Thank you, that helps.
-Original Message-
From: Jeff Newmiller [mailto:jdnew...@dcn.davis.ca.us]
Sent: Monday, March 5, 2018 3:36 PM
To: Sariya, Sanjeev
Cc: r-help@r-project.org; R Help
Subject: RE: [R] Help with apply and new column?
Comments interspersed, and some code at the end
t;:" ) # gets the whole column splits at once
# wildly guessing here
rs_chrmatrix <- do.call( rbind, temp )
rs_DF <- as.data.frame( rs_chrmatrix, stringsAsFactors = FALSE )
names( rs_DF ) <- c( "CHR", "P", "X1", "X2" )
rs_DF$P <- as.integer( rs_DF$P )
str( rs_DF )
##
data)<-c("SNP","P","CHR","BP")
-Original Message-
From: Jeff Newmiller [mailto:jdnew...@dcn.davis.ca.us]
Sent: Monday, March 5, 2018 1:48 PM
To: r-help@r-project.org; Sariya, Sanjeev ; R Help
Read the Posting Guide... (see message footer) ... some relevant things you can
find there:
a) Yes, this appears to be about how to use an R base function so it is on topic
b) Post a reproducible example (include some sample data, preferably using the
dput function)
c) Post using plain text so t
Hi
I would just use for cycle. Something like
lll <- vector("list", length(InputFiles))
for (i in InputFiles) {
everything what you want to do for each file
you can store it in a list
lll[[i]] <- whatever you want to store there
}
After populating list you can do anything with it.
Regards
Pet
M.Ribeiro wrote:
>
> Ok, I am sorry, I was building a reproducible example to post here and I
> made a mistake. Th second loop (in the variable k) updates ycorr.
> Therefore, it would be
>
> ycorr <- rnorm(10)
> Nindiv <- 10
> x <-matrix(sample(c(0:2),50,re = T),10)
> gtemp <- rnorm(10)
> for (i
Ok, I am sorry, I was building a reproducible example to post here and I made
a mistake. Th second loop (in the variable k) updates ycorr.
Therefore, it would be
ycorr <- rnorm(10)
Nindiv <- 10
x <-matrix(sample(c(0:2),50,re = T),10)
gtemp <- rnorm(10)
for (i in 1:Nindiv) {
for (k in 1:nco
Hi:
Are you *sure* that's what you want to do?
> y - x[,5] * gtemp[5]
[1] 0.10116205 2.46220119 3.04966544 0.56074569 -1.14872521 -0.89275228
[7] 3.52161381 0.64880894 0.70167063 -0.06120996
> identical(as.vector(ycorr), y - x[,5] * gtemp[5])
[1] TRUE
Perhaps you ought to explain what y
keley.edu]
> Sent: Monday, October 04, 2010 2:20 PM
> To: Doran, Harold
> Cc: Gabriela Cendoya; R-help
> Subject: Re: [R] Help with apply
>
> Harold -
> The first way that comes to mind is
>
> doit = function(i,j)exp(sum(log((exp(tmp[i,]*(qq$nodes[j]-b))) /
>
ld
Cc: Gabriela Cendoya; R-help
Subject: Re: [R] Help with apply
Harold -
The first way that comes to mind is
doit = function(i,j)exp(sum(log((exp(tmp[i,]*(qq$nodes[j]-b))) /
(factorial(tmp[i,]) * exp(exp(qq$nodes[j]-b)) *
((1/(s*sqrt(2*pi))) * exp(-((qq$nodes[j]-0)^2/(
PM
To: Doran, Harold
Cc: R-help
Subject: Re: [R] Help with apply
You are missing "s" in your definitions so I can't reproduce your code.
tmp <- data.frame(var1 = sample(c(0:10), 3, replace = TRUE), var2 =
sample(c(0:10), 3, replace = TRUE), var3 = sample(c(0:10), 3, repl
On Mon, Oct 4, 2010 at 10:44 AM, Gabriela Cendoya
wrote:
> You are missing "s" in your definitions so I can't reproduce your code.
When he uses apply(), he sets s = 1.
>
>> tmp <- data.frame(var1 = sample(c(0:10), 3, replace = TRUE), var2 =
>> sample(c(0:10), 3, replace = TRUE), var3 = sample(c
Sorry about that; s <- 1
-Original Message-
From: Gabriela Cendoya [mailto:gabrielacendoya.rl...@gmail.com]
Sent: Monday, October 04, 2010 1:44 PM
To: Doran, Harold
Cc: R-help
Subject: Re: [R] Help with apply
You are missing "s" in your definitions so I can't reproduce
You are missing "s" in your definitions so I can't reproduce your code.
> tmp <- data.frame(var1 = sample(c(0:10), 3, replace = TRUE), var2 =
> sample(c(0:10), 3, replace = TRUE), var3 = sample(c(0:10), 3, replace = TRUE))
>
> str(tmp)
'data.frame': 3 obs. of 3 variables:
$ var1: int 9 3 9
On 8/02/2010 4:33 PM, Jim Lemon wrote:
On 02/08/2010 12:26 PM, Nathan S. Watson-Haigh wrote:
I have a 2 column data.frame:
> d[1:5,]
a b
1 80015 C
2 80016 B
3 80023 C
4 80062 B
5 80069 B
I want to apply a function across each row:
> for(i in 1:nrow(d)) {
+ myFun(con, d[i,]$a, d[i,]
On 8/02/2010 2:08 PM, David Winsemius wrote:
On Feb 7, 2010, at 8:26 PM, Nathan S. Watson-Haigh wrote:
I have a 2 column data.frame:
d[1:5,]
a b
180015 C
280016 B
380023 C
480062 B
580069 B
I want to apply a function across each r
On 02/08/2010 12:26 PM, Nathan S. Watson-Haigh wrote:
I have a 2 column data.frame:
> d[1:5,]
a b
1 80015 C
2 80016 B
3 80023 C
4 80062 B
5 80069 B
I want to apply a function across each row:
> for(i in 1:nrow(d)) {
+ myFun(con, d[i,]$a, d[i,]$b)
+ }
How do I do this using apply()? I'm unsu
On Feb 7, 2010, at 8:26 PM, Nathan S. Watson-Haigh wrote:
I have a 2 column data.frame:
> d[1:5,]
a b
180015 C
280016 B
380023 C
480062 B
580069 B
I want to apply a function across each row:
> for(i in 1:nrow(d)) {
+myFun(con, d[i,]$a, d[i,]$b)
+
> datg=t(sapply(split(datgic, datgic$NPERMNO, drop=TRUE),
function(x){return(
> x[nrow(x),] )}))
>
>
> I get something like this...
>
> GVKEY NPERMNO GIC year
> 10001 12994 10001 55102010 2007
> 10002 19049 10002 40101015 2007
> 10009 16739 10009 40101010 1999
>
> Has this
What you are seeing are the row numbers of the original locations. If
datgic is really a data frame, this is probably what you want using
lapply and do.call:
> x <- data.frame(a=letters, b=sample(1:4, 26, TRUE))
> y <- lapply(split(x, x$b), tail, 1)
> do.call(rbind, y)
a b
1 z 1
2 y 2
3 w 3
4 x
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