Have you read the Intro to R tutorial or checked the relevant parts of
the R Language Definition.? Had you done so, I think you would have
found in one or the other or both that a matrix is just a vector with
a "dim" attribute. mapply() just sees that single argument -- a vector
of length 50. The conversion to a data frame makes the matrix a
(single) list with three elements/components, as again described in
the referenced docs, so that mapply() then does as its Help file says.
If you read the docs, you would not be so mystified... but that's a
tautology, I suppose.
-- Bert
On Thu, Apr 18, 2013 at 7:54 PM, C W <tmrs...@gmail.com> wrote:
> I agree! I would have thought of using as.data.frame() no matter how
> hard I had to think. ;)
> Thanks,
> Mike
>
> On Thu, Apr 18, 2013 at 10:18 PM, arun <smartpink...@yahoo.com> wrote:
>> Hi,
>> You can try:
>> par(mfrow=c(3, 2))
>> mapply(hist,split(a,col(a)),main=colnames(a),xlab="x")
>>
>>
>> If you look at the output,
>> mapply(hist,a,main=colnames(a),xlab="x")
>>
>> [,1] [,2] [,3]
>> breaks Numeric,2 Numeric,2 Numeric,2
>> counts 1 1 1
>> density 2 5 20
>> mids 0.75 0.3 0.025
>> xname "dots[[1L]][[1L]]" "dots[[1L]][[2L]]" "dots[[1L]][[3L]]"
>> equidist TRUE TRUE TRUE
>> [,4] [,5] [,6]
>> breaks Numeric,2 Numeric,2 Numeric,2
>> counts 1 1 1
>> density 10 5 1
>> mids 0.15 0.3 0.5
>>
>> _____________________________________
>> [,46] [,47] [,48]
>> breaks Numeric,2 Numeric,2 Numeric,2
>> counts 1 1 1
>> density 0.002 0.01 0.02
>> mids -6250 -950 275
>> xname "dots[[1L]][[46L]]" "dots[[1L]][[47L]]" "dots[[1L]][[48L]]"
>> equidist TRUE TRUE TRUE
>> [,49] [,50]
>> breaks Numeric,2 Numeric,2
>> counts 1 1
>> density 0.001 0.005
>> mids 10500 3300
>> xname "dots[[1L]][[49L]]" "dots[[1L]][[50L]]"
>> equidist TRUE TRUE
>>
>>
>> #it is going through every single observation rather than by columns.
>> A.K.
>>
>>
>>
>>
>> ----- Original Message -----
>> From: C W <tmrs...@gmail.com>
>> To: arun <smartpink...@yahoo.com>
>> Cc: R help <r-help@r-project.org>
>> Sent: Thursday, April 18, 2013 7:20 PM
>> Subject: Re: [R] How to keep plot title same as column name using
>> apply/sapply?
>>
>>> mapply(hist,as.data.frame(a),main=colnames(a),xlab="x")
>>
>> Why it does't work when I use "a" instead of "as.data.frame(a)"?
>> ?mapply says "arguments to vectorize over (vectors or lists of
>> strictly positive length, or all of zero length)."
>>
>> Thanks,
>> Mike
>>
>> On Thu, Apr 18, 2013 at 6:49 PM, arun <smartpink...@yahoo.com> wrote:
>>> Hi,
>>> Try:
>>> par(mfrow=c(3, 2))
>>> mapply(hist,as.data.frame(a),main=colnames(a),xlab="x")
>>> A.K.
>>>
>>>
>>>
>>>
>>> ----- Original Message -----
>>> From: C W <tmrs...@gmail.com>
>>> To: r-help <r-help@r-project.org>
>>> Cc:
>>> Sent: Thursday, April 18, 2013 6:30 PM
>>> Subject: [R] How to keep plot title same as column name using apply/sapply?
>>>
>>> Dear list,
>>> I am trying to plot histogram of a 10 by 5 matrix by columns.
>>>
>>> What is a good way to paste the column names?
>>>
>>> dput(a)
>>> structure(c(0.891856482875103, 0.249593821948295, 0.0385066520473322,
>>> 0.109098868876336, 0.238035894186719, 0.971470380855763, 0.168018536530906,
>>> 0.941457062296419, -0.285381460315397, -0.0229335863796271,
>>> -20138.175683257,
>>> 28190.7238887329, 8521.44473371867, 37565.8599592035, -61036.6139527803,
>>> -57278.42539968, 24284.2860193156, 57470.8275786857, -5035.05919672665,
>>> -48223.3060594833, 0.105003428645432, 0.0385466025676578, 0.130895042931661,
>>> 0.0682065344415605, 0.155879093753174, 0.0598988083656877,
>>> 0.121624303748831,
>>> 0.120134906983003, 0.159434637567028, 0.0144033632241189, 6133.33965969139,
>>> 5363.20408479208, 22432.8473630982, 10627.5697498379, 273.481263455154,
>>> 8326.07386291277, 133.117356408579, 4108.23352922685, 643.558745779806,
>>> 5532.10081652847, -4512.80115728638, -5839.1178791698, 7341.05048720629,
>>> -1299.37511688102, -13151.5558568303, -6442.78100296569, -975.590624649323,
>>> 298.440121450997, 10808.9562034117, 3210.22497745543), .Dim = c(10L,
>>> 5L), .Dimnames = list(NULL, c("Tmrw", "hRm2", "vm", "Fluid",
>>> "SM")))
>>> par(mfrow=c(3, 2))
>>> apply(a, 2, function(x){hist(x, main="column name here")})
>>>
>>> This plots, but doesn't plot title.
>>>
>>> But,
>>>> apply(a,2,function(x){colnames(x)})
>>> NULL
>>>
>>> Maybe the column names is lost? But, I see column names here.
>>>
>>>> apply(a,2,function(x){is.vector(x)})
>>> Tmrw hRm2 vm Fluid SM
>>> TRUE TRUE TRUE TRUE TRUE
>>>
>>> My understanding is that, apply() takes data matrix column by column,
>>> and pastes names back on at the end? But I need it for plot title,
>>> how should I change my code?
>>>
>>> Thanks,
>>> Mike
>>>
>>> ______________________________________________
>>> 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.
>>>
>>
>
> ______________________________________________
> 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.
--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
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