[Tutor] how to split/partition a string on keywords?

[Jared Nielsen]

Thanks everyone. As I'm learning programming what I find most interesting
is that there's always more than one way to solve a problem.

I implemented eryksun's suggestion and used the replace() method.
But, playing around with it, what I discovered is that it won't store the
change.
For example, when the input text is, "Ham and cheese or chicken and
waffles":

#!/usr/bin/python

text = raw_input("Enter text: ")

print text.replace("and", "\nand").replace("or", "\nor")

I get:
Ham
and cheese
or chicken
and waffles.

But if I run the following:

#!/usr/bin/python

text = raw_input("Enter text: ")

text.replace("and", "\nand")
text.replace("or", "\nor")

print text

I get the text as it was entered.
Is there a way to replace text in a string without splitting or
partitioning?

The bigger picture for this little project is a "poetry machine", in which
a user enters some prose and the program chops it up into modern poetry.

So, this is a long shot naive noob question, but is there any way to count
syllables in words in a string? Or at least approximate this procedure?



On Thu, Aug 23, 2012 at 3:08 PM, <akleider at sonic.net> wrote:

> This question seemed a good excercise so I banged out a little script
> (which worked) but latter I saw posts showing code that by using string
> method 'partition' provided a more elegant solution.
> I was previously unaware of this method.  My "bible" has been David M.
> Beazley's Python Essential Reference (3rdEd) in which this method is not
> mentioned (that I can see.)
> Should I switch "bibles?"
> (I often find myself wanting to hack in "off line environments" so
> something as old fashion as a book would be nice:-)
>
> Here's my script for what it's worth:
>
> #!/usr/bin/env python
>
> import sys
>
> usage = """test0 separator
> Requires one parameter, the text to be used to separate the input which
> will be requested by the program."""
>
> if len(sys.argv) != 2:
>     print usage
> separator = sys.argv[1]
>
> def separate(string, separator):
>     ret = []
>     i = string.find(separator)
>     l = len(separator)
>     while i > 0:
>         ret.append(string[:i])
>         ret.append(separator)
>         string = string[i+l:]
>         i = string.find(separator)
>     ret.append(string)
>     return ret
>
> def repart(string, separator):
>     """Does the same as separator but using string method 'partition'"""
>     parts = string.partition(separator)
>     if parts[0] == string:
>         return (parts[0], )
>     else:
>         return parts[:-1] + repart(parts[-1], separator)
>
> input_str = raw_input("Enter text to split on '%s': "%(separator, ))
>
> separated_array = separate(input_str, separator)
> for s in separated_array:
>     print s
> parted_array = repart(input_str, separator)
> for s in parted_array:
>     print s
>
>
>
>
>  > Hi all,
> > I'm new to programming and Python.
> > I want to write a script that takes a string input and breaks the string
> > at
> > keywords then outputs the pieces on separate lines.
> > I'm not sure how to break the string, though.
> > I looked through the docs and found split() and partition(), which come
> > close.
> > But split() doesn't retain the separator and partition() retains the
> white
> > space and returns a 3-tuple which I'll have to figure out how to rejoin
> > nor
> > does it partition on subsequent instances of the separator.
> >
> > Here's the script in its basic form:
> >
> > #!/usr/bin/python
> >
> > text = raw_input("Enter text: ")
> > print "You entered ", text
> >
> > objects = text.partition(' and')
> > print objects
> >
> > for object in objects:        # Second Example
> >
> >    print object
> >
> > For example, if I run this with the input:
> > "Ham and cheese omelette with hasbrowns and coffee."
> > I get:
> > Ham
> >  and
> >  cheese omelette with hashbrowns and coffee.
> >
> > Any help is greatly appreciated.
> > _______________________________________________
> > Tutor maillist  -  Tutor at python.org
> > To unsubscribe or change subscription options:
> > http://mail.python.org/mailman/listinfo/tutor
> >
>
>
>
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