Creating a "top list" from dictionary

Question

I have dictionary, something like:

x = {"John":15,"Josh":2,"Emily":50,"Joey":12}

And I need to create something like this:

Emily - 50

John - 15

Joey - 12

Josh - 2

What would be the best way to do this? I have already tried to sort the dictionary somehow, but then it converted to list and I wasn't able to get both values (name and number).

score 0 · Accepted Answer · answered Mar 30 '18 at 09:28

0

With simple sorting by dictionary values:

x = {"John":15, "Josh":2, "Emily":50, "Joey":12}

for i, t in enumerate(sorted(x.items(), key=lambda x: x[1], reverse=True), 1):
    print('{}. {} - {}'.format(i, t[0], t[1]))

The output:

1. Emily - 50
2. John - 15
3. Joey - 12
4. Josh - 2

answered Mar 30 '18 at 09:28

RomanPerekhrest

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Thank you, last question, how to limit it to just top 3? – Arbys Mar 30 '18 at 09:49
@Arbys, welcome, just add `if i == 3: break` as next line after `print` statement – RomanPerekhrest Mar 30 '18 at 10:00

score 0 · Answer 2 · answered Mar 30 '18 at 09:32

See the answer here: How do I sort a dictionary by value?

In your case, you'll want to reverse sorted_x as you want higher values first, so:

import operator


x = {"John":15,"Josh":2,"Emily":50,"Joey":12}
sorted_x = sorted(x.items(), key=operator.itemgetter(1), reverse=True)

score 0 · Answer 3 · answered Mar 30 '18 at 09:33

0

An easy way:

x = {"John":15,"Josh":2,"Emily":50,"Joey":12}

val = sorted([y for y in x.values()])
name = val[:]
for key in x.keys():
    id = val.index(x[key])
    name[id] = key

answered Mar 30 '18 at 09:33

Mathieu

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Creating a "top list" from dictionary

3 Answers3