Displaying 20 results from an estimated 6000 matches similar to: "Reading data"
2017 Dec 09
2
Reducing code size of Position Independent Executables (PIE) by shrinking the size of dynamic relocations section
* Rahul Chaudhry via gnu-gabi:
> The encoding used is a simple combination of delta-encoding and a
> bitmap of offsets. The section consists of 64-bit entries: higher
> 8-bits contain delta since last offset, and lower 56-bits contain a
> bitmap for which words to apply the relocation to. This is best
> described by showing the code for decoding the section:
>
> typedef
2008 Apr 18
2
Correspondence and detrended correspondence analysis
Hi,
I hope someone knows the answer to this or has a real good reference about it (I am using Legendre & Legendre, Numerical Ecology, 1998).... My data is a data.frame with locations as rows and vegetation assemblages / species as columns. I've done a PCA, a correspondance analysis (CA) using ca in ca package and a detrended correspondance analysis (DCA) using decorana from vegan package.
2006 Jun 08
3
Reading in a table with ISO-latin1 encoding in MacOS-X (Intel)
Dear colleages in R,
I have earlier been working with R in Linux, where reading in a table
containing Scandinavian letters ("?", "?", and "?") in the header as
part of variable names has not caused any problem whatsoever.
However, when trying to do the same in R running on new MacOS-X (with
an Intel processor) with the same original text table does not seem to
2017 Dec 06
2
Remove
Hi Ashta,
There are many ways to do it. Here is one:
vars <- sapply(split(DM$x, DM$GR), var)
DM[DM$GR %in% names(vars[vars > 0]), ]
Best
Ista
On Wed, Dec 6, 2017 at 6:58 PM, Ashta <sewashm at gmail.com> wrote:
> Thank you Jeff,
>
> subset( DM, "B" != x ), this works if I know the group only.
> But if I don't know that group in this case "B", how
2017 Dec 09
1
Remove
Hello,
Try the following.
keep <- list(A = c(15, 30), B = c(40, 50), C = c(60, 75))
sp <- split(DM$x, DM$GR)
inx <- unlist(lapply(seq_along(sp), function(i) keep[[i]][1] <= sp[[i]]
& sp[[i]] <= keep[[i]][2]))
DM[inx, ]
# GR x y
#1 A 25 125
#2 A 23 135
#5 B 45 321
#6 B 47 512
#9 C 61 521
#10 C 68 235
Hope this helps,
Rui Barradas
On 12/9/2017 12:48 AM, Ashta
2017 Dec 09
2
Remove
> On Dec 8, 2017, at 4:48 PM, Ashta <sewashm at gmail.com> wrote:
>
> Hi David, Ista and all,
>
> I have one related question Within one group I want to keep records
> conditionally.
> example within
> group A I want keep rows that have " x" values ranged between 15 and 30.
> group B I want keep rows that have " x" values ranged
2017 Dec 09
0
Remove
Hi David, Ista and all,
I have one related question Within one group I want to keep records
conditionally.
example within
group A I want keep rows that have " x" values ranged between 15 and 30.
group B I want keep rows that have " x" values ranged between 40 and 50.
group C I want keep rows that have " x" values ranged between 60 and 75.
DM <-
2017 Dec 07
0
Remove
Thank you Ista! Worked fine.
On Wed, Dec 6, 2017 at 5:59 PM, Ista Zahn <istazahn at gmail.com> wrote:
> Hi Ashta,
>
> There are many ways to do it. Here is one:
>
> vars <- sapply(split(DM$x, DM$GR), var)
> DM[DM$GR %in% names(vars[vars > 0]), ]
>
> Best
> Ista
>
> On Wed, Dec 6, 2017 at 6:58 PM, Ashta <sewashm at gmail.com> wrote:
>> Thank
2009 Oct 01
4
Color of graph
I am trying to plot a line graph for 3 or more regression lines
abline(m1)
abline(m2)
abline(m3)
Can I change the color of each line? if so how?
Thanks in advance
Ashta
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2017 Dec 07
4
Remove
> On Dec 6, 2017, at 4:27 PM, Ashta <sewashm at gmail.com> wrote:
>
> Thank you Ista! Worked fine.
Here's another (possibly more direct in its logic?):
DM[ !ave(DM$x, DM$GR, FUN= function(x) {!length(unique(x))==1}), ]
GR x y
5 B 25 321
6 B 25 512
7 B 25 123
8 B 25 451
--
David
> On Wed, Dec 6, 2017 at 5:59 PM, Ista Zahn <istazahn at gmail.com> wrote:
2017 Dec 09
1
Remove
library(dplyr)
DM <- read.table( text='GR x y
A 25 125
A 23 135
.
.
.
)
DM %>% filter((GR == "A" & (x >= 15) & (x <= 30)) |
(GR == "B" & (x >= 40) & (x <= 50)) |
(GR == "C" & (x >= 60) & (x <= 75)))
On Fri, Dec 8, 2017 at 4:48 PM, Ashta <sewashm at gmail.com>
2017 Dec 09
0
Remove
> On Dec 8, 2017, at 6:16 PM, David Winsemius <dwinsemius at comcast.net> wrote:
>
>
>> On Dec 8, 2017, at 4:48 PM, Ashta <sewashm at gmail.com> wrote:
>>
>> Hi David, Ista and all,
>>
>> I have one related question Within one group I want to keep records
>> conditionally.
>> example within
>> group A I want keep rows that
2017 Dec 09
1
Remove
You could make numeric vectors, named by the group identifier, of the
contraints
and subscript it by group name:
> DM <- read.table( text='GR x y
+ A 25 125
+ A 23 135
+ A 14 145
+ A 35 230
+ B 45 321
+ B 47 512
+ B 53 123
+ B 55 451
+ C 61 521
+ C 68 235
+ C 85 258
+ C 80 654',header = TRUE, stringsAsFactors = FALSE)
>
> GRmin <- c(A=15, B=40, C=60)
> GRmax <-
2023 Dec 08
2
regarding CCA plot
Hii rstudio members
I am learning rstudio, For my manuscript I am trying to plot CCA using
species and environmental data. But I am getting error like
Error in cca.default(sptrans, envtrans) :
all row sums must be >0 in the community data matrix
*My code is like *
library(vegan)
library(ggplot2)
library(dplyr)
rassspec<-read.csv("C:/Users/hp/Desktop/R_data/rassspec.csv",
2009 Oct 14
2
Survival and nonparametric
Hi all,
Has any body the exprience to iclude a nonparametric component into the
survival analysis using R
package? *Can someone recommend *me * some ** references? *
Thanks a lot
Ashta
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2009 Sep 25
1
Binomial
Dear R-users,
Suppose I have the following sample of data,
0 1 2 4 3
1 2 1 3 1
1 3 3 4 1
0 1 2 1 2
1 4 1 4 2
1 2 2 1 1
The first variable is the response variable where 0 is defective and 1
normal. The other four factors( x1,x2,x3,x4) that influence the outcome. I
want to fit a binomial model . How do I do that? I am guessing the response
variable
2009 Oct 10
1
Creating new variables
Hi all,
I have a data set called x with 200 rows and 12 columns. I want
create two more columns based on probability. ie
if p >0 .4 then v1 =1 else v1=0;
if p >0 .6 then v2 =1 else v2=0;
Finally x will have 14 variables.
Can any one show me how to do that?
Thanks
Ashta
.
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2010 Mar 25
1
how to deal with vector[0]?
Hi,
I have a vector with 4 elements, e.g., tau_i=c(100,200,300,400), but
potentially tau_i[0]=0. In a "for" loop,
tau_i=c(100,200,300,400)
m=4
tau_i[0]=0 # <------- ?
P_i=1
for(i in 2:m)
{
P_i = P_i*(tau_i[i-1]-tau_i[i-2])
}
Error in P_i = P_i * (tau_i[k - 1] - tau_i[k - 2]):
replacement has length zero
Unfortunately, I can add this potential element into
2012 Dec 12
1
data download
I am trying to download the tar files on the website below
filename<-"
http://rredc.nrel.gov/solar/old_data/nsrdb/1991-2010/SolarAnywhere/x.tar"
where x is one those tar files
I downloaded x using download.file(). But, the file was corrupt. Can
someone help me how to download and untar these files using R.
Thanks,
Alemu
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2017 Dec 06
0
Remove
Thank you Jeff,
subset( DM, "B" != x ), this works if I know the group only.
But if I don't know that group in this case "B", how do I identify
group(s) that all elements of x have the same value?
On Wed, Dec 6, 2017 at 5:48 PM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote:
> subset( DM, "B" != x )
>
> This is covered in the Introduction to