blocks|key|935480|text|def+my_find(haystack,+needle):
++++n_len+=+len(needle)
++++start+=+0
++++while+start+<=+(len(haystack)-n_len%2B1):
++++++++if+haystack[start:start%2Bn_len-1]+==+needle:
++++++++++++return+True
++++++++start+%2B=+1|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|935481|据我所知，这是你的算法。未测试，将测试并让您知道它是否工作。|unstyled|935482|经过测试，似乎可以正常工作。|935483|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>def my_find(haystack, needle):
 n_len = len(needle)
 start = 0
 while start &lt;= (len(haystack)-n_len+1):
 if haystack[start:start+n_len-1] == needle:
 return True
 start += 1
</code></pre>

This is, as far as I can understand, your algorithm. Not tested, will test and let you know if it works.

Tested and seems to work.

blocks|key|2837070|text|听起来您正在学习正则表达式，这里有一段代码片段可以帮助您入门。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2837071|myFileName+=+"abbababaaa"
patternToMatch+=+"ababa"

i+=+0
j+=+0
while+(i+<+len(myFileName)):
++++if+(patternToMatch[i:i]+==+myFileName[j:j]):
++++++++i%2B%2B
++++++++j%2B%2B
++++else:
++++++++i+=+0++++++++

if+len(patternToMatch)+==+i:
++++#+matched+a+pattern|code-block|syntax|javascript|2837072|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|$]]

Sounds like you are learning about regular expressions, here's a snippet that may help you get started.

<pre><code>myFileName = "abbababaaa"
patternToMatch = "ababa"

i = 0
j = 0
while (i &lt; len(myFileName)):
 if (patternToMatch[i:i] == myFileName[j:j]):
 i++
 j++
 else:
 i = 0 

if len(patternToMatch) == i:
 # matched a pattern
</code></pre>

blocks|key|2837102|text|我猜这就是你需要的，下面的代码进行逐个字符的比较。您还可以将对find的调用替换为text上的迭代，其中包括检查text的第一个字符是否与pattern的第一个字符匹配|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2837103|def+my_find(text,+pattern):
++++'''Find+the+start+index+of+a+pattern+string+in+a+text.
++++Return+-1+if+not+found,+and+assume+that+pattern+is+not+empty'''

++++found+=+False
++++current_start_index+=+text.find(pattern[0])
++++index_text+=+current_start_index
++++index_pattern+=+0

++++while+not+found+and+index_text+%2B+len(pattern)+-+1+<+len(text)+and+\
++++++++++++current_start_index+!=+-1:

++++++++index_text+%2B=+1
++++++++index_pattern+%2B=+1

++++++++while+index_text+<+len(text)+and+\
++++++++++++++++index_pattern+<+len(pattern)+and+\
++++++++++++++++text[index_text]+==+pattern[index_pattern]:

++++++++++++if+index_pattern+==+len(pattern)+-+1:
++++++++++++++++found+=+True
++++++++++++++++break
++++++++++++else:
++++++++++++++++index_text+%2B=+1
++++++++++++++++index_pattern+%2B=+1

++++++++if+not+found:
++++++++++++current_start_index+=+text.find(pattern[0],current_start_index+%2B+1)
++++++++++++index_text+=+current_start_index

++++if+found:
++++++++return+current_start_index
++++else:
++++++++-1|code-block|syntax|javascript|2837104|entityMap^0|V|4|15|4|1K|4|1X|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

I guess here's what you need, the following code does character-by-character comparison. You may also replace the calls to <code>find</code> by iterations over <code>text</code> which includes checks whether the first character of <code>text</code> matches the first character of <code>pattern</code>:

<pre><code>def my_find(text, pattern):
 '''Find the start index of a pattern string in a text.
 Return -1 if not found, and assume that pattern is not empty'''

 found = False
 current_start_index = text.find(pattern[0])
 index_text = current_start_index
 index_pattern = 0

 while not found and index_text + len(pattern) - 1 &lt; len(text) and \
 current_start_index != -1:

 index_text += 1
 index_pattern += 1

 while index_text &lt; len(text) and \
 index_pattern &lt; len(pattern) and \
 text[index_text] == pattern[index_pattern]:

 if index_pattern == len(pattern) - 1:
 found = True
 break
 else:
 index_text += 1
 index_pattern += 1

 if not found:
 current_start_index = text.find(pattern[0],current_start_index + 1)
 index_text = current_start_index

 if found:
 return current_start_index
 else:
 -1
</code></pre>

I'm currently stuck on trying to make a naive algorithm which given a piece of a pattern e.g aabba 
search for it in a text e.g abbbbaababaabbaaabbaa one letter at a time. It will compare a with the text if that is right then compares the next letter and if that's wrong the whole pattern will shift one and compare a with b etc 

we were give code example

<pre><code>print "Input text: ",
text = raw_input()
print "Input pattern: ",
pattern = raw_input()

index = text.find(pattern)
while index &gt; -1:
 print index
 index = text.find(pattern, index+1)
</code></pre>

but the find() function in python is too fast(I need a non optimized sort of algorithm, using while
and for loops statements I guess).

Any help appreciated,
Thanks

Pattern Matching 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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我目前一直在尝试制作一个朴素的算法，它给出一个模式的一部分，例如在文本中搜索它，例如abbbbaababaabbaaabbaa，一次一个字母。它将a与文本进行比较，如果正确，则比较下一个字母，如果错误，则整个模式将移位一，并将a与b进行比较，依此类推我们给出了代码示例print "Input text: ",text ...

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