CODEWITHSUNDEEP
pythonstringreplace
Terencio
Terencio

Reputation: 11

How do I use the .replace() function to change all but the last 4 characters of a string into the # symbol?

I'm trying to change all but the last four characters in a string to the # symbol on output.
Here's what I have currently

>>> v
'0123456789'
>>> for ch in v:
...     if ch in v:
...             v = v.replace(ch, '#')
...
>>> print v
##########

As you can see, ALL characters are replaced.

I've tried using all types of index slicing methods but nothing seems to work! I've also tried...

>>> v = '0123456789'
>>> for ch in v:
...     if ch in v:
...             v = v[0:-4].replace(ch, '#') + v[-3:]
...
>>> print v
789

But, as you can see here, it gets rid of all the other characters.
Please help.

Upvotes: 1

Views: 2604

Answers (5)

nigel222
nigel222

Reputation: 8202

That's not really what replace is for: a sensible usage is:

>>>  'abcbabcba'.replace('ba','zy')
'abczybczy'

What you want is best addressed by string indexing with negative subscripting and string duplication

assert len(v) >= 4
result = ( '#' * (len(v)-4) ) + v[-4:]

Note that assertion. You'll need to code specially for that case if v can ever be shorter than 4 characters.

Upvotes: 1

Lolgast
Lolgast

Reputation: 339

While the solution by @cricket_007 works just fine if you want to replace all characters, if you want to filter some you can use the approach you used. However, the last part in your new v assignment should be v[-4:] rather than v[-3:]

Upvotes: 0

SpoonMeiser
SpoonMeiser

Reputation: 20417

The .replace() method is not appropriate for what you're trying to do. It replaces all instances of a character with another:

>>> '123412341234'.replace('1', '#')
'#234#234#234'

So looping through calling replace for each character you find, is pointless.

Instead, just construct a new string from the right number of #s and the last 4 digits of the original.

'%s%s' % (
    '#' * (len(v) - 4),
    v[-4:]
)

Upvotes: 0

RichSmith
RichSmith

Reputation: 940

s = "0123456789"

result = "".join(['#' for x in s[:-4]]) + s[-4:]

Upvotes: 4

OneCricketeer
OneCricketeer

Reputation: 191728

Python strings are immutable, so something like this should work. No loops 

>>> v = '0123456789'
>>> '#'*(len(v) - 4)+v[-4:]
'######6789'

The thing about replace is that what if you had 333333 and you replace('3', '#')? Everything becomes a #, right?

Upvotes: 5

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