Very weird bug in Python 3.3

Question

Okay, so I made this little function that allows me to make a string into multiplier of 32 characters, but when I use String .replace I get some really, really weird bug. Since its making me pull my hair, can you guys take a look and see what I'm missing.

Variables:
        self.blockSize = 32
        self.interrupt = '$^EnD#Block^$'
        self.filler = '#'

Functions:
    def pad(self, data):
        joint1 = ''.join([data, self.interrupt])
        joint2 = self.filler * ((self.blockSize - len(joint1)) % self.blockSize) 
        return ''.join([joint1, joint2])

    def unpad(self, data):
        data = str(data).rstrip(self.interrupt)
        return data.replace(self.filler, '')

Call:
p = e.pad('this is not a very good idea  yo')
print(p)
print(e.unpad(p))


Output:
    Jans-MacBook-Pro:test2 jan$ ../../bin/python3 data.py 
    this is not a very good idea  yo123$^EnD#Block^$################
    this is not a very good idea  yo123
    Jans-MacBook-Pro:test2 jan$ ../../bin/python3 data.py 
    this is not a very good idea  yo$^EnD#Block^$###################
    this is not a very good idea  y
    Jans-MacBook-Pro:test2 jan$

It makes o in yo disappear. Ahhhh! But nothing disappears if I add some random numbers after.

SOLUTION - EDIT: My bad. I have misplaced self.filler and self.interrupt. I am so embarrassed now. The code should have been:

def unpad(self, data):
    data = str(data).rstrip(self.filler)
    return data.replace(self.interrupt, '')

As an aside, s1 + s2 is a shorter equivalent of ''.join([s1, s2]). — Lev Levitsky
– Lev Levitsky, Commented Jan 6, 2013 at 23:01
You should make the title of this question a bit more descriptive in case it is searched for later. — Fredrick Brennan
– Fredrick Brennan, Commented Jan 6, 2013 at 23:13
In [1]: %timeit "test" + "test" 100000000 loops, best of 3: 15.2 ns per loop In [2]: %timeit "".join(("test", "test")) 10000000 loops, best of 3: 98.9 ns per loop — Jakob Bowyer
– Jakob Bowyer, Commented Jan 6, 2013 at 23:26

BrenBarn · Accepted Answer · 2013-01-06 23:01:51Z

5

Read the documentation for rstrip:

The chars argument is a string specifying the set of characters to be removed.

rstrip removes all trailing characters present in the passed set of characters. It doesn't remove a trailing substring consisting of those characters in that order. 'abczyx'.rstrip('xyz') gives 'abc', and 'abczyx'.rstrip('zyx') also gives 'abc'.

answered Jan 6, 2013 at 23:01

BrenBarn

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3 Comments

if __name__ is None Over a year ago

Okay. But why does .replace and .rstrip remove o, where o is nowhere defined to be removed in first place?

BrenBarn Over a year ago

When your string ends with "yo", it ends with "o". "o" is present in your "interrupt" string, which means it is one of the characters to be stripped. So it is stripped.

if __name__ is None Over a year ago

Oh my bad, I figured out why the problem is there :) I misplaced self.filler and self.interrupt respectively. How stupid is that?

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