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
您可以使用python中提供的内置函数
l.count(l[i])
d=[]
for i in range(len(l)):
if l[i] not in d:
d.append(l[i])
print(l.count(l[i])
上面的代码自动删除列表中的重复项,并打印原始列表和没有重复项的列表中每个元素的频率。
一枪双鸟!X维
其他回答
from collections import Counter
a=["E","D","C","G","B","A","B","F","D","D","C","A","G","A","C","B","F","C","B"]
counter=Counter(a)
kk=[list(counter.keys()),list(counter.values())]
pd.DataFrame(np.array(kk).T, columns=['Letter','Count'])
通过遍历列表并计算它们,手动计算出现的数量,使用collections.defaultdict跟踪到目前为止看到的内容:
from collections import defaultdict
appearances = defaultdict(int)
for curr in a:
appearances[curr] += 1
在Python 2.7+中,可以使用集合。计数的计数器
>>> a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
>>>
>>> from collections import Counter
>>> c=Counter(a)
>>>
>>> c.values()
[4, 4, 2, 1, 2]
>>>
>>> c.keys()
[1, 2, 3, 4, 5]
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
seta = set(a)
b = [a.count(el) for el in seta]
a = list(seta) #Only if you really want it.