Issue in printing top 20 dictionary items by dictionary value

Shiva shivaji_tn at yahoo.com
Sun Oct 5 08:58:50 CEST 2014


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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