Sort dictionary of lists by key value pairs

Question

I am trying to sort values that are inside a dictionary of lists and create a new list out of them. Here is the data:

{
    'fbi': [229, 421, 586, 654, 947, 955, 1095, 1294, 1467, 2423, 3063, 3478, 3617, 3730, 3848, 3959, 4018, 4136, 4297, 4435, 4635, 4679, 4738, 5116, 5211, 5330, 5698, 6107, 6792, 6906, 7036], 
    'comey': [605, 756, 1388, 1439, 1593, 1810, 1959, 2123, 2506, 3037, 6848], 
    'hillary': [14, 181, 449, 614, 704, 1079, 1250, 2484, 2534, 2659, 3233, 3374, 3488, 3565, 4076, 4756, 4865, 6125, 7109]
}

What I am trying is to find the 20 smallest values in these and get a list of their corresponding keys. For example, the first three least values are 14(hillary), 181(hillary) and 229(fbi). Therefore, how can I get a list like this:

['hillary', 'hillary', 'fbi']

All values will always be different. Also, all the values in the list are sorted, from ascending to descending.

Here is what I have tried:

for m in range(1,20):
    for i in sort_vals.values():
        if i[0] < a[0]:
            a[0] = i[0]

This gives me the least value but not any other as after one iteration, the least value is always the same. I guess if I can delete that particular value it will be helpful. Can't think of anything else. Thanks!

yatu · Accepted Answer · 2019-08-26 12:36:50Z

17

You could flatten the dictionary (d here) into a list of tuples with the corresponding key/value pairs, and sort the tuples according to the values:

from operator import itemgetter

l = [(k,i) for k,v in d.items() for i in v]
# [('fbi', 229), ('fbi', 421), ('fbi', 586), ('fbi', 654),...
list(zip(*sorted(l, key=itemgetter(1))[:3]))[0]
# ('hillary', 'hillary', 'fbi')

answered Aug 26, 2019 at 12:36

yatu

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Jean-François Fabre · Accepted Answer · 2019-08-26 12:37:35Z

12

you could

invert your mapping, creating a dictionary with numbers => list of names
sort this dictionary (as tuple)
pick the 3 first items

like this:

import collections

d = collections.defaultdict(list)
data = {'fbi': [229, 421, 586, 654, 947, 955, 1095, 1294, 1467, 2423, 3063, 3478, 3617, 3730, 3848, 3959, 4018, 4136, 4297, 4435, 4635, 4679, 4738, 5116, 5211, 5330, 5698, 6107, 6792, 6906, 7036], 'comey': [605, 756, 1388, 1439, 1593, 1810, 1959, 2123, 2506, 3037, 6848], 'hillary': [14, 181, 449, 614, 704, 1079, 1250, 2484, 2534, 2659, 3233, 3374, 3488, 3565, 4076, 4756, 4865, 6125, 7109]}

for k,vlist in data.items():
    for v in vlist:
        d[v].append(k)

result = [v[0] for k,v in sorted(d.items())[:3]]

print(result)

this prints:

['hillary', 'hillary', 'fbi']

note that if there are several names attached to a value, this code will pick only the first one (v[0])

answered Aug 26, 2019 at 12:37

Jean-François Fabre♦

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2 Comments

DeepSpace Over a year ago

Hopefully a number is only in one list

Jean-François Fabre Over a year ago

in that case it will work. but I wanted to outline the other case.

Jamie · Accepted Answer · 2019-08-27 09:13:31Z

3

Just use lambda function in sorted().

l = [(k,i) for k,v in d.items() for i in v]
res = [v[0] for v in sorted(l, key=lambda x: x[1])][:20]

edited Aug 27, 2019 at 9:13

answered Aug 26, 2019 at 13:30

Jamie

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Comments

LiuXiMin · Accepted Answer · 2019-09-12 00:00:02Z

0

Since all the values in the list are sorted, you can use heapq.merge:

d = {
    'fbi': [229, 421, 586, 654, 947, 955, 1095, 1294, 1467, 2423, 3063, 3478, 3617, 3730, 3848, 3959, 4018, 4136, 4297, 4435, 4635, 4679, 4738, 5116, 5211, 5330, 5698, 6107, 6792, 6906, 7036], 
    'comey': [605, 756, 1388, 1439, 1593, 1810, 1959, 2123, 2506, 3037, 6848], 
    'hillary': [14, 181, 449, 614, 704, 1079, 1250, 2484, 2534, 2659, 3233, 3374, 3488, 3565, 4076, 4756, 4865, 6125, 7109]
}
import heapq
pairs = [[(k, i) for i in v] for k, v in d.items()]
sorted_pairs = heapq.merge(*pairs, key=lambda a: a[1])
smallest_num_you_want = 3
res = [next(sorted_pairs)[0] for i in range(smallest_num_you_want)]

answered Sep 12, 2019 at 0:00

LiuXiMin

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