CODEWITHSUNDEEP
pythondictionary
Superdooperhero
Superdooperhero

Reputation: 8096

Finding the dictionary keys whose values are numerically highest

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

Upvotes: 0

Views: 120

Answers (6)

jfs
jfs

Reputation: 414645

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.

Upvotes: 5

Padraic Cunningham
Padraic Cunningham

Reputation: 180481

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)]

Upvotes: 1

Shan Valleru
Shan Valleru

Reputation: 3121

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

Upvotes: 3

Lachezar
Lachezar

Reputation: 6703

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.

Upvotes: 1

Danstahr
Danstahr

Reputation: 4319

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).

Upvotes: 7

furas
furas

Reputation: 142859

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

Upvotes: 0

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