Python: How to sort a dictionary of X and Y coordinates by ascending X coordinate value?

Question

I have the following dictionary that I would like to sort based on their X coordinate in ascending fashion so that I can identify the "beacon" by the color arrangement (RGB in different orders). I keep trying to sort it like a list but that's not working out too well. Thanks in advance :)

Beacon2 = {
    'r': [998.9282836914062, 367.3825378417969],
    'b': [985.82373046875, 339.2225646972656], 
    'g': [969.539794921875, 369.2041931152344]
}

For this specific dictionary the expected result is

sortedBeacon = {
    'g': [969.539794921875, 369.2041931152344], 
    'b': [985.82373046875, 339.2225646972656],
    'r': [998.9282836914062, 367.3825378417969]
}

For this specific dictionary the expected result is sortedBeacon = {'g': [969.539794921875, 369.2041931152344], 'b': [985.82373046875, 339.2225646972656], 'r': [998.9282836914062, 367.3825378417969]} — Kara Schmidt, Dec 01 '17 at 02:19
Dictionaries being sorted is an implementation detail, and should not be relied on. You should use an `OrderedDict` instead, which will let you do this. — jhpratt, Dec 01 '17 at 02:32

score 3 · Answer 1 · answered Dec 01 '17 at 02:13

Note that dictionaries in general are not sortable. You can generate the internals sorted however without any lambdas by using itemgetter:

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

If you really want to maintain order, wrap the above in an OrderedDict

score 1 · Answer 2 · edited Dec 01 '17 at 02:15

1

The method sort() in Python is normally used on lists and tuples whereas sorted() is better for data structures like dictionaries.

In this case, using a simple lambda function can help you get what you want.

print(sorted(Beacon2.values(), key = lambda x: (x[0]))

edited Dec 01 '17 at 02:15

OneCricketeer

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answered Dec 01 '17 at 02:12

Shawna

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score 0 · Answer 3 · answered Dec 01 '17 at 02:10

0

If you just want the values, use this:

sorted(data.values())

If you want the keys associated with the sorted values, use this:

sorted(data, key=data.get)

Both key and values:

sorted(data.items(), key=lambda x:x[1])

Courtesy of: sort dict by value python

answered Dec 01 '17 at 02:10

jeevaa_v

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`sorted(data.values())` will sort the lists against each other , not the values within them – OneCricketeer Dec 01 '17 at 02:13

RoadRunner · Accepted Answer · 2017-12-01T02:43:41.977

0

You can try this:

from collections import OrderedDict

Beacon2 = {'r': [998.9282836914062, 367.3825378417969], 'b':
[985.82373046875, 339.2225646972656], 'g': [969.539794921875, 369.2041931152344]}

sorted_beacons = sorted(Beacon2.items(), key = lambda x: x[1][0])

>>> print(OrderedDict(sorted_beacons))
OrderedDict([('g', [969.539794921875, 369.2041931152344]), ('b', [985.82373046875, 339.2225646972656]), ('r', [998.9282836914062, 367.3825378417969])])

Where you first sort the list of tuples from Beacon2.items(), with a sorting key applied on the X coordinate located at [1][0] of each tuple.

Note that you need to wrap an OrderedDict to your result to preserve order of the dictionary.

edited Dec 01 '17 at 02:43

answered Dec 01 '17 at 02:11

RoadRunner

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Worked like a charm! Thanks :) – Kara Schmidt Dec 01 '17 at 02:17
@Kara But this is descending order within the list. Was that expected? – OneCricketeer Dec 01 '17 at 02:20
The first value within the items are ascending. The list themselves are descending – OneCricketeer Dec 01 '17 at 02:24
@cricket_007 I wanted the values the way they are here, ascending by X value maintaining corresponding keys – Kara Schmidt Dec 01 '17 at 02:27

Python: How to sort a dictionary of X and Y coordinates by ascending X coordinate value?

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