I see !

Thank you everyone for the responses.

-k



On Thu, Oct 8, 2009 at 4:55 PM, David Scott <d.scott@auckland.ac.nz> wrote:

> Henrique Dallazuanna wrote:
>
>> Change the breaks argument:
>>
>> t1 <- hist(1:5, 0:5)
>> t1$counts
>>
>> On Thu, Oct 8, 2009 at 4:47 PM, Khanh Nguyen <knguyen@cs.umb.edu> wrote:
>>
>>> Hi all,
>>>
>>> I have a question about hist()
>>>
>>> 1)
>>>
>>>> t1 <- hist(c(1,2,3,4,5))
>>>> t1
>>>>
>>> $breaks
>>> [1] 1 2 3 4 5
>>>
>>> $counts
>>> [1] 2 1 1 1
>>>
>>> why is there 2 counts for 1? And should the counts be '1 1 1 1 1' ?
>>>
>>> Is there any other function to count frequency of discrete data?
>>>
>>> Thanks.
>>>
>>> -k
>>>
>>>
> Nobody has mentioned what I think is the important point here, that
> histogram is not intended for the purpose of dealing with discrete data.  To
> expect that it will give you the counts you want is just wrong. As others
> have pointed out if you make things more explicit and don't take the
> defaults it will do so.
>
> Using hist to obtain counts like this is like using a hammer to drive in a
> screw.
>
> David Scott
>
> --
> _________________________________________________________________
> David Scott     Department of Statistics
>                The University of Auckland, PB 92019
>                Auckland 1142,    NEW ZEALAND
> Phone: +64 9 923 5055, or +64 9 373 7599 ext 85055
> Email:  d.scott@auckland.ac.nz,  Fax: +64 9 373 7018
>
> Director of Consulting, Department of Statistics
>
>

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