Function that takes a string and returns a list of 8-character strings

Question

I'm very new to programming. I need to create a function that takes any string and a fill character and returns a list of 8 character strings. For example:

>>>segmentString("Hello, World!","-")
['Hello, W', 'orld!---']

As you can see, the string has 13 characters. The function splits into 2 strings, where the first contains 8 characters, and the second contains 5 characters and 3 fill characters.

So far, I've figured out how to do this for strings less than 8 characters, but not for more than 8 characters, where the string is split up, which is what I am stuck on.

def segmentString(s1,x):
    while len(s1)<8:
        s1+=x
    return s1

How do you split a string up?

jpp · Accepted Answer · 2018-11-10T01:58:24.737

3

You can use a slightly modified version of this chunking recipe:

def chunks(L, n, char):
    for i in range(0, len(L), n):
        yield L[i:i + n].ljust(n, char)

res = list(chunks('Hello, World!', 8, '-'))

['Hello, W', 'orld!---']

str.ljust performs the necessary padding.

edited Nov 10 '18 at 01:58

answered Nov 10 '18 at 01:33

jpp

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Easier way to write the padding step: `yield x.ljust(n, char)`. No need to perform the calculations for required padding yourself over and over. And because it doesn't even need to know the original length of `x`, you could one-line the body of the loop to `yield L[i:i+n].ljust(n, char)` – ShadowRanger Nov 10 '18 at 01:56

score 1 · Answer 2 · answered Nov 10 '18 at 01:48

A slightly modified version of @jpp's answer is to first pad the string with your fill character, so that the length is cleanly divisible by 8:

def segmentString(s, c):
    s = s + c * (8 - len(s) % 8)
    return [s[i:i+8] for i in range(0, len(s), 8)]

>>> segmentString("Hello, World!","-")
['Hello, W', 'orld!---']

And if you needed the size to be variable, you can just add in a size argument:

def segmentString(s, c, size):
    s = s + c * (size - len(s) % size)
    return [s[i:i+size] for i in range(0,len(s),size)]

>>> segmentString("Hello, World!","-",4)
['Hell', 'o, W', 'orld', '!---']

score 1 · Answer 3 · answered Nov 10 '18 at 01:53

1

Yet another possibility:

def segmentString(s1, x):
    ret_list = []
    while len(s1) > 8:
        ret_list.append(s1[:8])
        s1 = s1[8:]
    if len(s1) > 0:
        ret_list.append(s1 + x*(8-len(s1)))
    return ret_list

answered Nov 10 '18 at 01:53

John Anderson

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score 0 · Answer 4 · edited Nov 10 '18 at 01:54

0

I am also new to programming but I think you could use this...

Word = "Hello, World!---"
print([Word[i:i+x] for i in range(0, len(Word), x)])

Just change x to whatever value you want.. Check out https://www.google.co.in/amp/s/www.geeksforgeeks.org/python-string-split/amp/ Hope it helps.

edited Nov 10 '18 at 01:54

sacuL

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answered Nov 10 '18 at 01:50

Toolazytothinkofaname

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I think that the issue is that the word is not `"Hello, World!---"`, but `"Hello, World!"`, and it needs to be split into 8 character chunks, with `-` as a filler when the last chunk is not 8 characters long – sacuL Nov 10 '18 at 01:53

score 0 · Answer 5 · answered Nov 10 '18 at 02:05

Here is the simple answer that you can understand that solves your problem with few simple ifs:

a= 'abcdefghijklmdddn'

def segmentString(stuff):
    c=[]
    i=0
    hold=""

    for letter in stuff:
        i+=1
        hold+=letter

        if i%8==0:
            c.append(hold)
            hold=""

    c.append(hold)

    if hold=="":
        print(c)

    else:
        if hold==c[-1]:
            print(c)

        else:
            c.append(hold)
            print(c)

Function that takes a string and returns a list of 8-character strings

5 Answers5