在Python中如何按索引中字母的频率计数

Question

假设你给出了5个输入，如下所示：

hello
hullo
zeros

然后它被存储到一个类似so['hello','hullo','zeros']的列表中 我希望它能返回一本字典，计算他们索引中字母的频率。它应该返回这样的结果

{h:2,z:1} {e:2,u:1} {l:2,r:1} {l:2,o:1} {o:2,s:1}

Answer 1

工作zip函数：

for t in zip('123','456','789'):
  print(t)

In the above example zip takes 3 strings as an argument.

1st iteration : when the first time "for" works, zip takes all first elements from each three strings and return it as a tuple => ('1', '4', '7')

2nd iteration zip returns = > ('2', '5', '8')

3rd iteration zip returns => ('3', '6', '9')

for t in zip('123','456','789'):
  print(t)



('1', '4', '7')
('2', '5', '8')
('3', '6', '9')

转到你的问题

s=['你好'，'你好'，'零']

*s => will unpack the list into 3 items. 

zip(*s) => zip('hello','hullo','zeros')

for t in zip(*s):

('h', 'h', 'z')
('e', 'u', 'e')
('l', 'l', 'r')
('l', 'l', 'o')
('o', 'o', 's')

Answer 2

你可以试试这个：

s = ['hello','hullo','zeros']
final_data = [{d:''.join(i).count(d) for d in ''.join(i)} for i in zip(*s)]

输出：

[{'h': 2, 'z': 1}, {'u': 1, 'e': 2}, {'r': 1, 'l': 2}, {'l': 2, 'o': 1}, {'s': 1, 'o': 2}]

编辑：函数方法：

def get_count(the_index, the_list):
   listing = [i[the_index] for i in the_list]
   return {a:listing.count(a) for a in listing}