このプログラムが存在しないリスト要素を呼び出さないようにするにはどうすればよいですか？

Question

import string 

## This part of the code initializes the program by recieving inputs and saving 
## them as strings. 
## It also currently works! 

intext = str.lower(raw_input("Please input the text you are going to use. ")) 
##print intext 
instring = str.lower(raw_input("Please input your string. ")) 
##print instring 
inlengthminus = int(raw_input("Please input the number of words you would like before the matched word. ONLY PUT AN INTEGER HERE!")) 
inlengthplus = int(raw_input("Please input the number of words you would like after the matched word. ONLY PUT AN INTEGER HERE!")) 
## This part of the code splits the text into searchable lists. 
## It only works when the cases match, which is why the search lists 
## convert the text to lowercase. 

searchtext=str.split(intext) 
##print searchtext 
searchstring=list(instring+1) 
##print searchstring 

## This is the actual search process. 
## It works, mostly. 

length = len(searchtext) 
stringlength = len(searchstring) 
##print stringlength 

x=0 
for a in range(0,length): 
    print a 
    print searchstring[x] 
    print x 
    if searchstring[x] in searchtext[a]: 
     if (a-inlengthminus)<0: 
      print "You almost caused an error, please pick a number of words before your matched word that won't call text that doesn't exist." 
      break 
     else: 
      print searchtext[a-inlengthminus:a+inlengthplus+1]  
      x+=1 
    else: 
     pass 

私はこのプログラムがsearchstring [x]の値をsearchstringの長さより長く呼び出すのを止める方法を知りません。 xがこのオーバーフローエラーを起こさないようにする方法はありますか？

Answer 1

は、モジュラス演算子が除算を行い、剰余を返し

x = (x + 1) % len(searchstring) 

で

x+=1 

を交換します。 xが0からsearchstringの長さになるように、モジュラスは何もしません（それ自身を返します）。 x == searchstringの長さの場合、分割時に余りがないため、0にリセットされます。