Hi Jingxia,
May be this helps:
dat1 <- read.table(text="fatfreemilk fatmilk halfmilk 2fatmilk
A A A A
A B B A
B A A A
C C C C
D . A A
A E A E
C A B A
A . A A
A B . A
A A B
E",sep="",header=TRUE,stringsAsFactors=FALSE,check.names=FALSE,na.strings=".")
dat2 <- dat1
dat2$id <- 1:nrow(dat2)
librar
Also,
You can do the same with the previous solution:
result1 <- result[,-6]
vec1 <- unique(unlist(dat1))
result2 <- as.data.frame(t(sapply(dat1,function(x) {counts<-
table(factor(x,levels=vec1));
percentage<-sprintf("%.1f",(counts/sum(counts))*100);
c(paste0(counts,paste0("(",percentage,")")),
Hi,
Another way is:
vec1 <- unique(unlist(dat1))
res3 <- as.data.frame(t(sapply(dat1,function(x) {counts<-
table(factor(x,levels=vec1)); percentage<-(counts/sum(counts))*100;
paste0(counts,paste0("(",percentage,")"))})))
colnames(res3) <- vec1
identical(res3,as.data.frame(res2))
#[1] TRUE
A
Hi,
Try:
dat1 <- read.table(text="fatfreemilk fatmilk halfmilk 2fatmilk
A A A A
A B B A
B A A A
C C C C
D A A A
A E A E
C A B A
A A A A
A B B A
A A B E",sep="",header=TRUE,stringsAsFactors=FALSE,check.names=FALSE)
dat2 <- dat1
dat2$id <- 1:nrow(dat2)
library(reshape2)
res <- dcast(melt(dat2,id.v
Is this homework? (We generally don't do homework here).
However, hint: ?table and links therein.
Also, as you can see below, post in plain text, not HTML, which is
stripped and can lead to hard-to-read gobbledygook.
Cheers,
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data
Dear R helpers,
I have a data sheet (milk) with four types of milk from five brands (A,
B, C, D, E), the column shows the brands that each customer chose for each
type of the milk they bought. The data sheet goes like below. You can see
for some type of milk, no brand is chosen.
fatfreemilk fat
Also, you could try:
library(plyr)
ddply(dat1,.(A),mutate, D=length(A))
#or
library(data.table)
dt1 <- data.table(dat1,key='A')
dt1[,D:=.N,by='A']
A.K.
On , arun wrote:
Hi,
Try:
dat1 <- read.table(text="A B C
1 a 1 ff
2 b 1 re
3 c 1 sd
5 a 2 as
6 c 4 fe
7
Hi,
Try:
dat1 <- read.table(text="A B C
1 a 1 ff
2 b 1 re
3 c 1 sd
5 a 2 as
6 c 4 fe
7 d 5 tt
8 d 5 tt
9 d 9 oi",sep="",header=TRUE,stringsAsFactors=FALSE)
within(dat1,D<-ave(seq_along(A),A,FUN=length))
A.K.
On Thursday, December 26,
Hi everyone,
My data looks like this one:
AB C
1 a 1 ff
2 b 1 re
3 c 1 sd
5 a 2 as
6 c 4 fe
7 d 5 tt
8 d 5 tt
9 d 9 oi
I want to add a new column D, which will contain how many an element from
the column A is repeated,
got it, did it, worked ! thanks a lot !
--
View this message in context:
http://r.789695.n4.nabble.com/counts-on-bars-on-barplot-tp4630060p4630114.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https:
I want to label each bar with the values of each bar (counts) on top of each
of them.
--
View this message in context:
http://r.789695.n4.nabble.com/counts-on-bars-on-barplot-tp4630060p4630107.html
Sent from the R help mailing list archive at Nabble.com.
_
If you use simple barplot graphic try this, e.g.:
a<-barplot(some.data)
text(a, some.data, some.data)
The "text" command needs x and y coordinates, by assigning your barplot to
"a" (here), you have the x coordinats and by using the y-values of the
barplot, the values will be written ontop of the
:34 -0700 (PDT)
> To: r-help@r-project.org
> Subject: [R] counts on bars on barplot
>
> How can I have the corresponding counts shown on each bar?
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/counts-on-bars-on-barplot-tp4630060.html
> Sent fro
How can I have the corresponding counts shown on each bar?
--
View this message in context:
http://r.789695.n4.nabble.com/counts-on-bars-on-barplot-tp4630060.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing
Hello,
Just look at your second 'apply': the 'function(x)' is not using the 'x' (!)
Solution:
function(x) hist(x, breaks=0:nrow(test), plot=FALSE)$counts)
Note that instead of 'nrow' you could also use 'length(x)'.
Hope this helps,
Rui Barradas
--
View this message in context:
http://r.789
Excellent! Thank you!
ben
On Wed, Oct 5, 2011 at 9:18 PM, Dennis Murphy wrote:
> Hi:
>
> Here's one way:
>
> m <- matrix(rpois(100, 8), nrow = 5)
> f <- function(x) {
>q <- quantile(x, c(0.1, 0.9), na.rm = TRUE)
>c(sum(x < q[1]), sum(x > q[2]))
>}
>
> t(apply(m, 1, f))
>
> HTH,
> De
Hi:
Here's one way:
m <- matrix(rpois(100, 8), nrow = 5)
f <- function(x) {
q <- quantile(x, c(0.1, 0.9), na.rm = TRUE)
c(sum(x < q[1]), sum(x > q[2]))
}
t(apply(m, 1, f))
HTH,
Dennis
On Wed, Oct 5, 2011 at 8:11 PM, Ben qant wrote:
> Hello,
>
> I'm trying to get the count of value
Hello,
I'm trying to get the count of values in each row that are above and below
quantile thresholds. Thanks!
Example:
> x = matrix(1:30,5,6)
> x
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]16 11 16 21 26
[2,]27 12 17 22 27
[3,]38 13 18 23 28
[4,]4
Thank you all!
Particulary Wu Gong whose answer is very compact
--
View this message in context:
http://r.789695.n4.nabble.com/counts-of-a-vector-tp2232047p2232353.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org m
### Special cumsum
### Do cumsum when TRUE, and reset to 0 when FALSE
x <- c( TRUE, TRUE, TRUE, FALSE, FALSE, TRUE, TRUE, FALSE, TRUE )
## The rle function computes the lengths of equal values of a vector
## And gives two vecters: $lengths and $values
## The sequence function creates and concaten
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of speretti
> Sent: Wednesday, May 26, 2010 10:22 AM
> To: r-help@r-project.org
> Subject: [R]
Check out this thread for numerous solutions:
https://stat.ethz.ch/pipermail/r-help/2007-June/134557.html
On Wed, May 26, 2010 at 1:22 PM, speretti wrote:
>
> Hi,
>
> I need help to find an efficient way to transform a vector like:
>
> a<-c(1,1,0,1,0,0,0,1,1,1,1,0,1,0,1,1)
>
> in a vector that co
speretti wrote:
Hi,
I need help to find an efficient way to transform a vector like:
a<-c(1,1,0,1,0,0,0,1,1,1,1,0,1,0,1,1)
in a vector that counts only di 1 elements, like:
b<-c(1,2,0,1,0,0,0,1,2,3,4,0,1,0,1,2)
Thank you!
One way:
rl <- rle(a)
unlist(mapply("*", lapply(rl$lengths, fun
Hi,
I need help to find an efficient way to transform a vector like:
a<-c(1,1,0,1,0,0,0,1,1,1,1,0,1,0,1,1)
in a vector that counts only di 1 elements, like:
b<-c(1,2,0,1,0,0,0,1,2,3,4,0,1,0,1,2)
Thank you!
--
View this message in context:
http://r.789695.n4.nabble.com/counts-of-a-vector-tp
Try using aggregate( )
--
View this message in context:
http://n4.nabble.com/counts-of-elements-in-vector-tp1704036p1710369.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailma
On Mar 29, 2010, at 6:52 PM, Ali Tofigh wrote:
Assume you have a vector of characters x:
x
[1] "a" "b" "a" "d" "d" "c"
I use a function that counts the number of times each string occurs
in x:
sapply(unique(x), function(s) {sum(x == s)})
a b d c
2 1 2 1
Is there a more efficient way
Assume you have a vector of characters x:
> x
[1] "a" "b" "a" "d" "d" "c"
I use a function that counts the number of times each string occurs in x:
> sapply(unique(x), function(s) {sum(x == s)})
a b d c
2 1 2 1
Is there a more efficient way of doing this?
Cheers,
/Ali
Gabor Csardi wrote:
>On Mon, Feb 04, 2008 at 03:21:10PM +0800, Ng Stanley wrote:
>
>
>>Hi,
>>
>>Given a test matrix, test <- matrix(c(1,2,3,NA,2,3,NA,NA,2), 3,3)
>>
>>A) How to compute the counts of each column (excluding the NA) i.e., 3, 2, 1
>>
>>
>
>apply(test, 2, function(x) sum(!is.na(x
hits=-2.5 tests�YES_00,FORGED_RCVD_HELO
X-USF-Spam-Flag: NO
check the following options:
test <- matrix(c(1,2,3,NA,2,3,NA,NA,2), 3, 3)
# A
colSums(!is.na(test))
# B
mat <- test > rep(colMeans(test, na.rm = TRUE), each = nrow(test))
colSums(!is.na(mat) & mat)
apply(test, 2, function(x) {
m
On Mon, Feb 04, 2008 at 03:21:10PM +0800, Ng Stanley wrote:
> Hi,
>
> Given a test matrix, test <- matrix(c(1,2,3,NA,2,3,NA,NA,2), 3,3)
>
> A) How to compute the counts of each column (excluding the NA) i.e., 3, 2, 1
apply(test, 2, function(x) sum(!is.na(x)))
> B) How to compute the counts of e
Hi,
Given a test matrix, test <- matrix(c(1,2,3,NA,2,3,NA,NA,2), 3,3)
A) How to compute the counts of each column (excluding the NA) i.e., 3, 2, 1
B) How to compute the counts of each column (excluding the NA) that are
greater than the column means ? i.e., 1, 1, 0
I could write a for loop, but
31 matches
Mail list logo
