how to remove the same words in the paragraph

[kylin-3]

I need to remove the word if it appears in the paragraph twice. could
some give me some clue or some useful function in the python.
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[Andre Engels]

On Tue, Nov 3, 2009 at 11:13 PM, kylin <huili.song@...> wrote:
> I need to remove the word if it appears in the paragraph twice. could
> some give me some clue or some useful function in the python.

Well, it depends a bit on what you call 'the same word' (In the
paragraph "Fly fly, fly!" does the word fly occur 0, 1, 2 or 3
times?), but the split() function seems a logical choice to use
whatever the answer to that question.


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[Peter Otten]

kylin wrote:

> I want to remove all the punctuation and no need words form a string
> datasets for experiment.

> I need to remove the word if it appears in the paragraph twice. could
> some give me some clue or some useful function in the python.

>>> para = u"""I need to remove the word if it appears in the paragraph
twice. could
... some give me some clue or some useful function in the python.
... """
>>> print "\n".join(sorted(set(para.translate(dict.fromkeys(map(ord,
".:,-"))).split())))
I
appears
clue
could
function
give
if
in
it
me
need
or
paragraph
python
remove
some
the
to
twice
useful
word


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[Tim Chase-4]

kylin wrote:
> I need to remove the word if it appears in the paragraph twice. could
> some give me some clue or some useful function in the python.

Sounds like homework.  To fail your class, use this one:

 >>> p = "one two three four five six seven three four eight"
 >>> s = set()
 >>> print ' '.join(w for w in p.split() if not (w in s or s.add(w)))
one two three four five six seven eight

which is absolutely horrible because it mutates the set within
the list comprehension.  The passable solution would use a
for-loop to iterate over each word in the paragraph, emitting it
if it hadn't already been seen.  Maintain those words in set, so
your words know how not to be seen. ("Mr. Nesbitt, would you
please stand up?")

This also assumes your paragraph consists only of words and
whitespace.  But since you posted your previous homework-sounding
question on stripping out non-word/whitespace characters, you'll
want to look into using a regexp like "[\w\s]" to clean up the
cruft in the paragraph.  Neither solution above preserves non
white-space/word characters, for which I'd recommend using a
re.sub() with a callback.  Such a callback class might look
something like

 >>> class Dedupe:
...     def __init__(self):
...             self.s = set()
...     def __call__(self, m):
...             w = m.group(0)
...             if w in self.s: return ''
...             self.s.add(w)
...             return w
...
 >>> r.sub(Dedupe(), p)

where I leave the definition of "r" to the student.  Also beware
of case-differences for which you might have to normalize.

You'll also want to use more descriptive variable names than my
one-letter tokens.

-tkc




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[S.Selvam Siva]

On Wed, Nov 4, 2009 at 4:27 AM, Tim Chase <python.list@...> wrote:

kylin wrote:

I need to remove the word if it appears in the paragraph twice. could
some give me some clue or some useful function in the python.

Sounds like homework.  To fail your class, use this one:

>>> p = "one two three four five six seven three four eight"
>>> s = set()
>>> print ' '.join(w for w in p.split() if not (w in s or s.add(w)))
one two three four five six seven eight

which is absolutely horrible because it mutates the set within the list comprehension.  The passable solution would use a for-loop to iterate over each word in the paragraph, emitting it if it hadn't already been seen.  Maintain those words in set, so your words know how not to be seen. ("Mr. Nesbitt, would you please stand up?")

 

  Can we use inp_paragraph.count(iter_word) to make it simple ?

This also assumes your paragraph consists only of words and whitespace.  But since you posted your previous homework-sounding question on stripping out non-word/whitespace characters, you'll want to look into using a regexp like "[\w\s]" to clean up the cruft in the paragraph.  Neither solution above preserves non white-space/word characters, for which I'd recommend using a re.sub() with a callback.  Such a callback class might look something like

>>> class Dedupe:
...     def __init__(self):
...             self.s = set()
...     def __call__(self, m):
...             w = m.group(0)
...             if w in self.s: return ''
...             self.s.add(w)
...             return w
...
>>> r.sub(Dedupe(), p)

where I leave the definition of "r" to the student.  Also beware of case-differences for which you might have to normalize.

You'll also want to use more descriptive variable names than my one-letter tokens.

-tkc

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--
Yours,
S.Selvam

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[Tim Chase-4]

>   Can we use inp_paragraph.count(iter_word) to make it simple ?

It would work, but the performance will drop off sharply as the
length of the paragraph grows, and you'd still have to keep track
of which words you already printed so you can correctly print the
first one.  So you might as well not bother with counting.

-tkc


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[S.Selvam Siva]

On Wed, Nov 4, 2009 at 4:27 AM, Tim Chase <python.list@...> wrote:

kylin wrote:

I need to remove the word if it appears in the paragraph twice. could
some give me some clue or some useful function in the python.

Sounds like homework.  To fail your class, use this one:

>>> p = "one two three four five six seven three four eight"
>>> s = set()
>>> print ' '.join(w for w in p.split() if not (w in s or s.add(w)))
one two three four five six seven eight

which is absolutely horrible because it mutates the set within the list comprehension.  The passable solution would use a for-loop to iterate over each word in the paragraph, emitting it if it hadn't already been seen.  Maintain those words in set, so your words know how not to be seen. ("Mr. Nesbitt, would you please stand up?")

This also assumes your paragraph consists only of words and whitespace.  But since you posted your previous homework-sounding question on stripping out non-word/whitespace characters, you'll want to look into using a regexp like "[\w\s]" to clean up the cruft in the paragraph.  Neither solution above preserves non white-space/word characters, for which I'd recommend using a re.sub() with a callback.  Such a callback class might look something like

>>> class Dedupe:
...     def __init__(self):
...             self.s = set()
...     def __call__(self, m):
...             w = m.group(0)
...             if w in self.s: return ''
...             self.s.add(w)
...             return w
...
>>> r.sub(Dedupe(), p)

where I leave the definition of "r" to the student.  Also beware of case-differences for which you might have to normalize.

You'll also want to use more descriptive variable names than my one-letter tokens.

-tkc

I think simple regex may come handy,
 

  p=re.compile(r'(.+) .*\1')    #note the space
  s=p.search("python and i love python")
  s.groups()
  (' python',)

But that matches for only one double word.Someone else could light up here to extract all the double words.Then they can be removed from the original paragraph.
 

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--
Yours,
S.Selvam
Sent from Bangalore, KA, India
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[Tim Chase-4]

> I think simple regex may come handy,
>
>   p=re.compile(r'(.+) .*\1')    #note the space
>   s=p.search("python and i love python")
>   s.groups()
>   (' python',)
>
> But that matches for only one double word.Someone else could light up here
> to extract all the double words.Then they can be removed from the original
> paragraph.

This has multiple problems:

 >>> p = re.compile(r'(.+) .*\1')
 >>> s = p.search("python one two one two python")
 >>> s.groups()
('python',)
 >>> s = p.search("python one two one two python one")
 >>> s.groups() # guess what happened to the 2nd "one"...
('python one',)

and even once you have the list of theoretical duplicates (by
changing the regexp to r'\b(\w+)\b.*?\1' perhaps), you still have
to worry about emitting the first instance but not subsequent
instances.

-tkc




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