Finding the dictionary keys whose values are numerically highest

Question

Given a Python dict of the form:

dict = {'Alice': 2341, 'Beth': 9102, 'Cecil': 3258, ......}

Is there an easy way to print the first x keys with the highest numeric values? That is, say:

Beth   9102
Cecil  3258

Currently this is my attempt:

max = 0
max_word = ""
for key, value in w.word_counts.iteritems():
    if value > max:
        if key not in stop_words:
            max = value
            max_word = key

print max_word

possible duplicate of [Python: Sort a dictionary by value](http://stackoverflow.com/questions/613183/python-sort-a-dictionary-by-value) — Zero Piraeus, May 26 '14 at 22:52
You might consider using a [`Counter`](https://docs.python.org/2/library/collections.html#collections.Counter) instead of a dictionary initially. Then you have `word_counts.most_common(x)` — jscs, May 26 '14 at 22:55

score 7 · Accepted Answer · answered May 26 '14 at 22:51

7

I'd simply sort the items by the second value and then pick the first K elements :

d_items = sorted(d.items(), key=lambda x: -x[1])
print d_items[:2]
[('Beth', 9102), ('Cecil', 3258)]

The complexity of this approach is O(N log N + K), not that different from optimal O(N + K log K) (using QuickSelect and sorting just the first K elements).

answered May 26 '14 at 22:51

Danstahr

4,060
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1

Thanks. That works. Have no idea how or why, but none the less it works. – Superdooperhero May 26 '14 at 22:59
1

`d_items = sorted(d.items(), key=lambda x: x[1], reverse=True)`, the `reverse=True` kwarg is missing. – brunsgaard May 26 '14 at 23:02
@brunsgaard: note the minus in the lambda function. That's exactly what `reverse=True` does for numeric keys, it's just a few characters shorter. – Danstahr May 26 '14 at 23:04
@Danstahr, doh, of cause... You are the boss of the python!, ignore my comment above. – brunsgaard May 26 '14 at 23:07

score 5 · Answer 2 · answered May 26 '14 at 23:03

Using collections.Counter.most_common:

>>> from collections import Counter
>>> d = {'Alice': 2341, 'Beth': 9102, 'Cecil': 3258}
>>> c = Counter(d)
>>> c.most_common(2)
[('Beth', 9102), ('Cecil', 3258)]

It uses sorted (O(n*log n)), or heapq.nlargest(k) that might be faster than sorted if k << n, or max() if k==1.

score 3 · Answer 3 · answered May 26 '14 at 22:58

3

>>> (sorted(dict.items(), key=lambda x:x[1]))[:2]
[('Alice', 2341), ('Cecil', 3258)]

answered May 26 '14 at 22:58

Shan Valleru

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score 1 · Answer 4 · answered May 26 '14 at 22:58

1

items = sorted(w.word_counts.items(), lambda x, y: cmp(x[1], y[1]), None, True) 
items[:5]

Replace 5 with the number of elements you want to get.

answered May 26 '14 at 22:58

Lachezar

6,284
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score 1 · Answer 5 · answered May 26 '14 at 23:03

1

d = {'Alice': 2341, 'Beth': 9102, 'Cecil': 3258}

vs = sorted(d, key=d.get,reverse=True)

l = [(x,d.get(x)) for x in vs[0:2]]
n [4]: l
Out[4]: [('Beth', 9102), ('Cecil', 3258)]

answered May 26 '14 at 23:03

Padraic Cunningham

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score 0 · Answer 6 · answered May 26 '14 at 22:51

0

Convert dict to list of tuples [(2341, 'Alice'), ...] then sort it (without key=lambda ...).

answered May 26 '14 at 22:51

furas

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Finding the dictionary keys whose values are numerically highest

6 Answers6