From data frame to list object

On Jan 31, 2011, at 2:18 PM, Bogaso Christofer wrote:

Sorry if I did not clarify that. Here I have a data frame with many  
columns,
which was taken from some outside DB. Now I want to split that data  
frame
and create a "list" object (to make my further calculation easier),  
on basis
of a typical column of that DB. I cannot post my original DB here  
(due to
some security reason and ofcourse it's huge size), therefore I  
posted an
artificial DB.

Here my artificial DB with 3 columns:

dfrm <- data.frame(x=rnorm(18), y=rep(c("a", "b", "c"), each=6),
z=rep(c("x", "y", "z"), each=2))

I would like to create a list object where each element now is a  
matrix or
data frame, based on that "y" column. 1st element of that list will  
be a
data frame with observations of "x" and "z" columns, that address the
attribute "y = a". Similarly other two.

It is a job for split. The result is a data.frame but that is, after  
all, just a list with certain attributes.

split(dfrm, dfrm$y)

$a
           x y z
1 -1.16790385 a x
2 -0.84831139 a x
3 -0.64312051 a y
4 -1.66841121 a y
5  0.03737404 a z
6 -0.42165643 a z

$b
           x y z
7   1.1045024 b x
8   1.4787933 b x
9   0.5278083 b y
10  0.1770083 b y
11 -0.5054573 b z
12 -0.6512499 b z

$c
            x y z
13  0.61225420 c x
14 -0.45032691 c x
15  0.36502921 c y
16  0.33505288 c y
17  0.02189088 c z
18 -0.53893624 c z

str(split(dfrm, dfrm$y)$a)

'data.frame':	6 obs. of  3 variables:
$ x: num  -1.1679 -0.8483 -0.6431 -1.6684 0.0374 ...
$ y: Factor w/ 3 levels "a","b","c": 1 1 1 1 1 1
$ z: Factor w/ 3 levels "x","y","z": 1 1 2 2 3 3

-- 
David

Hope I could be able to make my intentions clearer.

Any idea how I can achieve that?

Thanks,



-----Original Message-----
From: David Winsemius [mailto:dwinsemius at comcast.net]
Sent: 01 February 2011 00:13
To: Bogaso Christofer
Cc: r-help at r-project.org
Subject: Re: [R] From data frame to list object


On Jan 31, 2011, at 1:56 PM, Bogaso Christofer wrote:

Thanks David for this reply. However if my data frame has only 2
columns
then it is working fine. It is not working for a general setting:

dfrm <- data.frame(x=rnorm(18), y=rep(c("a", "b", "c"), each=6),
z=rep(c("x", "y", "z"), each=2))
tapply(dfrm[,1], dfrm$y, c) # this is working fine

tapply(dfrm[,c(1,3)], dfrm$y, c)  # this is giving error!

Error in tapply(dfrm[, c(1, 3)], dfrm$y, c) :
arguments must have same length

Can you please help me how to modify that?

You will need to specify what you goals are. What to you want to
happen to those two columns referred to by dfrm[, c(1,3)]? It's
possible that split() may be the answer, but clarify the goals first.
You should provide an example that represents the complexity of the
task.

Thanks,

-----Original Message-----
From: David Winsemius [mailto:dwinsemius at comcast.net]
Sent: 31 January 2011 23:26
To: Bogaso Christofer
Cc: r-help at r-project.org
Subject: Re: [R] From data frame to list object


On Jan 31, 2011, at 1:03 PM, Bogaso Christofer wrote:

Dear all, let say I have following data frame:

dfrm <- data.frame(x=rnorm(18), y=rep(c("a", "b", "c"), each=6))  >

tapply(dfrm$x, dfrm$y, c) $a [1]  0.9711995  1.4018345 -1.4355713
-0.5106138
-0.8470171 [6]  1.1634586

$b
[1] -0.8058164  0.4977112  1.1556391  0.8158588  0.2549273 [6]
3.0758260

$c
[1]  0.437345128 -0.415874363  0.003230285 -0.737117910 [5]
1.247972964
0.903001077

data.frame(x=rnorm(18), y=rep(c("a", "b", "c"), each=6))

          x y

1  -1.072152537 a

2   0.382985265 a

3   0.058877377 a

4  -0.006911939 a

5  -2.355269051 a

6  -0.303095553 a

7   0.484038422 b

8   0.733928931 b

9  -1.136014346 b

10  0.503552090 b

11  1.708609658 b

12 -0.294599403 b

13  1.239308497 c

14  0.754081946 c

15 -0.237346858 c

16 -0.051011439 c

17 -0.618675146 c

18  0.537612359 c

From this data frame I want to create a "list" of length 3, where
each

element of this list will be a vector corresponding to the value of
y.
For example, 1st element will be all "x" values corresponding to  
the
"y=a", and similarly the other elements of this list. Can somebody
point me how to do this without having some "for" loop?



Thanks and regards,


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David Winsemius, MD
West Hartford, CT

David Winsemius, MD
West Hartford, CT

David Winsemius, MD
West Hartford, CT