在 python 中按日期对字典值列表进行排序

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'

Answer 1

You can do it this way:

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

Answer 2

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))

Answer 3

I actually had this almost exact question yesterday and solved it using search . The best answer applied to your question is this:

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

Answer 4

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)

Answer 5

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)