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
a=[1,2,3,4,5,1,2,3]
b=[0,0,0,0,0,0,0]
for i in range(0,len(a)):
b[a[i]]+=1
其他回答
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]
num=[3,2,3,5,5,3,7,6,4,6,7,2]
print ('\nelements are:\t',num)
count_dict={}
for elements in num:
count_dict[elements]=num.count(elements)
print ('\nfrequency:\t',count_dict)
我迟到了,但这也有用,也会帮助到其他人:
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
freq_list = []
a_l = list(set(a))
for x in a_l:
freq_list.append(a.count(x))
print 'Freq',freq_list
print 'number',a_l
会产生这个…
Freq [4, 4, 2, 1, 2]
number[1, 2, 3, 4, 5]
from collections import OrderedDict
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
def get_count(lists):
dictionary = OrderedDict()
for val in lists:
dictionary.setdefault(val,[]).append(1)
return [sum(val) for val in dictionary.values()]
print(get_count(a))
>>>[4, 4, 2, 1, 2]
删除副本并维持秩序:
list(dict.fromkeys(get_count(a)))
>>>[4, 2, 1]
seta = set(a)
b = [a.count(el) for el in seta]
a = list(seta) #Only if you really want it.