Python: words in a string get stripped in the output

Question

I'm writing a text adventure in Python 2.7 and now want to make nice-looking output. I already tried to make a GUI with Tkinter, but it didn't work, because the user's text input could not be decoded in Tk, and I need it for German Umlauts. So now I'm trying to get good output straight in the Python shell or the Jupyter notebook. The problem is, when I do a print statement, I get output like:

This is a text and just a little exa

mple to show the problem.

But of course I don't want to get words stripped, like 'example'. It should look like this:

This is a text and just a little

example to show the problem.

I tired to work with the width of a display and thought that it would be good to split the string after 80 characters, but with the condition that it should only be split at a whitespace and not in a word. I wanted to do something like this:

def output(string):
    for pos, char in enumerate(string):
        if pos <= 80 and pos >=70 and char == " ":
            templist = string.split(char)
            ...

Strings are immutable, so I think I have to convert it into a list, but then I don't know how to put the split-position of the string and the list together. Maybe I'm thinking too intricately.

Is there a way to say: If the string is longer than 80 characters, split the string at the nearest whitespace to the 80th character?

post the actual input string (longer than 80 chars) and expected result — RomanPerekhrest, Aug 04 '17 at 15:38

score 3 · Accepted Answer · answered Aug 04 '17 at 15:41

3

Use the textwrap module.

For example, if you want 40 character line width:

import textwrap
a = '''This is a text and just a little example to show the problem.'''
print("\n".join(textwrap.wrap(a,40)))
##This is a text and just a little example
##to show the problem.

answered Aug 04 '17 at 15:41

Sam Craig

875
5
16

This is exactly, what I was looking for (more than than plenty hours of frustrating coding :D) Maybe I searched with incongruous keywords. Thank you very much, I appreciate it. – FabianPeters Aug 04 '17 at 17:19

score 2 · Answer 2 · answered Aug 04 '17 at 15:43

2

I would look at the first 80 characters of a string and find the last occurrence of the space character, and then split your string there. You can use rfind() for this:

string = "This is a text and just a little example to show the problem"
lines = []
while string:
    index = string[:80].rfind(' ')
    if index == -1:
        index = 80
    lines.append(string[:index])
    string = string[index:].lstrip()

answered Aug 04 '17 at 15:43

TallChuck

1,725
11
28

This is an interesting way to do it. I will try this way too. Thank you! – FabianPeters Aug 04 '17 at 17:23
well I wrote this without knowing about the `textwrap` library. That is probably a better way to go – TallChuck Aug 04 '17 at 17:25

score 0 · Answer 3 · answered Aug 04 '17 at 15:41

0

I think you're looking for the textwrap module.

text_wrap.fill(your_text, width=your_width)

answered Aug 04 '17 at 15:41

leegenes

31
2

That is the module, what I was looking for. Thank you very much for your answer, I appreciate it. – FabianPeters Aug 04 '17 at 17:20

Python: words in a string get stripped in the output

3 Answers3