列表中一个列表的计数器对象的总和

Question

我试图找到多个列表中列表中单词的出现次数之和。list中的list对象非常大所以我只使用了一个虚拟实例

multiple=[['apple','ball','cat']['apple','ball']['apple','cat'].......]
words=['apple','ball','cat','duck'......]
word = 'apple' 
cnt = Counter()
total = 0
for i in multiple:
        for j in i:
            if word in j:
                cnt[word] +=1
                total += cnt[word]

我想要这样的输出:

{'apple':3,'ball':2,'cat':2}

Answer 1

您可以直接为Counter提供生成器表达式:

cnt = Counter(word for sublist in multiple for word in sublist)
cnt
Out[40]: Counter({'apple': 3, 'ball': 2, 'cat': 2})
sum(cnt.values())
Out[41]: 7

我真的不明白你的words列表的意义。你没有使用它。

如果需要过滤掉不在words中的单词，则使words为set， 不为list。

words = {'apple','ball','cat','duck'}
cnt = Counter(word for sublist in multiple for word in sublist if word in words)

否则你会得到O(n**2)个行为，而本应是O(n)个操作

Answer 2

这适用于Python 2.7和Python 3.x:

from collections import Counter
multiple=[['apple','ball','cat'],['apple','ball'],['apple','cat']]
words=['apple','ball','cat','duck']
cnt = Counter()
total = 0
for i in multiple:
        for word in i:
            if word in words:
                cnt[word] +=1
                total += 1
print cnt  #: Counter({'apple': 3, 'ball': 2, 'cat': 2})
print dict(cnt)  #: {'apple': 3, 'ball': 2, 'cat': 2}
print total  #: 7
print sum(cnt.values())  #: 7

在Python 2中。你应该使用.itervalues()而不是.values()，即使两者都可以工作。

根据roippi的回答给出一个更短的解决方案:

from collections import Counter
multiple=[['apple','ball','cat'],['apple','ball'],['apple','cat']]
cnt = Counter(word for sublist in multiple for word in sublist)
print cnt  #: Counter({'apple': 3, 'ball': 2, 'cat': 2})