Changing parts of a string with '#'

Question

Wanted to see if I'm going in the right direction. I have to change everything but the last four characters of a string into #. I've got two ideas so far.

First one:

def maskify(cc):
   cc = raw_input("Enter passcode: ")
   n = len(cc)
   cc.replace(cc[0:n-4], #)  # this one gives me the unexpected EOF while parsing error

Second one (I think this one's closer because it supposedly needs an algorithm):

def maskify(cc):
   cc = raw_input("Enter passcode: ")  
   n = len(cc)
   for i in range (0, n-4):  # i think a for loop would be good but i don't know how i'm going to use it yet
      cc.replace(                  #not entirely sure what to put here
   pass                           

Answer 1

# return masked string
def maskify(cc):

    maskify = cc
    maskifyL= len(maskify) - 4

    char = ""
    a=0

    if maskifyL <= 2:
       c2 = cc
    else:
        for a in range(maskifyL):
            char += "#"
            a += 1
        c2 = maskify.replace(maskify[:maskifyL], char, maskifyL)

    return c2

Answer 2

cc = raw_input("Enter passcode: ")
cc = ''.join(('#' * (len(cc) - 4), cc[-4:]))

Answer 3

First of all strings are immutable(they can't be changed) once created--so you can't change the value of cc with replace(). To change all parts of the string except the last four, do this:

def doso(text):
  assert len(str(text))>4, 'Length of text should be >4'
  ctext=str(text).replace(str(text)[:(len(str(text))-4)],'#'*(len(str(text))-4)) 
  print(ctext)

>>>doso(123) prints 'Length of text should be >4' because len(text)== 3 which is what we expect

Hovever,

 >>>doso(12345678) prints #####6789 which is exactly what we expect

Note: Doing '#' * (len(str(text))-4)) accounts for the number of characters we want to replace

Answer 4

How about the following?

def maskify() :
    cc = input("Enter passcode: ")
    mask = '#'*(len(cc)-4)
    return mask + cc[-4:]

I'm not sure how the flow of the rest of your program works, but I doubt whether you should be prompting for raw_input inside of this function. You can decide that depending on your needs. The alternative would look something like this:

def maskify(cc) :
    return '#'*(len(cc)-4) + cc[-4:]

myInput = input("Enter passcode: ")
maskedInput = maskify( myInput )

• NB: python2 uses raw_input instead of input

Answer 5

The following solution makes the assumption that this would have a security type use, as such passcodes of 4 or fewer characters should just be hashed, otherwise someone would know the whole passcode.

def maskify(cc):
    if len(cc) < 9:
        split = [0,1,2,3,4,4,4,4,4][len(cc)]
    else:
        split = len(cc) - 4

    return "#" * split + cc[split:]

for length in range(1,12):
    test = string.lowercase[:length]
    print "%s > %s" % (test, maskify(test))

Giving the following results:

a > #
ab > ##
abc > ###
abcd > ####
abcde > ####e
abcdef > ####ef
abcdefg > ####efg
abcdefgh > ####efgh
abcdefghi > #####fghi
abcdefghij > ######ghij
abcdefghijk > #######hijk

If the short hash is not required, then simply change the array as follows to get the other results:

def maskify(cc):
    if len(cc) < 9:
        split = [0,0,0,0,0,1,2,3,4][len(cc)]
    else:
        split = len(cc) - 4

    return "#" * split + cc[split:]

Giving:

a > a
ab > ab
abc > abc
abcd > abcd
abcde > #bcde
abcdef > ##cdef
abcdefg > ###defg
abcdefgh > ####efgh
abcdefghi > #####fghi
abcdefghij > ######ghij
abcdefghijk > #######hijk

Answer 6

As others have mentioned # starts a comment. If you want a string containing a hash you need to do '#'.

As André Laszlo mentioned, Python strings are immutable, so it's impossible for a string operation to change a string's content. Thus the str.replace() method can't change the original string: it needs to create a new string which is a modified version of the original string.

So if you do

cc = 'cat'
cc.replace('c', 'b')

then Python would create a new string containing 'bat' which would get thrown away because you're not saving it anywhere.

Instead, you need to do something like

cc = 'cat'
cc = cc.replace('c', 'b')

This discards the original string object 'cat' that was bound to the name cc and binds the new string 'bat' to it.

---

The best approach (AFAIK) to solving your problem is given in bebop's answer. Here's a slightly modified version of bebop's code, showing that it handles short strings (including the empty string) correctly.

def maskify(s) :
    return '#' * (len(s) - 4) + s[-4:]

alpha = 'abcdefghij'
data = [alpha[:i] for i in range(len(alpha)+1)]

for s in data:
    print((s, maskify(s)))

output

('', '')
('a', 'a')
('ab', 'ab')
('abc', 'abc')
('abcd', 'abcd')
('abcde', '#bcde')
('abcdef', '##cdef')
('abcdefg', '###defg')
('abcdefgh', '####efgh')
('abcdefghi', '#####fghi')
('abcdefghij', '######ghij')

Answer 7

The string literal '#' is not the same as the character # which starts an inline comment.

def maskify(cc):
   cc = raw_input("Enter passcode: ")  
   mask = '#'*(len(cc)-4)
   return mask + cc[-4:]

Answer 8

Just a little change to your own code:

cc = raw_input("Enter passcode: ")  
n = len(cc)
c=""
for i in range(0, n-4):  # i think a for loop would be good but i don't know how i'm going to use it yet
    c+="#"               #not entirely sure what to put here
cc= c+cc [-4:]
print cc

output:

Enter passcode: kased
#ased

Answer 9

The problem in the first example is that the # is unquoted. You need to change it to '#' otherwise it is parsed as the start of a comment and the enclosing parenthesis is a part of that comment. Although, this will only fix the parsing error.

The problem with strings is that you can't change characters inside of them (they are immutable). A common way to get around this is to create an array of the string, change the characters you want to change and then convert the array back to a string (often using ''.join(character_array)). Try that!