如何在一个列表中获得数字频率，然后索引到另一个列表中？

Question

Python 2.7.5（不要评判我）

我所拥有的：

numbers = [1,3,5,1,3,5,7,0,2,2,9,1]

frequencies = [0,0,0,0,0,0,0,0,0,0]

我需要一个for循环，它通过遍历数字的元素来计数，并通过索引相应地修改频率。频率应如下所示：

[1,3,2,2,0,2,0,1,0,1]

我试过使用len（）函数…出于某种原因…我只是没有找到一种方法来正确地合并count。你知道吗

Answer 1

for n in numbers:
    frequencies[n] += 1

Answer 2

假设数字的每一个元素都在0和频率长度之间，包括-排除：

for i in range(len(frequencies)):
    frequencies[i] = numbers.count(i)

Answer 3

这个柜台很好用。你知道吗

from collections import Counter

numbers = [1,3,5,1,3,5,7,0,2,2,9,1]
freq = Counter(numbers)
# Counter({1: 3, 2: 2, 3: 2, 5: 2, 0: 1, 7: 1, 9: 1})

# dictionary
d = dict(freq)
#{0: 1, 1: 3, 2: 2, 3: 2, 5: 2, 7: 1, 9: 1}

# tuples
t = freq.items()
# [(0, 1), (1, 3), (2, 2), (3, 2), (5, 2), (7, 1), (9, 1)]

# list
L = [freq[n] for n in xrange(max(freq.keys()) + 1)]
# [1, 3, 2, 2, 0, 2, 0, 1, 0, 1]

# add more
freq.update(numbers)
# Counter({1: 6, 2: 4, 3: 4, 5: 4, 0: 2, 7: 2, 9: 2})  

collections.Counter