给定一个无序的值列表,比如

a = [5, 1, 2, 2, 4, 3, 1, 2, 3, 1, 1, 5, 2]

我怎样才能得到出现在列表中的每个值的频率,就像这样?

# `a` has 4 instances of `1`, 4 of `2`, 2 of `3`, 1 of `4,` 2 of `5`
b = [4, 4, 2, 1, 2] # expected output

当前回答

seta = set(a)
b = [a.count(el) for el in seta]
a = list(seta) #Only if you really want it.

其他回答

我将简单地以以下方式使用scipy.stats.itemfreq:

from scipy.stats import itemfreq

a = [1,1,1,1,2,2,2,2,3,3,4,5,5]

freq = itemfreq(a)

a = freq[:,0]
b = freq[:,1]

您可以在这里查看文档:http://docs.scipy.org/doc/scipy-0.16.0/reference/generated/scipy.stats.itemfreq.html

seta = set(a)
b = [a.count(el) for el in seta]
a = list(seta) #Only if you really want it.

你可以这样做:

import numpy as np
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
np.unique(a, return_counts=True)

输出:

(array([1, 2, 3, 4, 5]), array([4, 4, 2, 1, 2], dtype=int64))

第一个数组是值,第二个数组是具有这些值的元素的数量。

所以如果你想要得到一个数字数组,你应该使用这个:

np.unique(a, return_counts=True)[1]

计算元素的频率可能最好使用字典:

b = {}
for item in a:
    b[item] = b.get(item, 0) + 1

要删除重复项,请使用set:

a = list(set(a))
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
counts = dict.fromkeys(a, 0)
for el in a: counts[el] += 1
print(counts)
# {1: 4, 2: 4, 3: 2, 4: 1, 5: 2}