IndentationError in Python list

Question

Trying to write a python code to encrypt a string.

Encrypts the string and output is an encrypted string.

print "Enter the string "
a=raw_input()
e=''
i=0
while i<len(a):
  c=''
  c+=a[i]
  f=ord(c)
  if i%3==0:
    if f>21:
      e+=chr(f-21)
    else:
      e+=chr(f+5)
  elif i%3==1:
    if ord(c)>24:
      e+=chr(f-24)
    else:  
      e+=chr(f+2)   
  else:
    if ord(c)>21:
      e+=chr(f-20)
    else:
      e+=chr(f+6)     


  i=i+1
  del c

print e

But when on running this script, error comes.

e+=chr(f-24)
           ^
IndentationError: unindent does not match any outer indentation level

It works for me. Take what you've got here and copy paste it back into your script. — Alex Hall, May 12 '16 at 17:55
You can remove `c` completely and write directly `f=ord(a[i])` — Daniel, May 12 '16 at 18:00
Use proper variable names! It is perhaps the most important thing in writing good code. — Alex Hall, May 12 '16 at 18:06
Use an IDE (I recommend PyCharm) and it will help greatly in keeping your code properly indented, as well as many other things. — Alex Hall, May 12 '16 at 18:21

score 0 · Answer 1 · edited May 12 '16 at 18:19

This indentation error is due to the mixing of tabs and spaces in your script. The fix is to go through each line in this file, making sure to use four spaces for each level of indentation. It looks like you are only using two spaces currently, so I'm not exactly sure how you've ended up in this position, but removing all indentation and using four spaces instead of tabs should resolve your problem.

Try to make your code look more like this and see if you still have these issues:

while i<len(a):
    c=''
    c+=a[i]
    f=ord(c)
    if i%3==0:
        if f>21:

Notice how there are four spaces instead of two for each level of indentation. This means the line c='' is four spaces to the right of the while statement. Additionally, the if f>21 line is four spaces to the right of if i%3==0, and eight spaces to the right of the while statement, since it's two levels of indentation under the while statement.

Alex Hall · Answer 2 · 2016-05-12T18:29:24.703

0

I cleaned up your code:

plaintext = raw_input("Enter the string ")
encrypted = ''
for index, char in enumerate(plaintext):
    char_code = ord(char)
    index_mod = index % 3
    if index_mod == 0:
        if char_code > 21:
            offset = -21
        else:
            offset = 5
    elif index_mod == 1:
        if char_code > 24:
            offset = -24
        else:
            offset = 2
    else:
        if char_code > 21:
            offset = -20
        else:
            offset = 6
    encrypted += chr(char_code + offset)

print encrypted

For fun, it could also be done like this:

offsets = [{True: -21, False: 5}, {True: -24, False: 2}, {True: -20, False: 6}]
upper_limits = [21, 24, 21]

plaintext = raw_input("Enter the string ")
encrypted = ''
for index, char in enumerate(plaintext):
    char_code = ord(char)
    index_mod = index % 3
    offset = offsets[index_mod][char_code > upper_limits[index_mod]]
    encrypted += chr(char_code + offset)

print encrypted

You could even have

offsets = [[5, -21], [2, -24], [6, -20]]

but it's less clear what's going on there.

However now that I see the pattern in the offsets (the second always being the first minus 26) that should be done in code:

offsets = [5, 2, 6]
upper_limits = [21, 24, 21]

plaintext = raw_input("Enter the string ")
encrypted = ''
for index, char in enumerate(plaintext):
    char_code = ord(char)
    index_mod = index % 3
    offset = offsets[index_mod]
    if char_code > upper_limits[index_mod]:
        offset -= 26
    encrypted += chr(char_code + offset)

print encrypted

edited May 12 '16 at 18:29

answered May 12 '16 at 18:04

Alex Hall

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Whether or not this is helpful, it isn't an answer to the question. – Tadhg McDonald-Jensen May 12 '16 at 18:08
Error 'line 21, in encrypted += char(char_code + offset) TypeError: 'str' object is not callable' – Shubham Chahal May 12 '16 at 18:12
@ShubhamChahal sorry, made a typo (char instead of chr). Fixed. – Alex Hall May 12 '16 at 18:13
@AlexHall try to give input='a'. – Shubham Chahal May 12 '16 at 18:17
@ShubhamChahal I just did and it gave me `L`. Again, I edited the code. – Alex Hall May 12 '16 at 18:19
@AlexHall yeah it gave `L` but output should be `f`. – Shubham Chahal May 12 '16 at 18:22
Ah, I didn't know what the output was supposed to be. When I run your code in the question though it still gives me `L`. I guess you have some numbers to fix up? – Alex Hall May 12 '16 at 18:24
@ShubhamChahal the problem is probably that you expect `ord` to give the position in the alphabet, when in fact it gives the unicode codepoint (for ASCII characters such as `a`, that's just the position in the ASCII table). For example `ord('a')` is 97 and `ord('A')` is 65. You need to correct for that. – Alex Hall May 12 '16 at 18:32
@AlexHall for that `ord('a')%32` or `ord('A')%32` should be used. – Shubham Chahal May 12 '16 at 18:46
@ShubhamChahal if you say so. I can't tell what the output of the code is supposed to be, so this kind of thing is up to do. I just wanted to show you what good clean code is like. – Alex Hall May 12 '16 at 18:48
@TadhgMcDonald-Jensen I'd rather be helpful than do exactly what is expected. This won't hurt anyone. – Alex Hall May 12 '16 at 18:50

IndentationError in Python list

2 Answers2