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
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}