本文介绍了如何获取项目的计数格,但保持它们出现的顺序?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
例如,我需要计算一个单词在列表中出现的次数,不是按频率排序,而是按照单词出现的顺序(即插入顺序)进行计数.
For example, I need to count how many times a word appears in a list, not sorted by frequency but with the order in which the words appear, i.e. insertion order.
from collections import Counter
words = ['oranges', 'apples', 'apples', 'bananas', 'kiwis', 'kiwis', 'apples']
c = Counter(words)
print(c)
所以代替:{'apples': 3, 'kiwis': 2, 'bananas': 1, 'oranges': 1}
我宁愿得到:{'oranges': 1, 'apples': 3, 'bananas': 1, 'kiwis': 2}
我真的不需要这个Counter
方法,任何可以产生正确结果的方法对我来说都是可以的.
And I don't really need this Counter
method, any way that will produce correct result is OK for me.
推荐答案
您可以使用:
You can use the recipe that uses collections.Counter
and collections.OrderedDict
:
from collections import Counter, OrderedDict
class OrderedCounter(Counter, OrderedDict):
'Counter that remembers the order elements are first encountered'
def __repr__(self):
return '%s(%r)' % (self.__class__.__name__, OrderedDict(self))
def __reduce__(self):
return self.__class__, (OrderedDict(self),)
words = ["oranges", "apples", "apples", "bananas", "kiwis", "kiwis", "apples"]
c = OrderedCounter(words)
print(c)
# OrderedCounter(OrderedDict([('oranges', 1), ('apples', 3), ('bananas', 1), ('kiwis', 2)]))
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