Question

我正在尝试编写代码，用符号和重复次数替换字符串中的重复符号（例如：“ aaaaggggtt”->“ a4g4t2”）。但我的字符串索引超出范围错误（（

seq = input()
i = 0
j = 1
v = 1
while j<=len(seq)-1:
  if seq[i] == seq[j]:
    v += 1
    i += 1
    j += 1
  elif seq[i] != seq[j]:
    seq.replace(seq[i-v:j], seq[i] + str(v))
    v = 1
    i += 1
    j += 1
print(seq)

第6行，在如果seq [i] == seq [j]： IndexError：字符串索引超出范围

UPD：将len（seq）更改为len（seq）-1后，不再有字符串索引错误，但是代码仍然无法正常工作。输入：aaaaggggtt
输出：aaaaggggtt（相同）

Answer 1

您可以遍历字符串，保持计数器运行并随时创建字符串

s = 'aaaaggggtt'

res = ''
counter = 1

#Iterate over the string
for idx in range(len(s)-1):
    #If the character changes
    if s[idx] != s[idx+1]:
        #Append last character and counter, and reset it
        res += s[idx]+str(counter)
        counter = 1
    else:
        #Else increment the counter
        counter+=1

#Append the last character and it's counter
res += s[-1]+str(counter)
print(res)

或者您可以使用itertools.groupby

from itertools import groupby

s = 'aaaaggggtt'

#Count numbers and associated length in a list
res = ['{}{}'.format(model, len(list(group))) for model, group in groupby(s)]

#Convert list to string
res = ''.join(res)

print(res)

输出将为

a4g4t2

Answer 2

简单方法：

str1 = 'aaaaggggtt'

set1 = set(str1)

res = ''

for i in set1:

    res+=i+str(str1.count(i))

print(res)

如何修复'String index out of range'错误

2 个答案: