blocks|key|1334000|text|是的，您不需要任何库：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1334001|list1+=+list("aaaacccc")
letters+=+[]
for+i+in+list1:
++++if+i+not+in+letters:
++++++++letters.append(i)
string+=+""
for+i+in+letters:
++++string+%2B=+str(list1.count(i))
++++string%2B=str(i)
print(string)|code-block|syntax|javascript|1334002|基本上，它循环遍历列表，找到唯一的字母，然后打印字母本身的计数。反转是相同的功能，只需打印金额即可。|1334003|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

yes, you do not need any libraries:
<pre><code>list1 = list(&quot;aaaacccc&quot;)
letters = []
for i in list1:
 if i not in letters:
 letters.append(i)
string = &quot;&quot;
for i in letters:
 string += str(list1.count(i))
 string+=str(i)
print(string)
</code></pre>
Basically, it loops through the list, finds the unique letters and then prints the count with the letter itself. Reversing would be the same function, just print the amount.

blocks|key|1333992|text|这将在一次迭代中完成工作|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1333993|保留两个temp变量，一个用于当前字符，另一个用于字符计数，另一个用于结果。|1333994|只需迭代该字符串，如果它与前一个匹配，则不断增加计数。|1333995|如果不是，则使用字符的计数和值来更新结果，并更新字符和计数。|1333996|最后，将最后一个字符和计数添加到结果中。完成了！|1333997|input_str+=+"aaaacccc"
if+input_str.isalpha():
++current_str+=+input_str[0]
++count+=+0
++final_string+=+""
++for+i+in+input_str:
++++if+i==current_str:
++++++count%2B=1
++++else:
++++++final_string%2B=str(count)%2Bcurrent_str
++++++current_str+=+i
++++++count+=+1
++final_string%2B=str(count)%2Bcurrent_str

++print+(final_string)|code-block|syntax|javascript|1333998|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|T|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|U|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|V|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

This will do the work in one iteration
Keep two temp variable one for current character, another for count of that character and one variable for the result.
Just iterate through the string and keep increasing the count if it matches with the previous one.
If it doesn't then update the result with count and value of character and update the character and count.
At last add the last character and the count to the result. Done!
<pre><code>input_str = &quot;aaaacccc&quot;
if input_str.isalpha():
 current_str = input_str[0]
 count = 0
 final_string = &quot;&quot;
 for i in input_str:
 if i==current_str:
 count+=1
 else:
 final_string+=str(count)+current_str
 current_str = i
 count = 1
 final_string+=str(count)+current_str

 print (final_string)
</code></pre>

blocks|key|1317274|text|另一个解决方案，我甚至包括了一个拼凑的反向操作，就像你在帖子中提到的那样。它们的运行速度都是O(n)，并且都很容易理解。编码基本上与Akanasha发布的相同，他只是在我编写decode()时发布他的答案快了一点。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1317275|def+encode(x):
++++if+not+x.isalpha():
++++++++raise+ValueError()
++++output+=+""
++++current_l+=+x[0]
++++counter+=+0
++++for+pos+in+x:
++++++++if+current_l+!=+pos:
++++++++++++output+%2B=+str(counter)+%2B+current_l
++++++++++++counter+=+1
++++++++++++current_l+=+pos
++++++++else:
++++++++++++counter+%2B=+1

++++return+output+%2B+str(counter)+%2B+current_l


def+decode(x):
++++output+=+""
++++i+=+0
++++while+i+<+len(x):
++++++++if+x[i].isnumeric():
++++++++++++n+=+i+%2B+1
++++++++++++while+x[n].isnumeric():
++++++++++++++++n+%2B=+1
++++++++++++output+%2B=+int(x[i:n])*x[n]
++++++++++++i+=+n
++++++++i+%2B=+1
++++return+output


test+=+"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaasasggggggbbbbdd"
test1+=+encode(test)
print(test1)

test2+=+decode(test1)
print(test2)

print(test+==+test2)|code-block|syntax|javascript|1317276|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Another solution and I included even a patchwork reverse operation like you mentioned in your post. Both run in O(n) and are fairly simple to understand. The encode is basically identical one posted by Akanasha, he was just a bit faster in posting his answer while i was writing the decode().
<pre><code>def encode(x):
 if not x.isalpha():
 raise ValueError()
 output = &quot;&quot;
 current_l = x[0]
 counter = 0
 for pos in x:
 if current_l != pos:
 output += str(counter) + current_l
 counter = 1
 current_l = pos
 else:
 counter += 1

 return output + str(counter) + current_l


def decode(x):
 output = &quot;&quot;
 i = 0
 while i &lt; len(x):
 if x[i].isnumeric():
 n = i + 1
 while x[n].isnumeric():
 n += 1
 output += int(x[i:n])*x[n]
 i = n
 i += 1
 return output


test = &quot;aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaasasggggggbbbbdd&quot;
test1 = encode(test)
print(test1)

test2 = decode(test1)
print(test2)

print(test == test2)
</code></pre>

For example: string = aaaacccc, then I need the output to be 4a4c. Is there a way to do this without using any advanced methods, such as libraries or functions?
Also, if someone knows how to do the reverse: turning &quot;4a4c: into aaaacccc, that would be great to know.

How to group consecutive letters in a string in Python?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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例如: string = aaaacccc，那么我需要输出为4a4c。有没有办法做到这一点，而不使用任何高级方法，如库或函数？另外，如果有人知道如何反其道而行之:将"4a4c:变成aaaacccc“，那就太好了。

问如何在Python中对字符串中的连续字母进行分组？
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问如何在Python中对字符串中的连续字母进行分组？EN