我有一个看起来有点像这样的计数器:
Counter: {('A': 10), ('C':5), ('H':4)}
我想专门按字母顺序对键进行排序,而不是按
counter.most_common()
有什么办法可以实现这个目标吗?
只需使用已排序:
>>> from collections import Counter
>>> counter = Counter({'A': 10, 'C': 5, 'H': 7})
>>> counter.most_common()
[('A', 10), ('H', 7), ('C', 5)]
>>> sorted(counter.items())
[('A', 10), ('C', 5), ('H', 7)]
>>> from operator import itemgetter
>>> from collections import Counter
>>> c = Counter({'A': 10, 'C':5, 'H':4})
>>> sorted(c.items(), key=itemgetter(0))
[('A', 10), ('C', 5), ('H', 4)]
sorted(counter.items(),key = lambda i: i[0])
例如:
arr = [2,3,1,3,2,4,6,7,9,2,19]
c = collections.Counter(arr)
sorted(c.items(),key = lambda i: i[0])
外: [(1, 1), (2, 3), (3, 2), (4, 1), (6, 1), (7, 1), (9, 1), (19, 1)] 如果你想获得字典格式,只需
dict(sorted(c.items(),key = lambda i: i[0]))
获取按排序顺序列出的值
array = [1, 2, 3, 4, 5]
counter = collections.Counter(array)
sorted_occurrences = list(dict(sorted(counter.items())).values())
在Python 3中,可以使用集合的most_common函数。计数器:
x = ['a', 'b', 'c', 'c', 'c', 'd', 'd']
counts = collections.Counter(x)
counts.most_common(len(counts))
这使用了集合中可用的most_common函数。Counter,它允许您查找
n
最常用键的键和计数。
请参考我下面的评论
# input data
counter = Counter({'A': 10, 'C': 5, 'H': 7})
# Sorting the Counter by keys
sorted_counter = sorted(counter.items())
# change it back into a counter
result_counter = Counter(sorted_counter)
结果应该是这样的:
Counter({'A': 10, 'C': 5, 'H': 7})