获取数组中最不常见的元素

Question

为了找到最常见的,我知道我可以使用这样的东西:

most_common = collections.Counter(array).most_common(to_find)

然而,我似乎无法找到任何可比的,因为找到最不常见的元素.

我可以获得有关如何做的建议.

Answer 1

most_common没有任何参数返回所有条目,从最常见到最少排序.

因此,要找到最不常见的,只需从另一端开始查看它.

Answer 2

借用collections.Counter.most_common适当的来源和反转:

from operator import itemgetter
import heapq
import collections
def least_common_values(array, to_find=None):
    counter = collections.Counter(array)
    if to_find is None:
        return sorted(counter.items(), key=itemgetter(1), reverse=False)
    return heapq.nsmallest(to_find, counter.items(), key=itemgetter(1))

>>> data = [1,1,2,2,2,2,3,3,3,3,3,3,3,4,4,4,4,4,4,4]
>>> least_common_values(data, 2)
[(1, 2), (2, 4)]
>>> least_common_values([1,1,2,3,3])
[(2, 1), (1, 2), (3, 2)]
>>>

Answer 3

关于什么

least_common = collections.Counter(array).most_common()[-1]

Answer 4

抱歉，这个帖子迟到了...发现文档非常有帮助： https://docs.python.org/3.7/library/collections.html

搜索“最少”，您将看到此表，它有助于获取超过列表中最后一个 (-1) 的元素：

c.most_common()[:-n-1:-1]       # n least common elements

这是一个例子：

n = 50

word_freq = Count(words)
least_common = word_freq.most_common()[:-n-1:-1]

Answer 5

只获取最不常见的元素，仅此而已：

>>> from collections import Counter
>>> ls = [1, 2, 3, 3, 2, 5, 1, 6, 6]
>>> Counter(ls).most_common()[-1][0]
5

Answer 6

def least_common_values(array, to_find):
    """
    >>> least_common_values([1,1,2,2,2,2,3,3,3,3,3,3,3,4,4,4,4,4,4,4], 2)
    [(1, 2), (2, 4)]
    """
    counts = collections.Counter(array)
    return list(reversed(counts.most_common()[-to_find:]))