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
Python 2.7+引入了字典理解。从列表中构建字典将获得计数并去除重复项。
>>> a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
>>> d = {x:a.count(x) for x in a}
>>> d
{1: 4, 2: 4, 3: 2, 4: 1, 5: 2}
>>> a, b = d.keys(), d.values()
>>> a
[1, 2, 3, 4, 5]
>>> b
[4, 4, 2, 1, 2]
其他回答
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]
简单的解决方法就是用字典。
def frequency(l):
d = {}
for i in l:
if i in d.keys():
d[i] += 1
else:
d[i] = 1
for k, v in d.iteritems():
if v ==max (d.values()):
return k,d.keys()
print(frequency([10,10,10,10,20,20,20,20,40,40,50,50,30]))
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))
def frequencyDistribution(data):
return {i: data.count(i) for i in data}
print frequencyDistribution([1,2,3,4])
...
{1: 1, 2: 1, 3: 1, 4: 1} # originalNumber: count
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}