如何修復'String index out of range'錯誤

Question

我正在嘗試編寫代碼，用符號和重復次數替換字符串中的重復符號（例如：“ aaaaggggtt”->“ a4g4t2”）。 但是我得到的字符串索引超出范圍錯誤（（

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，如果seq [i] == seq [j]：IndexError：字符串索引超出范圍

UPD：將len（seq）更改為len（seq）-1之后，不再出現字符串索引錯誤，但是代碼仍然無法正常工作。 輸入：aaaaggggtt
輸出：aaaaggggtt（相同）

Answer 1

您可以遍歷字符串，保持運行中的計數器並在運行時創建字符串

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來解決這個問題

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)

輸出將是

a4g4t2

Answer 2

簡單方法：

str1 = 'aaaaggggtt'

set1 = set(str1)

res = ''

for i in set1:

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

print(res)