Python Counter：计数为x的打印键

Question

Say I have a Counter object representing a collection of words:

>>> words = ['hello', 'hello', 'hello', 'world']
>>> counter = Counter(words)

One way to find out which words have count 1 would be to iterate over counter :

for word, count in counter.items():
    if count == 1:
        print(word)

Is there an easier/better way to do this? That is, can one "invert" counter to give the word whose count is x ?

Answer 1

To reverse any mapping—whether a Counter , a dict , or anything else:

rev = {v: k for k, v in d.items()}

Then you use that like any other dictionary:

key_whose_count_is_10 = rev[10]

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In the case where there are two keys with the same value, the value will map to one of them, arbitrarily. But that's pretty much inherent in your problem. You're asking for "the" key whose count is x ; what do you want to do if there are three keys whose count is x ?

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If you're only going to be making one query, rather than multiple queries, it's more efficient to just iterate. Which one is clearer (which is almost always more important) is arguable. Here's one way to do it, for comparison:

key_whose_count_is_10 = next(k for k, v in d.items() if v==10)

Answer 2

It would be far better to put every element with value 1 in a list, I think. Here's a Pythonic way to do this:

new_list = [w for w in words if counter[w] == 1]

Like this, you will store every word in words that has value of 1 in your counter.

So, for example, if you have in your list another string, say the string test :

words = ['hello', 'hello', 'hello', 'world', 'test']

then, the new list will have the values world and test .

Answer 3

Your Counter object uses each word as the key and stores the number of occurrences as the value.

To do what you want to do, you need to use the number of occurrences as the key and the list of words as the value:

wordDict = {}
for word, count in counter.items():
    if count in wordDict:
        wordDict[count].append(word)
    else:
        wordDict[count] = [word]

Then you can use wordDict[2] to grab the list of words that appear twice.

Answer 4

You can use a list comprehension to check each element's count

>>> words = ['hello', 'hello', 'hi', 'hi', 'world', 'foo', 'bar']
>>> from collections import Counter
>>> counter = Counter(words)
>>> [i for i in counter if counter[i] == 1]
['world', 'bar', 'foo']

You can also use the count() function on the original list

>>> [i for i in words if words.count(i) == 1]
['world', 'foo', 'bar']

Answer 5

You can use defaultdict :

import collections
d = collections.defaultdict(list)
for word, count in counter.items():
    d[count].append(word)

You can then do:

d[1]

to get all words with count one (as there can be one or more words).

Answer 6

Using filter (or ifilter if itertools is imported)

n = 1
for word in filter(lambda w: counter[w] == n, words):
  print word