
[Row(title='b', date=2), Row(title='a', date=1)]

Or

>>> from operator import attrgetter
>>> lst = [Row('a', 1), Row('b', 2)]
>>> lst.sort(key=attrgetter('date'), reverse=True)
>>> lst
[Row(title='b', date=2), Row(title='a', date=1)]

Here’s how namedtuple looks inside:

>>> Row = namedtuple('Row', 'title date', verbose=True)

class Row(tuple):
        'Row(title, date)'

        __slots__ = ()

        _fields = ('title', 'date')

        def __new__(cls, title, date):
            return tuple.__new__(cls, (title, date))

        @classmethod
        def _make(cls, iterable, new=tuple.__new__, len=len):
            'Make a new Row object from a sequence or iterable'
            result = new(cls, iterable)
            if len(result) != 2:
                raise TypeError('Expected 2 arguments, got %d' % len(result))
            return result

        def __repr__(self):
            return 'Row(title=%r, date=%r)' % self

        def _asdict(t):
            'Return a new dict which maps field names to their values'
            return {'title': t[0], 'date': t[1]}

        def _replace(self, **kwds):
            'Return a new Row object replacing specified fields with new values'

            result = self._make(map(kwds.pop, ('title', 'date'), self))
            if kwds:
                raise ValueError('Got unexpected field names: %r' % kwds.keys())

            return result

        def __getnewargs__(self):
            return tuple(self)

        title = property(itemgetter(0))
        date = property(itemgetter(1))

Answered By: jfs

Answer 4

If you’re into the whole brevity thing:

data = "data from database"
sorted_data = sorted(
    [{'title': x.title, 'date': x.created_on} for x in data], 
    key=operator.itemgetter('date'),
    reverse=True)

Answered By: recursive

Answer 5

You can do it like this:

from operator import itemgetter
list.sort(key=itemgetter('date'), reverse=True)

See Also: How do I sort a list of dictionaries by a value of the dictionary?

Answered By: Dave

sorting a list of dictionary values by date in python

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