Thanks a lot!!!
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hello , I am new user in R . I have datafile (class = data.frame) which has
825 columns with unique column name i want extract 200 selected column from
datafile how can I do this?
my datafile look like..
Mi RBN RBFnDB nX
3 2.6225979 0.53132756
Is this what you want:
x - read.table(text = Mi RBN RBFnDB
nX
+ 3 2.6225979 0.53132756 -0.80599902 -1.4471864 -0.5705269
+ 10 0.4818746 -1.72143092 -2.19579027 2.0118824 -0.5705269
+ 12 2.8519611 1.88298265 0.09614617 0.6282549 -0.5705269
+
A look at the tutorial might help here, but anyway:
Say you have that dataframe down there with the name myData (you should use
dput() to give us the data btw),
then you can subset that by using myData[rows,columns], where left of the comma
you define which rows you want, and right of the
Please read the Introduction to R tutorial that ships with R to get
started with R. Quoting Rolf Turner:
Learn something about R; don't just hammer and hope. Read the
introductory manuals and scan the FAQ.
Cheers,
Bert
On Tue, Oct 30, 2012 at 7:09 AM, alex_123 deepak.j...@gmail.com wrote:
Dear R-experts!
I found a great post on the getting genetics done blog which describes how
to make use of NCBI E-utilities in Python to extract data from PubMed:
http://gettinggeneticsdone.blogspot.com/2011/05/using-ncbi-e-utilities.html
I was wondering whether there would be a way to directly
Hi R-users,
I have a set of data from 1958-2009, how do I extract the data from 1927 and
2007?
beechworth.dt
Year Month Rain
1 1858 3 21.8
2 1858 4 47.0
3 1858 5 70.1
4 1858 6 78.7
5 1858 7 101.6
6 1858 8 129.8
7 1858 9 80.8
8
Roslina Zakaria wrote:
Hi R-users,
I have a set of data from 1958-2009, how do I extract the data from 1927 and
2007?
beechworth.dt
Year Month Rain
11858 3 21.8
21858 4 47.0
31858 5 70.1
41858 6 78.7
51858 7 101.6
61858 8 129.8
7
If Year is numeric data, then you can use:
beechworth.dt.2 - subset(beechworth.dt, Year=1997 Year =2008)
If it is character, then you can use:
beechworth.dt.2 - subset(beechworth.dt, Year %in% as.character(1997:2008))
2009/4/20 Roslina Zakaria zrosl...@yahoo.com:
Hi R-users,
I have a set of
Hi,
Try this one:
beechworth.dt.2 - beechworth.dt[beechworth.dt$Year=1927
beechworth.dt$Year=2007,]
Martins
On Apr 20, 9:59 am, Roslina Zakaria zrosl...@yahoo.com wrote:
Hi R-users,
I have a set of data from 1958-2009, how do I extract the data from 1927 and
2007?
beechworth.dt
Hi,
maybe the following line works for you:
beechworth.dt.2 - beechworth.dt[as.numeric(beechworth.dt$Year) %in% 1927:2007,
]
(using as.numeric() to make sure that Year is numeric, maybe not needed?).
Christian
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On Sun, 19 Apr 2009 23:59:43 -0700 (PDT) Roslina Zakaria
zrosl...@yahoo.com wrote:
RZ I have a set of data from 1958-2009, how do I extract the data from
RZ 1927 and 2007?
RZ beechworth.dt
RZ Year Month Rain
how about:
beech.cut-subset(beechworth.dt,(Year1926Year2008))
hth
Stefan
Hi,
Thanks for your reply. However, this didn't work exactly as I needed
it to since the expression is dynamically built as a character vector
i.e. not executed as
e - expression(Sepal.Width 4)
but as
e - expression(Sepal.Width 4)
in which case subset() throws an error (must evaluate to
On 12 Aug 2008, at 17:00, Gabor Grothendieck wrote:
Executing strings is probably not a very
R-ish thing to do
I know - but as far as I can see there was no other way around in
this case...
but if that's your aim use eval and parse:
s - iris[iris$Sepal.Width 4,]
eval(parse(text = s))
Dries Knapen-2 wrote:
Hi,
Thanks for your reply. However, this didn't work exactly as I needed
it to since the expression is dynamically built as a character vector
i.e. not executed as
e - expression(Sepal.Width 4)
but as
e - expression(Sepal.Width 4)
in which case
The code I posted does work if you use it as I explained, not
as you changed it. Executing strings is probably not a very
R-ish thing to do but if that's your aim use eval and parse:
s - iris[iris$Sepal.Width 4,]
eval(parse(text = s))
On Tue, Aug 12, 2008 at 10:35 AM, Dries Knapen [EMAIL
eval(parse(text = iris[iris$Sepal.Width 4,]))
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
16 5.7 4.4 1.5 0.4 setosa
33 5.2 4.1 1.5 0.1 setosa
34 5.5 4.2 1.4 0.2 setosa
On
Hi,
Based on user input, I wrote a function that creates a list which
looks like:
str(list)
List of 4
$ varieties: chr [1:12] temp.26_time.5dagen_biorep.1 time.
5dagen_temp.26_biorep.2 temp.18_time.5dagen_biorep.1 temp.18_time.
5dagen_biorep.2 ...
$ temp : Factor w/ 2 levels 18,26:
Try this:
e - expression(Sepal.Width 4)
subset(iris, eval(e), select = Sepal.Length)
Sepal.Length
16 5.7
33 5.2
34 5.5
subset(iris, eval(e))
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
16 5.7 4.4 1.5 0.4 setosa
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