question string manipulation

Question

output of my code is:

extr @ ate @ rre @ str @ ial @

but it should be

ext @ rat @ err … and so on

How can i do that?
https://code.sololearn.com/cqqAbZz6efUs/?ref=app

Answer 1

Per Bratthammar Thank you for these information brother! I love learning these.

Answer 2

Hi, Dragon RB !

Slicing is good to know. The basic is easy. Suppose:

>>> s = "abcdefghij"

Then s[i: j: k] creates a copy of that part of s that starts with index i and end with index j-1. k is the step length, and can be ignored if k=1.

i, j are non negativ integers, if couning the index from left to rigth (otherwise the opposit) k is a non zero integer:

>>> print(s[1:8:2])
bdfh

print() has two keyword arguments: sep and end.

As default end='\n ', that is the new line character. When you change end, first everything in the print function is printed, and after that it will end the output with what you assigned end to:

>>> print(s, end='hello world\n')
abcdefghijhello world

if s is as above.

And sep is a separator and separats what every you assign sep to, after every coma in print the print function:

>>> print(1, 2, 3, 4, 5, sep='\n')
1
2
3
4
5

You can compare using sep keyword argument in print with the string method .join():

>>> print('\n'.join('12345'))
1
2
3
4
5

Answer 3

Here is the modified code:

https://code.sololearn.com/cH3Q8aoHibXv/?ref=app

Always remember that index always start at 0, so I added the index by 1 (in the if condition) to get the expected output.
I also declared a string variable(for string builder) outside the for loop, because it is a good practice for that.

Answer 4

# Hi, Angela !
# Here’s another one:

msg, step = "extraterrestrial", 3
for k in range(0, len(msg), step):
print(msg[k: k+step], end=' @ ')

Answer 5

Per Bratthammar I think index slicing and end = "..." is a little difficult for me to understand..
Edit: nevermind, I got it!