python - 计算每个数字的出现次数

Question

我有一长串用逗号分隔的数字.我可以搜索和计算大多数数字的出现次数,或者更准确地说是2位数字.

如果我有一个数字序列如下: 1,2,3,4,5,1,6,7,1,8,9,10,11,12,1,1,2 我想计算这个数字1出现的次数我真的应该得到5.

但是,因为它正在计算1中10,11而且12,我正在得到9.

有谁知道如何使下面的代码只匹配整个"字符串"？

def mostfreq(numString):
    import json 
    maxNum=45
    count=1
    list={}
    while count <= maxNum:
        list[count] = 0
        count+=1
    #numString is the array with all the numbers in it
    count=1
    topTen = ""
    while count <= maxNum:
        list[count]=numString.count(str(count))
        topTen = topTen+json.dumps(
        {count: list[count]},
        sort_keys=True,
        indent=4)+","
        count+=1
    response_generator = ( "["+topTen[:-1]+"]" )
    return HttpResponse(response_generator)

Answer 1

在2.7+上,只需split使用collections.Counter:

from collections import Counter
numstring = "1,2,3,4,5,1,6,7,1,8,9,10,11,12,1,1,2"
numcount = Counter(numstring.split(','))

或者,2.7之前:

from collections import defaultdict
numstring = "1,2,3,4,5,1,6,7,1,8,9,10,11,12,1,1,2"
numcount = defaultdict(int)
for num in numstring.split(','):
    numcount[num] += 1

如果你想使用count:

numstring = "1,2,3,4,5,1,6,7,1,8,9,10,11,12,1,1,2"
numlist = numstring.split(',')
numcount = dict((num, numlist.count(num)) for num in set(numlist))

但它是O(m*n)而不是O(n),因为它为每个唯一的数字迭代一次数字列表.