I have a list of words:
words = ['all', 'awesome', 'all', 'yeah', 'bye', 'all', 'yeah']
And I want to get a list of tuples:
[(3, 'all'), (2, 'yeah'), (1, 'bye'), (1, 'awesome')]
where each tuple is...
(number_of_occurrences, word)
The list should be sorted by the number of occurrences.
What I've done so far:
def popularWords(words):
dic = {}
for word in words:
dic.setdefault(word, 0)
dic[word] += 1
wordsList = [(dic.get(w), w) for w in dic]
wordsList.sort(reverse = True)
return wordsList
The question is...
Is it Pythonic, elegant and efficient?
Are you able to do it better?
Thanks in advance.
You can use the counter for this.
import collections
words = ['all', 'awesome', 'all', 'yeah', 'bye', 'all', 'yeah']
counter = collections.Counter(words)
print(counter.most_common())
>>> [('all', 3), ('yeah', 2), ('bye', 1), ('awesome', 1)]
It gives the tuple with reversed columns.
From the comments: collections.counter is >=2.7,3.1. You can use the counter recipe for lower versions.
The defaultdict collection is what you are looking for:
from collections import defaultdict
D = defaultdict(int)
for word in words:
D[word] += 1
That gives you a dict where keys are words and values are frequencies. To get to your (frequency, word) tuples:
tuples = [(freq, word) for word,freq in D.iteritems()]
If using Python 2.7+/3.1+, you can do the first step with a builtin Counter class:
from collections import Counter
D = Counter(words)
Is it Pythonic, elegant and efficient?
Looks good to me...
Are you able to do it better?
"better"? If it's understandable, and efficient, isn't that enough?
Maybe look at defaultdict to use that instead of setdefault.
