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Python 2.7.5(不要评判我)

我有什么(例如):

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,通过遍历数字元素并通过索引相应地修改频率来计数。频率应如下所示:

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

我尝试使用 len() 函数...出于某种原因...我只是还没有找到正确合并计数的方法。

4

3 回答 3

6
for n in numbers:
    frequencies[n] += 1
于 2013-10-19T05:02:25.680 回答
2

假设数字的每个元素都在 0 和频率长度之间,包括 - 不包括:

for i in range(len(frequencies)):
    frequencies[i] = numbers.count(i)
于 2013-10-19T05:03:03.280 回答
0

计数器非常适合这个。

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})  

集合.计数器

于 2013-10-19T06:36:13.820 回答