Python programming - beginner

Question

so i have to create a code in which it reads every third letter and it creates a space in between each letter, my code creates the spaces but it also has a space after the last letter, this is my code:

msg = input("Message? ")
length = len(msg)
for i in range (0, length, 3):
 x = msg[i]
 print(x, end=" ")

My output was:

Message?

I enter:

 cxohawalkldflghemwnsegfaeap

I get back

c h a l l e n g e

when the output isn't meant to have the last " " after the e.

I have read by adding print(" ".join(x)) should give me the output i need but when i put it in it just gives me a error. Please and Thank you

Can you show us the second version that used join but failed? join requires a list or sequence and you likely didn't build the list first. — tdelaney
– tdelaney, Commented Mar 6, 2016 at 5:44
Use the title to describe the specific problem. The current title has little information. — Yu Hao
– Yu Hao, Commented Mar 6, 2016 at 5:47

Antti Haapala · Accepted Answer · 2016-03-06 05:53:34Z

In Python, strings are one kind of data structures called sequences. Sequences support slicing, which is a simple and fancy way of doing things like "from nth", "to nth" and "every nth". The syntax is sequence[from_index:to_index:stride]. One does not even a for loop for doing that.ago

We can get every 3th character easily by omitting from_index and to_index, and have stride of 3:

>>> msg = input("Message? ")
cxohawalkldflghemwnsegfaeap
>>> every_3th = msg[::3]
>>> every_3th
'challenge'

Now, we just need to insert spaces after each letter. separator.join(iterable) will join elements from iterable together in order with the given separator in between. A string is an iterable, whose elements are the individiual characters.

Thus we can do:

>>> answer = ' '.join(every_3th)
>>> answer
'c h a l l e n g e'

For the final code we can omit intermediate variables and have still a quite readable two-liner:

>>> msg = input('Message? ')
>>> print(' '.join(msg[::3]))

2 revs, 2 users 89% user6022341 · Accepted Answer · 2016-03-06 07:50:02Z

1

Try

>>> print " ".join([msg[i] for i in range(0, len(msg), 3)])
'c h a l l e n g e'

edited Mar 6, 2016 at 7:50

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