python string index out of range

Question

Ok. So I am probably the most stupidest person on this planet, but can someone tell me why this gives me an error (The program is for searching a character in an alphabetically ordered string of characters using bisection search):

def isInIter(char, aStr):
  print('char :'+char)
  print('aStr :'+aStr)
  isin = False
  while(len(aStr) > 0):
    if aStr[len(aStr)/2] == char:
      isin = True
      break
    else:
      if aStr[len(aStr)/2] > char:
        aStr = aStr[0 : len(aStr)/2]
        continue
      if aStr[len(aStr)/2] < char:
        aStr = aStr[(len(aStr)/2 + 1) :]
        continue
return isin

Error:

isInIter('d','cddfggjkkqtwyy')
---------------------------------------------------------------------------
IndexError                                Traceback (most recent call last)
<ipython-input-66-1de0eb4793fd> in <module>()
----> 1 isInIter('d','cddfggjkkqtwyy')

/home/user/Study/Ex/lec5_BisectionSearch.py in isInIter(char, aStr)
      5   isin = False
      6   while(len(aStr) > 0):
----> 7     if aStr[len(aStr)/2] == char:
      8       isin = True
      9       break

IndexError: string index out of range

But this works as expected:

char = raw_input('char :')
aStr = raw_input('aStr :')

isin = False
while(len(aStr) > 0):
 if aStr[len(aStr)/2] == char:
  isin = True
  break
 else:
  if aStr[len(aStr)/2] > char:
    aStr = aStr[0 : len(aStr)/2]
    continue
  if aStr[len(aStr)/2] < char:
    aStr = aStr[len(aStr)/2 + 1 :]
    continue

print isin

I appreciate your help.

you can directly search without looping it. Python provides you 'in' keyword which helps you to search for a item. You can use "if char in string" in a single statement which tells you if char is present in string or not.. — user2109788
– user2109788, Commented Jul 11, 2014 at 6:16
isInIter('d','cddfggjkkqtwyy') doesn't give error for me ? and works fine ? — myildirim
– myildirim, Commented Jul 11, 2014 at 6:24
that looks like a very bad way to search a character but your code runs just fine on my machine — xor
– xor, Commented Jul 11, 2014 at 6:24
You are returning the value at the wrong place.Check-out your code again and be sure that your line indentations are correct ! — myildirim
– myildirim, Commented Jul 11, 2014 at 6:25
Thanks for the response, but this is a homework problem for a MOOC course that I'm taking. The requirement is to implement the thing using a bisection search, not any other methods. — kchak
– kchak, Commented Jul 11, 2014 at 6:32

kilojoules · Accepted Answer · 2014-07-11 07:04:37Z

1

When python goes through a while loop, it waits until the end of the loop to check the exit criteria. It might be easier if you condense your function:

def isItIn(streen,carroter):
  while len(streen)>0:
    if streen[len(streen)/2]==carroter:
       return True
    if streen[len(streen)/2]>carroter:
       streen=streen[0:len(streen)/2]
    if len(streen)==0:return False
    if streen[len(streen)/2]<carroter:
        streen=streen[len(streen)/2+1:]

print isItIn('asdilk','a')
print isItIn('kdsoe','a')

edited Jul 11, 2014 at 7:04

answered Jul 11, 2014 at 6:59

kilojoules

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Comments

user1907906 · Accepted Answer · 2014-07-11 06:17:20Z

0

Why don't you use str.index(sub\[, start\[, end\]\])? You can bet that it has an optimized implementation.

In [1]: s = "cddfggjkkqtwyy"

In [2]: s.index("d")
Out[2]: 1

In [3]: s.index("d") > -1
Out[3]: True

answered Jul 11, 2014 at 6:17

user1907906

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