Replace a specific character at a specific position without using loops - Python

I guess, if you strictly want to avoid for loops, the easiest way is converting to a list, replacing items, and then converting back.

This would be:

s = "0123456789012345689"
lst = list(s)   # result: ['0', '1', '2', ..., '9']
lst[0:-1:5] = ["x"]*len(lst[0:-1:5])

Result then is:

In [43]: lst                                                                                                                                                                                                         
Out[43]: 
['x',
 '1',
 '2',
 '3',
 '4',
 'x',
 '6',
 '7',
 '8',
 '9',
 'x',
 '1',
 '2',
 '3',
 '4',
 'x',
 '6',
 '8',
 '9']

For getting it back to string you would simply use a join:

In [44]: "".join(lst)                                                                                                                                                                                                
Out[44]: 'x1234x6789x1234x689

The part lst[0:-1:5] select every 5th element in the list, beginning with the very first entry, denonted by 0:. If you want it to start at the 5th element, then simply do lst[5:-1:5]. The -1 part means "until the end", the last :5 stands for "every fifth".

Assigning values with ["x"]*len(lst[0:-1:5]) is needed, since "x" here would try to assign a single value to a slice of the original list with the length len(lst[0:-1:5]), because this is exactly how many items we selected in the first place.

EDIT: Giving it a second look the expected outcome is actually to change every 6th character, not every 5th (while preserving 5 characters in between the changed ones from the original string).

One would then, of course, need to adjust the slicing to select every 6th character:

lst[5:-1:6] = ["x"]*len(lst[5:-1:6])
         ^                       ^

Result:

In [12]: "".join(lst)                                                                                                                                                                                                
Out[12]: '01234x67890x23456x9'

Every sixth character now is being replaced, while preserving the 5 ones in between from the original string.

My solution to this would be this:

def replace_position(message, replacement, n):
    lis = list(message)  # get a list of all characters.
    lis.pop(n)  # delete the chosen character from the list.
    lis.insert(n, replacement)  # replace it at the given position.

    return ''.join(lis)

Example use:

>>> a = 'Hello world, the 6th character is an x!'
>>> replace_postion(a, 'x', 6)
'Hello xorld, the 6th character is an x!'

However, this only gets the first one in that position. In order to do that, this is what I would do:

def replace_position(message, replacement, n):
    lis = list(message)
    for i in range(n, len(message), n + 1):
        lis.pop(i)
        lis.insert(i, replacement)

    return ''.join(lis) 

The only difference is that we are now iterating over the entire message, and replacing each of them as we go. Similar use:

>>> a = 'Hello world, every 6th character is an x!'
>>> replace_position(a, 'x', 6)
'Hello xorld, xvery 6xh charxcter ix an x!'

Hope this helps.

Try

s = "01234567890123456789012345"

pieces = []
for i in range((len(s)-1)//5):
    pieces.append(s[i*6:i*6+5])

result = 'x'.join(pieces)

assert(result ==  '01234x67890x23456x89012x45')

We iterator over every window of six characters in the string, and then collect the first five characters in each window. We then join all of the windows together, using the string x as a separator.

To avoid using loops, you can convert the above into a list comprehension. I'll leave this step for you to complete.