CODEWITHSUNDEEP
pythonsortingdictionary
Federer
Federer

Reputation: 34745

Ascending Sort of Float Dictionary Values

I have a dictionary of the form:

d = { 'someText': floatNumber }

Where floatNumber is an epoch timestamp. I'm trying to organise this such that time is in ascending order.

Example: {'someText':0000001, 'someText1':0000002, and so on}

Only way I can think of doing it is by looping with for k,v in dict.items() and then manually sorting it into a list but that may take a long time. Any help would be greatly appreciated.

Upvotes: 1

Views: 3335

Answers (4)

Raymond Hettinger
Raymond Hettinger

Reputation: 226346

An order dictionary can be used to store the entries in sorted order:

>>> from collections import OrderedDict
>>> d = OrderedDict(sorted(dict.items(), key=lambda item: item[1]))

Upvotes: 1

Spencer Rathbun
Spencer Rathbun

Reputation: 14900

First, by default dictionaries are unsorted. You may want to use a list and insert into the appropriate location as you build the dataset. Otherwise, use the sorted function.

sorted(dict.iteritems(), key=lambda (x, y): y)

Upvotes: 0

user647772
user647772

Reputation: 

from operator import itemgetter

d = {'foo':1, 'bar':3, 'baz':2}
l = [(k, v) for k, v in d.items()]
s = sorted(l, key=itemgetter(1))

# s == [('foo', 1), ('baz', 2), ('bar', 3)]

More on sorting: http://wiki.python.org/moin/HowTo/Sorting

Edit:

Improved version (thanks for the comment):

from operator import itemgetter

d = {'foo':1, 'bar':3, 'baz':2}
s = sorted(d.items(), key=itemgetter(1))

# s == [('foo', 1), ('baz', 2), ('bar', 3)]

Upvotes: 0

Felix Kling
Felix Kling

Reputation: 816472

Maybe you want:

import operator
values = sorted(d.items(), key=operator.itemgetter(1))

which would generate a sorted list of tuples, like

[('someText', 1), ('someText', 2), ...]

Dictionaries cannot be sorted, so you have to use another data structure to store your key-value pairs.

Upvotes: 4

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