Python create sorted list based on attribute

Question

I need to create a list of dictionary values, sorted based on whether or not the value is empty, my attempt gets the desired output but has to iterate twice over the dictionary, I was wondering if there was a way to do this only iterating once? My actual dictionary contains like 50,000 values and just iterating once would save some time when doing it repeatedly

d = {'id1':[],
     'id2':[1],
     'id3':[1]}
l = [v for v in d.values() if v]+[v for v in d.values() if not v]
print (l)

this is my desired output:

[[1], [1], []]

Daniel · Accepted Answer · 2017-06-05 10:12:12Z

3

You can sort with respect to length:

l = sorted(d.values(), key=len, reverse=True)

answered Jun 5, 2017 at 10:12

Daniel

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khelili miliana · Accepted Answer · 2017-06-05 10:51:02Z

1

I suggest SortedContainers : Fast, pure-Python implementation of SortedList, SortedDict, and SortedSet types.

from sortedcontainers import SortedList
d = {'id1':[],
     'id2':[1],
     'id3':[1]}

res=SortedList(d.values())

edited Jun 5, 2017 at 10:51

answered Jun 5, 2017 at 10:15

khelili miliana

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