You are given an input string.

For each symbol in the string if it's the first character occurence, replace it with a '1', else replace it with the amount of times you've already seen it…

But will your code be performant enough?

---

Examples:

input   =  "Hello, World!"
result  =  "1112111121311"

input   =  "aaaaaaaaaaaa"
result  =  "123456789101112"

题目大意：将字符出现的次数打印出来，以数字1,2,3，，，表示

def numericals(s):
# code
dic = {}
L = []
for i in s:
if i not in dic.keys():
dic[i] = 1
L.append(dic[i])
else:
dic[i] += 1
L.append(dic[i])
return ''.join([str(x) for x in L])

用时：1小时3分51秒，，，真长。。。

看下别人的解法：

1、使用defaultdict

from collections import defaultdict

def numericals(s):
d, lst = defaultdict(int), []
for c in s:
d[c] += 1
lst.append(d[c])
return ''.join(map(str, lst))

2、使用生成器

from collections import Counter

def numericals(s):
def f(s):
cnts = Counter()
for c in s:
cnts[c] += 1
yield cnts[c]
return ''.join(map(str, f(s)))

1、使用计数count

2、字典可以使用特殊字符作为key

3、将一个list转化为str：

　　3.1、如果list中有数值，需要转为str，可以使用遍历：

str(x) for x in L

　　3.2、可以使用map

map(str, L)

''.join(str(x) for x in L)

''.join(map(str, L))