I see !
Thank you everyone for the responses.
-k
On Thu, Oct 8, 2009 at 4:55 PM, David Scott <d.sc...@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 <kngu...@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.sc...@auckland.ac.nz, Fax: +64 9 373 7018
>
> Director of Consulting, Department of Statistics
>
>
[[alternative HTML version deleted]]
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
