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Groups > comp.lang.python > #78713
| From | Shiva <shivaji_tn@yahoo.com> |
|---|---|
| Subject | Re: Issue in printing top 20 dictionary items by dictionary value |
| Date | 2014-10-05 06:58 +0000 |
| References | <mailman.14596.1412413918.18130.python-list@python.org> <m0oi4p$7ir$1@speranza.aioe.org> <mailman.14602.1412444228.18130.python-list@python.org> <cZadnUgaZver963JnZ2dnUU7-IGdnZ2d@giganews.com> |
| Newsgroups | comp.lang.python |
| Message-ID | <mailman.14611.1412492351.18130.python-list@python.org> (permalink) |
Larry Hudson <orgnut <at> yahoo.com.dmarc.invalid> writes:
>
> On 10/04/2014 10:36 AM, Shiva wrote:
> >
> > What I don't understand is:
> >
> > for w in eachword:
> > textstorage[w]=textstorage.get(w, 0) + 1
> >
> > How does textstorage.get(w,0)+1 give the count of the word??
> >
>
> Very long-winded explanation: (But to shorten it a bit, I'm going to use
ts in place of
> textstorage. Also lhs = left-hand-side and rhs = right-hand-side.)
>
> What we're trying to do here is to update the word count. We could
(erroneously) write it as:
>
> ts[w] = ts[w] + 1
>
> If w already exists in the ts dictionary, this works fine. But if it does
not it will abort
> with a KeyError when it comes to the ts[w] on the rhs of the assignment.
>
> The get() method is an alternate way of accessing the value of a key in a
dictionary, but with a
> default value given as well. Now let's break down the statement
>
> ts[w] = ts.get(w, 0) + 1
>
> Case 1: w already exists in the ts dictionary:
>
> ts.get(w, 0) gets the value of ts[w] (the current word count), adds 1,
which is then used to
> update the word-count value of ts[w] (on the lhs of the assignment).
>
> case2: w does not exist in the ts dictionary:
>
> ts.get(w, 0) gives the default value of 0, and 1 is added to that. ts[w]
on the lhs of the
> assignment does not exist, so a new entry is created in the ts dictionary
with the given w as
> the key, and the value is initialized with the 1 from the get()+1.
>
> Make sense?
>
> -=- Larry -=-
>
>
Hi Larry,
Thanks for the explanation - I was a bit confused as get() operation in this
case would have got None for words occurring the first time.
Now I understand by writing a small example in the interpreter:
>>> dt={}
>>> splitw=('aa','bb','cc')
>>> for w in splitw:
... dt[w]=dt.get(w,0)
...
>>> dt
{'cc': 0, 'bb': 0, 'aa': 0}
So we just increment 0 to 1 for count.
Thanks,
Pradeep
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Issue in printing top 20 dictionary items by dictionary value Shiva <shivaji_tn@yahoo.com> - 2014-10-04 09:11 +0000
Re: Issue in printing top 20 dictionary items by dictionary value Alexander Blinne <news@blinne.net> - 2014-10-04 12:28 +0200
Re: Issue in printing top 20 dictionary items by dictionary value Shiva <shivaji_tn@yahoo.com> - 2014-10-04 17:36 +0000
Re: Issue in printing top 20 dictionary items by dictionary value Larry Hudson <orgnut@yahoo.com> - 2014-10-04 14:44 -0700
Re: Issue in printing top 20 dictionary items by dictionary value Shiva <shivaji_tn@yahoo.com> - 2014-10-05 06:58 +0000
Re: Issue in printing top 20 dictionary items by dictionary value MRAB <python@mrabarnett.plus.com> - 2014-10-04 20:40 +0100
Re: Issue in printing top 20 dictionary items by dictionary value Denis McMahon <denismfmcmahon@gmail.com> - 2014-10-05 01:11 +0000
Re: Issue in printing top 20 dictionary items by dictionary value MRAB <python@mrabarnett.plus.com> - 2014-10-05 02:41 +0100
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