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Hi,
I would rather have a Statistics related question hope experts here can provide
some suggestions. I have posted this request in some other forum but failed to
generate meaningful response
I am looking for some technical document on deriving the Distribution function
for sum of 2 ReLU(𝑋)=max{
= "%d%b%Y%H%M%OS")
[1] "2018-01-03 16:00:00 GMT"
_
Arun Kumar Saha, FRM
QUANTITATIVE RISK AND HEDGE CONSULTING SPECIALIST
LinkedIn: http://in.linkedin.com/in/ArunFRM
Personal : http://WWW.ARUNSAHA.IN <http://WWW.ARUNSAHA.IN>
___
,
country, city, ad type, browser, postal code etc. How do I approach the
analysis and modelling in this regard.
Thanks,
Arun
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by = ID], ID+Visit~ Diff, value.var='Diff', length)
ID Visit 136 255 857
1: 1 2 1 0 0
2: 2 3 0 1 1
On Wednesday, February 11, 2015 5:47 PM, farnoosh sheikhi
wrote:
Hi Arun,
I have a data set that look s like below. I wanted to get a
Try
indx <- which(!!mat, arr.ind=TRUE)
v1 <-unname(sapply(split(indx[,2], indx[,1]),toString))
cat(paste(v1, collapse="\n"), sep="\n")
1, 2, 3, 6, 7, 8, 9
1, 2, 3, 6, 8, 9
1, 3, 4, 6, 7, 8, 9
1, 8
1, 3, 6, 7, 8, 9
1, 3, 4, 6, 8, 9
1, 3, 5, 9
A.K.
Hi,
Is there a way to map data from Bina
Hi Juvin,
The error "dim(X) must have a positive length" usually shows when you are
passing a vector to "apply", ie.
apply(1:5,2,mean)
#Error in apply(1:5, 2, mean) : dim(X) must have a positive length
Also, if your dataset originally has "1206" columns, it is not clear why you
n
[1:3], pch=1, col=1:3)
A.K.
On Friday, December 5, 2014 5:45 PM, farnoosh sheikhi
wrote:
Hi Arun,
I hope you are doing well.
I have a data set as follow:
my.df <- data.frame(ID=rep(c("A","B","C"), 5), TIME=rep(1:5, each=3), X=1:5)
I would like to get a scat
Try
range(data2$oiddate[complete.cases(data2$oiddate) & is.finite(data2$oiddate)])
#[1] "2006-09-01" "2011-11-04"
If you look at the `dput` output, it is `Inf` for oiddate
dput(data2$oiddate)
structure(c(14078, -Inf, 15260, 13796, 13392, 15252, 15282), class = "Date")
A.K.
On Monday, Nov
ith that warning.
data1 <- data1[rowSums(is.na(data1[,-1]))!=4,]
data1 %>%
rowwise()%>%
mutate(oldflag= as.Date(max(mrjdate, cocdate, inhdate, haldate,
na.rm=TRUE), origin='1970-01-01')
A.K.
On Sunday, November 9, 2014 9:16 AM, "Muhuri, Pradip (SAMHSA/CB
You could try
library(dplyr)
data1 %>%
rowwise() %>%
mutate(oldflag=as.Date(max(mrjdate,cocdate, inhdate, haldate,
na.rm=TRUE), origin='1970-01-01'))
Source: local data frame [7 x 6]
Groups:
idmrjdatecocdateinhdatehaldateo
If the `ids` are ordered as shown in the example, perhaps you need
tbl <- table(df$id)
rep(seq(as.Date("2000-01-01"), length.out=length(tbl), by=1), tbl)
[1] "2000-01-01" "2000-01-01" "2000-01-01" "2000-01-01" "2000-01-01"
[6] "2000-01-02" "2000-01-02" "2000-01-02" "2000-01-02"
Hi,
Use ?format
format(d, "%m/%d/%Y")
#[1] "09/01/2014"
A.K.
On Monday, September 1, 2014 5:26 AM, Velappan Periasamy
wrote:
d=Sys.Date()
"2014-09-01"
How to convert this "2014-09-01" to "09/01/2014" format?
(ie y-m-d to m/d/y format)
thanks
veepsirtt
Try:
format(as.Date("05/07/2014", "%m/%d/%Y"), "%m")
#[1] "05"
#or
strptime("05/07/2014", "%m/%d/%Y")$mon+1
#[1] 5
A.K.
How to extract a Month from Date object?
almost 13 peoples visited my Question with out replying in New to R , i have
task yaar
don't mind plz could you HELP ME
How t
Hi Tom,
You could try:
library(stringr)
str_extract(x, perl("(?<=[A-Za-z]{4}/).*(?=/[0-9])"))
#[1] "S01-012"
A.K.
On Friday, August 15, 2014 12:20 PM, Tom Wright wrote:
Hi,
Can anyone please assist.
given the string
> x<-"/mnt/AO/AO Data/S01-012/120824/"
I would like to extract "S01-012"
A similar post was found in stackoverflow
(http://stackoverflow.com/questions/25328311/how-to-avoid-change-string-to-number-automaticlly-in-r),
which already got an accepted reply.
A.K.
On Friday, August 15, 2014 2:18 PM, Wenlan Tian wrote:
I was trying to save some string into a matrix, bu
You could try:
x1$V2[1] <- "TCLA1"
x[outer(rownames(x), colnames(x), FUN=paste) %in%
as.character(interaction(x1, sep=" "))] <- 1
x
TCLA1 VPS41 ABCA13 ABCA4
AKT3 1 0 0 0
AKTIP 0 1 0 0
ABCA13 0 0 0 0
ABCA4 0 0 0 0
A.
You could try:
lv <- levels(unique(unlist(df)))
as.data.frame(t(apply(df, 2, function(x) table(factor(x, levels=lv)
+ - 0
i1 10 0 0
i2 10 0 0
i3 0 10 0
i4 0 9 1
i5 10 0 0
i6 1 9 0
i7 9 0 1
i8 4 2 4
i9 7 1 2
A.K.
On Tuesday, August 5, 2014 5:36 AM, Alain D. wrote:
Dear R-
You could try data.table
#dat is the dataset
library(data.table)
v1 <- setNames(c("HT", "A", "B", "Aht", "Bht"), c("11", "10", "01", "1-", "-1"))
dat2 <- setDT(dat1)[, lapply(.SD, function(x) v1[paste(x, collapse="")]),
by=CloneID]
A.K.
On Monday, August 4, 2014 5:55 AM, raz wrote:
Dear a
You could use:
library(dplyr)
library(tidyr)
x.df %>% group_by(Year, Group, Eye_Color) %>% summarize(n=n()) %>%
spread(Eye_Color,n, fill=0)
Source: local data frame [6 x 5]
Year Group blue brown green
1 2000 1 2 1 0
2 2000 2 0 0 2
3 2001 1 1
Forgot about as.numeric.
sapply(str_extract_all(xx, perl('(?<=[A-Z]|\\:)\\d+')),as.numeric)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 24 24 24 24 24 24
[2,] 57 86 119 129 138 163
On Friday, August 1, 2014 10:59 AM, arun wrote:
You could try:
li
You could try:
library(stringr)
simplify2array(str_extract_all(xx, perl('(?<=[A-Z]|\\:)\\d+')))
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] "24" "24" "24" "24" "24" "24"
[2,] "57" "86" "119" "129" "138" "163"
A.K.
On Friday, August 1, 2014 10:49 AM, "Doran, Harold" wrote:
I have done an
Use ?split()
split(dat[,-4], dat$Year_Month) #dat is the dataset.
A.K.
Country Product Price Year_Month
AE 1 20 201204
DE 1 20 201204
CN 1 28 201204
AE 2 28 201204
DE 2
Try:
If dat is the dataset.
library(stringr)
res <- str_extract(dat$Gene.Symbol, perl('[[:alnum:]]+(?= \\/)'))
res[!is.na(res)]
#[1] "CDH23"
A.K.
On Thursday, July 31, 2014 9:54 PM, Stephen HK Wong wrote:
Dear All,
I appreciate if you can help me out this. I have a data frame cont
Try:
gsub(".*\\.(.*)\\..*","\\1", my.cache.list)
[1] "subject_test" "subject_train" "y_test" "y_train"
#or
library(stringr)
str_extract(my.cache.list, perl('(?<=\\.).*(?=\\.)'))
[1] "subject_test" "subject_train" "y_test" "y_train"
A.K.
On Thursday, July 31, 2014 11:05 AM,
If you have some variations of the order of numbers followed by chars,
library(stringr)
v1 <- c("absdfds0213451ab", "123abcs4145")
pattern=c("[A-Za-z]+", "\\d+")
do.call(`Map`,c(c,lapply(pattern, function(.pat) str_extract_all(v1, .pat
#[[1]]
#[1] "absdfds" "ab" "0213451"
#[[2]]
#[1] "
y, July 30, 2014 1:42 PM, farnoosh sheikhi
wrote:
Hi Arun,
I have two questions, I have a data like below:
dat1<-read.table(text="
Unit q1q2q3
A312
A2NA1
B224
BNA25
C32NA
C414
A32NA
",sep="",header=T,stringsAsFactors=F)
I want to get the average of each row by the numbe
1 0.4617737
13 CB27A 201307-201309 1 0.4513274
14 CB27A 201308-201310 1 0.4613779
A.K.
On Thursday, July 31, 2014 12:34 AM, arun wrote:
For the example, you gave:
x ##dataset
indx <- t(sapply(min(x$MTH_SUPPORT):(max(x$MTH_SUPPORT) - 2), function(x) c(x,
x +
For the example, you gave:
x ##dataset
indx <- t(sapply(min(x$MTH_SUPPORT):(max(x$MTH_SUPPORT) - 2), function(x) c(x,
x +
2)))
res <- do.call(rbind, apply(indx, 1, function(.indx) {
x1 <- x[x$MTH_SUPPORT >= .indx[1] & x$MTH_SUPPORT <= .indx[2], ]
Period <- paste(.indx[1], .indx[2],
Not able to reproduce the problem.
str(q)
# POSIXlt[1:1], format: "2014-03-09 02:00:00"
is.na(q)
#[1] FALSE
sessionInfo()
R version 3.1.0 (2014-04-10)
Platform: x86_64-unknown-linux-gnu (64-bit)
A.K.
On Wednesday, July 30, 2014 1:10 PM, John McKown
wrote:
"I'm so confused!" Why does is.na()
Try:
rbind(v2,unname(setNames(v1[,1],v1[,2])[v2]))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
v2 "a" "a" "a" "a" "a" "c" "c" "c" "c" "c" "c" "c" "c" "c"
"1" "1" "1" "1" "1" "3" "3" "3" "3" "3" "3" "3" "3" "3"
[,15] [,16]
Hi,
If `dat` is the dataset:
dat[!(dat$ID==2 & as.numeric(gsub("-.*","",dat$Month))<5),]
ID Month Value
1 1 03-2014 1
2 1 04-2014 10
3 1 05-2014 50
6 2 05-2014 4
7 2 06-2014 2
A.K.
hello together, i have a short question, maybe you can help me.
I have a data.frame li
Or
rapply(l,function(x) x[1])
#[1] 1 3 7
set.seed(42)
l1 <- replicate(1e6, list(sample(1:5,sample(8),replace=T)))
system.time(r1 <- sapply(l1, `[`, 1))
# user system elapsed
# 1.324 0.000 1.326
system.time(r2 <- rapply(l1, function(x) x[1]))
# user system elapsed
# 0.736 0.004
2 A x 2000-10-02
where f2 is your code wrapped up in a function (to make testing and use easier)
f2 <- function (data)
{
library(dplyr)
data %>% group_by(id, value) %>% arrange(date = as.Date(date)) %>%
filter(any(c(abs(diff(as.Date(date))), NA) > 31) & d
Hi,
If `dat` is the dataset
library(dplyr)
dat%>%
group_by(id,value)%>%
arrange(date=as.Date(date))%>%
filter(any(c(abs(diff(as.Date(date))),NA)>31)& date == min(date))
#Source: local data frame [3 x 3]
#Groups: id, value
#
# id date value
#1 a 2000-01-01 x
#2 c 2000-09-10 y
#3
You may also try:
library(reshape2)
A2 <- melt(A1)
split(A2[,2],A2[,1])
A.K.
On Tuesday, July 8, 2014 12:57 PM, Lorenzo Alfieri wrote:
Hi,
I'm trying to find a way to reorder the elements of a list.
Let's say I have a list like this:
A1<-list(c(1:4),c(2,4,5),23,c(4,5,13))
> A1
[[1]]
[1] 1 2
Hi,
Not sure about the expected output.
If `dat` is the dataset:
res <- dat[rep(1:nrow(dat), dat$score),]
head(res,7)
team year time score out top goals host format formed culture wcups cholder
1 ARG 1986 1 6 0 1 4 0 0 1893 93 8 0
1.1 ARG 1986 1
1 2.780045e-02 # SEC
A.K.
On Friday, July 4, 2014 10:27 AM, Gang Chen wrote:
I really your kind help! This is exactly what I was looking for except that I
need to get rid of the numbered row names.
On July 3, 2014 9:57:00 PM EDT, arun wrote:
Hi,
>May be this helps:
>nC <
Hi,
Try:
dat1 <- read.table(text="'1 > TC' 'WC'
'2 > 0' 'Instruments & Instrumentation; Nuclear Science & Technology;Physics,
Particles & Fields; Spectroscopy'
'3 > 0' 'Nanoscience & Nanotechnology; Materials Science,Multidisciplinary;
Physics, Applied'
'4 > 2' 'Physics, Nuclear; Physics, P
Hi,
May be this helps:
nC <- max(nchar(row.names(dd)))
term <- formatC(row.names(dd), width=-nC)
#or
term <- sprintf("%-11s", row.names(dd))
dd1 <- setNames(data.frame(unname(dd), term,stringsAsFactors=F),
c(colnames(dd), formatC("term",width=-nC)))
dd1
# # Chisq DF Pr(>Chisq) term
)
>> str(dat1)
>> --
>>
>>
>> I got this error,while running the above code under RStudio.,
>> how to correct it?.
>>
>> > dat1 <- readHTMLTable(tableNodes[[4]],
>> colClasses=c("numeric","character"
Hi veepsirtt,
If `dat` is the dataset
library(dplyr)
dat %>%
group_by(DATE) %>%
summarize(PROFIT=sum(PROFIT)) %>%
arrange(desc(as.Date(DATE,format="%d/%m/%Y")))
Source: local data frame [4 x 2]
DATE PROFIT
1 02/07/2014 -1350
2 01/07/2014 9400
3 30/06/2014 11325
4 27/06/2014
#or
res <- mapply(`%in%`, accel_data, v.to.match)
res1 <- sapply(seq_len(ncol(accel_data)),function(i)
accel_data[i]<=tail(v.to.match[[i]],1) & accel_data[i] >=v.to.match[[i]][1])
all.equal(res, res1,check.attributes=F)
#[1] TRUE
A.K.
On Tuesday, July 1, 2014 10:56 PM,
e-the-stack-size-using-ulimit-or-per-process-on-mac-os-x-for-a-c-or
A.K.
On Monday, June 30, 2014 10:08 PM, Jorge I Velez
wrote:
Hi Arun,
Thank you very much for your suggestion.
While running some tests, I came across the following:
# sample data
n <- 2000
p <- 1000
x2 <- data.
Hi Izhak,
If the position of the elements to be replaced follow the pattern below:
seq(1,length(t), by=7)
#[1] 1 8 15
t[seq(1,length(t), by=7)] <- c(50,90,100)
A.K.
On Monday, June 30, 2014 4:19 PM, "Adams, Jean" wrote:
t[1, 1] <- 50
t[3, 2] <- 90
t[5, 3] <- 100
Jean
On Mon, Jun 30, 20
Hi Jim,
May be this helps:
library(stringr)
paste(str_extract_all(initString,
perl('(?<=\\:)[[:alnum:]]+,?'))[[1]],collapse="")
#[1] "value1,value2"
A.K.
On Monday, June 30, 2014 9:58 AM, Jim Lemon wrote:
Hi all,
I have managed, with the help of glob2rx() to get two parts of a text
manipula
Hi Jorge,
You may try:
library(dplyr)
library(tidyr)
#Looks like this is faster than the other methods.
system.time({wide1 <- x2%>%
select(-rate) %>%
mutate(variable=factor(variable, levels=unique(variable)),id=factor(id,
levels=unique(id))) %>%
s
Hi,
You may try:
test[!grepl("(?<=AARSD1)[-\\d]", test, perl=T)]
A.K.
On Friday, June 27, 2014 6:43 AM, C Lin wrote:
Hi Duncan,
Thanks for trying to help. Sorry for not being clear.
The string I'd like to get is 'AARSD1'
It can be followed or preceded by white space or // or nothing
so, fro
You could use:
mapply(`==`, sapply(foo,`[`,1),1)
# A B C
# TRUE TRUE FALSE
A.K.
On Thursday, June 26, 2014 10:50 PM, ce wrote:
Dear all,
I have a list of arrays :
foo<-list(A = c(1,3), B =c(1, 2), C = c(3, 1))
> foo
$A
[1] 1 3
$B
[1] 1 2
$C
[1] 3 1
I want to use all foo$A
Hi,
May be you can use ?cut or ?findInterval for the range
dat1 <- read.table(text="100, 100, 200
250, 300, 350
100, 350, 100
400, 250, 300
200, 450, 200
150, 501, 300
150, 250, 300",sep=",",header=F)
sapply(dat1, findInterval, c(400,500))==1
# V1 V2 V3
#[1,] FALSE FALSE FALSE
#[2,]
I am very new to R. I am using R package "pscl" for hurdle regression (Binary
with "cloglog" link and Poisson with "log" link). My problem is to model road
crash frequencies at given locations as a function of road geometry and surface
condition. All of my dependent variable are covariates. I wa
Also, you can do:
library(dplyr)
dat%>%group_by(ID)%>%filter(length(unique(Disease))>1)%>%arrange(Disease,ID)
A.K.
On Wednesday, June 25, 2014 3:45 AM, arun wrote:
Forgot about:
library(reshape2)
On , arun wrote:
Hi,
Check if this works:
set.seed(495)
dat <
Hi,
May be this helps:
dat <- read.table(text="7.35600813091
7.35600956856
7.35601305921
7.35601654985
7.35602004049
7.35602353113
7.35602702178
7.35603397179
7.35604092182
7.35604787183
7.35605482185
7.35606177187
7.35606
Hi,
Check if this works:
set.seed(495)
dat <- data.frame(ID=sample(1:10,20,replace=TRUE),
Disease=sample(LETTERS[1:6], 20, replace=TRUE) )
subset(melt(table(dat)[rowSums(!!table(dat))>1,]), !!value,select=1:2)
ID Disease
1 2 A
3 4 A
4 6 A
6 10 A
8 3
To get the vector
setNames(stack(test)[,1],stack(test)[,2])
#ABCC2 ABCC5 ABCC5
#37280 12268 13308
#or
setNames(unlist(test),stack(test)[,2])
#ABCC2 ABCC5 ABCC5
#37280 12268 13308
A.K.
On , arun wrote:
You could do:
stack(test)[,2:1]
# ind values
#1 ABCC2 37280
#2 ABCC5 12268
#3
Hi,
Try:
set.seed(42)
X <- as.data.frame(matrix(sample(0:1, 4*50,replace=TRUE), ncol=4))
table(X[1:2])[4]
#[1] 15
sum(rowSums(X[1:2])==2)
#[1] 15
A.K.
On Saturday, June 21, 2014 10:59 AM, Kate Ignatius
wrote:
I have 4 columns, and about 300K plus rows with 0s and 1s.
I'm trying to count h
You could try:
library(plyr)
res <- ddply(TestData[,-2],.(ID),numcolwise(max))
colnames(res)[-1] <- paste0(colnames(res)[-1],".max")
A.K.
On Friday, June 20, 2014 3:43 PM, Jun Shen wrote:
Hi all,
Here is my situation. I have a dataframe, the structure would be something
like this,
TestData<
To get the maximum:
lapply(split(df, df$Year), function(x) max(sapply(1:(nrow(x)-2), function(i)
with(x, mean(Amount[i:(i+2)],na.rm=TRUE)
#$`1985`
#[1] 4.17
A.K.
On , arun wrote:
Hi,
You may try:
df <- structure(list()
lapply(split(df, df$Year), function(x) sapply(1:(nro
Hi,
You may try:
df <- structure(list()
lapply(split(df, df$Year), function(x) sapply(1:(nrow(x)-2), function(i)
with(x, mean(Amount[i:(i+2)],na.rm=TRUE
#$`1985`
# [1] 4.167 3.833 0.833 0.333 0.333 0.000 0.000
# [8] 0.000 0.000 0.000 0.000 0.0
III Speci BN 7/13/2011 Positive
A.K.
Thanks a lot for the reply Arun.
I would like to pull the quarter for that particular year. say i want Q2 for
year 2014 r Q2 for year 2013.
You assistance is very much appreciated.
Thanks Raghu
On Wednesday, June 11, 2014 10:19 AM, arun wrote:
homework ... but
just for a graph I'm trying do plot.
@Arun: Yes I've done a mistake, it is 1-100 , 101-200 etc .. or even
1-50,51-100 etc. The range is not important.
Thanks
> Date: Thu, 12 Jun 2014 02:04:26 -0700
> From: smartpink...@yahoo.com
> Subject: Re: [R] Aver
Hi,
If your working directory is "data 3" which has only two folders C1 and C2.
May be this helps:
lsF1 <- list.files(recursive=TRUE)
lst1 <- lapply(split(lsF1,gsub(".*\\_(\\d+)\\..*","\\1",lsF1)),function(x)
do.call(rbind,lapply(x,function(y) read.csv(y, header=TRUE
A.K.
Hello,
I am tryi
Hi,
Please Check ?as.yearqtr() fromlibrary(zoo)
dat <- read.table(text="ID Phase RESEARCH Area Date Result
100 IV S_Care A&P 7/23/2013 Positive
1001 IV P_Care EU 12/20/2012 Positive
2001 IV Car AS 9/13/2012 Positive
20003 III Spec IN 10/6/2011
then it is more usual to use n-1 as the
denominator (this makes the variance an unbiased estimator of the
population variance). That is what the R sd function does since it is
much more common to use it on a sample rather than an entire
population.
On Sun, Jun 8, 2014 at 1:17 AM, arun
Hi,
Please use ?dput() to show the data.
Assuming that the data is:
dat <- structure(list(X = c("ks", "ks", "tk", "tk", "tk", "ks", "eq",
"eq", "ks"), Y = c("dr", "zw", "dr", "", "zw", "zw", "", "zw",
"zw"), Z = c("tq", "tq", "tq", "", "tz", "", "tz", "tz", "tz"
)), .Names = c("X", "Y", "Z"), cl
Hi Alexsandro,
Suppose if you have strings
nw.str1 <- "[D][A|D]A:F[T|A:D]N[C|T]"
nw.str2 <- "[D][A|D]A[T|A:D][C|T]NA{DG]P"
you could use:
library(qdap)
as.vector(bracketXtract(nw.str1,"square",T))
#[1] "[D]" "[A|D]" "[T|A:D]" "[C|T]"
as.vector(bracketXtract(nw.str2,"square",T))
#[1] "
Hi,
If you have library(qdap) installed:
library(qdap)
as.vector(bracketXtract(nw.str,"square",with=T))
#[1] "[D]" "[A|D]" "[T|A:D]" "[C|T]"
A.K.
On Sunday, June 8, 2014 4:31 PM, Alexsandro Cândido de Oliveira Silva
wrote:
Hi,
I have a string something like that:
nw.str <- "[D][A|D
Hi VP,
Not sure what you wanted.
Perhaps:
library(XML)
URL <-
"http://money.securebank.in/index.php?option=com_dashboard&view=history&Itemid=56&startdate=01/01/2013&enddate=6/9/2014&exchange=MCX&sid=1";
doc <- htmlParse(URL)
tableNodes <- getNodeSet(doc, "//table")
dat1 <- readHTMLTable(tableN
Hi,
You may try:
fun1 <- function(len, low, high, mean, sd) {
x <- rnorm(len * 2, mean, sd)
x <- x[x < high & x > low]
x[1:len]
}
##slow
fun2 <- function(len, low, high, mean, sd) {
i <- 1
while (i < len) {
x <- rnorm(1, mean, sd)
if (x < high & x > low) {
Hi,
Please check this link:
http://stats.stackexchange.com/questions/25956/what-formula-is-used-for-standard-deviation-in-r
A.K.
It is my understanding that the R function SD finds the standard deviation of a
random variable or a list. Please consider the following list: { 1, 2, 3 }. I
claim t
Hi,
This is not clear.
If this is a combination of rows using a specific formula as you showed, use
?combn
dat <- read.table(text="0.7, 0.3, 0.6, 0.9
1.0, 0.1, 0.4, 0.7
1.2, 0.8, 0.3, 0.1",sep=",",header=FALSE)
indx <- combn(seq(dim(dat)[1]), 2)
vec1 <- c(0.86, 0.46, 0.5, 0.63)
indx1 <- sapply(
(NA, roa[-.N]), c(NA, diff(eta, by =
identif]
# or
dt3[, `:=`(roa1 = c(NA, roa[-.N]), eta1 = c(NA, diff(eta))), by = identif]
identical(dt1, dt2)
# [1] TRUE
identical(dt1,dt3)
#[1] TRUE
identical(dat2, as.data.frame(dt1))
# [1] TRUE
A.K.
On Friday, June 6, 2014 10:47 PM, arun wrote:
Hi,
Hi,
May be this helps:
dat1 <- read.table(text="identif roa eta
1 7 5
2 8 9
2 9 8
2 10 7
3 11 6
3 1 4
3 2 2
4 3 3
4 6 5",sep="",header=TRUE)
dat2 <-within(dat1, {
eta1 <- ave(eta, identif, FUN = function(x) c(NA, diff(x)))
roa1 <- ave(roa, identif, FUN = function(x) c(NA, x[-length(x)]))
_times_bought_3_days <- ave(x$items_bought, indx[indx1], FUN = cumsum)
indx2 <- seq(0, length(indx), 4)
indx3 <- c(FALSE, diff(indx[indx1]) > 0)
x[indx3, 4] <- x[indx3, 4] + indx[indx2]
x
}), dat1$user)
##Here also, the same confusion persists.
A.K.
On Thursday,
Hi,
The expected output is confusing.
dat1 <- read.table(text="date, user, items_bought
2013-01-01, x, 2
2013-01-02, x, 1
2013-01-03, x, 0
2013-01-04, x, 0
2013-01-05, x, 3
2013-01-06, x, 1
2013-01-01, y, 1
2013-01-02, y, 1
2013-01-03, y, 0
2013-01-04, y, 5
2013-01-05, y, 6
2013-01-06, y, 1",sep=",
Hi York,
Using "dat" (assuming that the data is ordered by "ID")
dat$NEWID <- cumsum(c(TRUE,with(dat, ID[-1]!= ID[-length(ID)])))
A.K.
Hi arun,
Thanks for your reply. I think you misunderstood my question, for the new ID, I
would like to have the following pattern,
I
Hi,
May be this helps:
dat <- read.table(text="DATE Price
01.01.2010 2
01.01.2010 3
01.01.2010 2
01.01.2010 7
02.01.2010 3
02.01.2010 9
02.01.2010 0
03.01.2010 2
03.01.2010 4
03.01.2010 3
03.01.2010 6
03.01.2010 8",sep="",header=TRUE,strin
Hi,
Using the example data,
library(zoo)
z1 <- read.zoo("dfRaw_20140509.csv",sep="",index.column=2,format="%Y%m%d")
z1
# P_num P_ID Totalvalue
#2013-05-10 25193 1230238 1.203
#2013-05-24 25190 1230238 1.201
#2013-07-05 25191 1230238 1.208
#2013-08-02 25194 1230238
Hi Wayne,
As John mentioned, it is not clear about your expectations. May be this helps:
library(xts)
x2New <- xts(test[,-1], order.by=test[,1])
library(xtsExtra)
plot(x2New,screens=1,auto.legend=TRUE)
A.K.
On Tuesday, June 3, 2014 9:29 AM, John Kane wrote:
I know essentially nothing abou
1 5 0% 100% 0%
#6 2 1 60% 20% 20%
A.K.
I think I want to have something like this:
Unit, MD, Response 1, Response 2, Response 3
11 90%8%2%
1280%10% 10%
2395%3%2%
Does this make sense to you?
Regards, Farnoosh Sheikhi
On Tuesday, May 2
.3 13.67 11.67 11.0
#or
res1[!grepl("SCHOOLID", colnames(res1))]
A.K.
I tried to explain all the things that I want to do in this picture :) Sorry,
if it's not so understandable, but I tried :)
On Monday, June 2, 2014 4:02 AM, arun wrote:
Hi,
Regarding
Hi,
Please check this link:
http://r.789695.n4.nabble.com/meaning-of-asymmetric-on-help-page-for-intersect-td877408.html
union(setdiff(v,w), setdiff(w,v))
#or in this case
setdiff(union(v,w),intersect(v,w))
#or
setdiff(c(v,w),c(v,w)[duplicated(c(v,w))])
A.K.
.pae...@gmail.com> wrote:
Hello
abet). And according to the result it counts not an avreage
values of the factors.
On Sunday, June 1, 2014 8:37 PM, arun wrote:
Hi,
May be this helps:
set.seed(42)
rev1 <- data.frame(SCHOOLID=sample(LETTERS[1:4],20,replace=TRUE),
matrix(sample(25, 20*5,replace=TRUE), ncol=5, dimnames=lis
Hi,
May be this helps:
set.seed(42)
rev1 <- data.frame(SCHOOLID=sample(LETTERS[1:4],20,replace=TRUE),
matrix(sample(25, 20*5,replace=TRUE), ncol=5, dimnames=list(NULL, c("MATH",
"AGE", "STO2Q01", "BFMJ", "BMMJ"))),stringsAsFactors=FALSE)
res1 <- aggregate(rev1[,-1], list(SCHOOLID=rev1[,1]), m
Hi,
May be this helps:
vec1 <- c("Humanities, Multidisciplinary; Social Sciences, Interdisciplinary;
History",
"History","Sociology; Religion","Humanities, Multidisciplinary;","Social
Sciences, Interdisciplinary","Literature","Sociology; Religion")
res1 <- as.data.frame(table(unlist(strsplit(v
Hi,
You may try:
##Assuming the dataset is a matrix
mat <- structure(c("FBgn0037249", "FBgn0036389", "FBgn0014002", "FBgn0034201",
"FBgn0029860", "FBgn0028526", "FBgn0003486", "FBpp0312226", "FBpp0312225",
"FBpp0312224", "FBpp0312223", "FBpp031", "FBpp0312221", "FBpp0312220",
"FBtr0346646"
Hi,
I need to merge two data sets but I need them to match two columns, not just
one. For example:
Hi,
If 'dat1` and 'dat2` are the two datasets.
merge(dat1,dat2,by=c("plant.species","insect.species"),all=TRUE)
# plant.species insect.species visit.freq no.grains
#1 p.A i.1
Hi,
You can try ?merge() or ?join() from library(plyr)
merge(tempr, pr, by="date",all=TRUE)
A.K.
Hi, all!
I have a problem. I have 2 data frame. Each contains column: "date" and one
unique column (temperature and pressure correspondingly). But dates are not a
same. First one since 1st Jan 20
Hi,
If it ordered by the variable "x", you could also try:
within(DF1, new_c<-sequence(table(x)))
A.K.
On Wednesday, May 28, 2014 6:14 PM, Andrija Djurovic
wrote:
Here is another approach:
x <- rep(c("A", "B", "C"), c(3,1,2))
DF1 <- data.frame(x)
cbind(DF1, new_c=ave(as.numeric(DF1$x),
Hi,
May be this helps:
data1 <- data[with(data, order(col1, col2,1*is.na(col3))),]
data1[!duplicated(data1[,1:2]),]
A.K.
On Wednesday, May 28, 2014 11:28 AM, jeff6868
wrote:
Hi everybody,
I have a little problem in my R-code which seems be easy to solve, but I
wasn't able to find the soluti
Hi Eliza,
No problem.
You can also do:
lst1 <- vector("list",12)
for(i in seq_along(lst1)) lst1[[i]] <- colMeans(AAA[seq(i,nrow(AAA), by=12),])
A.K.
On Wednesday, May 28, 2014 4:17 AM, eliza botto wrote:
Thankyou very much dennis and arun,
The codes worked as ever.
:D
HI,
data6$Gen=paste("G",data6$Gen,sep="")
#Error in paste("G", data6$Gen, sep = "") : object 'data6' not found
Please check this link:
http://stackoverflow.com/questions/7027288/error-could-not-find-function-in-r
A.K.
R-Community,
I am new to r. I have used SAS all my life. SO, here I am try
Hi,
You can also try:
dat <- read.table(text="Name C1 C2 C3
1 A 3 3 5
2 B 2 7 4
3 C 4 3 3
4 C 4 4 6
5 D 5 5 3",sep="",header=TRUE,stringsAsFactors=FALSE)
library(plyr)
ddply(dat,.(Name),numcolwise(mean,na.rm=TRUE))
A.K.
On Tuesday, May 27, 2014 4:08 PM, Verena We
Forgot, about the mean:
lapply(split(seq_len(nrow(AAA)),((seq_len(nrow(AAA))-1)%%12)+1),function(i)
colMeans(AAA[i,]))
A.K.
On , arun wrote:
Hi Eliza,
May be this helps:
lapply(split(seq_len(nrow(AAA)),((seq_len(nrow(AAA))-1)%%12)+1),function(i)
AAA[i,])
A.K.
On Tuesday, May 27, 2014
Hi Eliza,
May be this helps:
lapply(split(seq_len(nrow(AAA)),((seq_len(nrow(AAA))-1)%%12)+1),function(i)
AAA[i,])
A.K.
On Tuesday, May 27, 2014 6:48 PM, eliza botto wrote:
Dear R family,
I have this matrix say
AAA<-matrix(sample(1:240),ncol=2)
I first want to combine every 13th row in both co
Hi,
Please post in plain text and show the example using ?dput.
Assuming that you wanted a "data.frame" and not a "list"
May be this helps:
set.seed(428)
dat1 <- data.frame(`block identifier`=rep(1:40,each=12),`block
size`=12,`sequence within block`=rep(1:12,40), treatment=sample(paste("Group"
Hi,
I guess ?cast should work on the original dataset as it is in the long format
cast(data, siteS ~ species,value="abundance")
siteS sa sb sc sd se sg
1 11a 11 NA 37 NA NA 51
2 12d 15 NA NA NA NA NA
3 1a 31 55 62 NA NA NA
4 2v 42 NA NA 40 NA NA
5 6a 30 23 74 84 10 NA
#if you ar
Neighbouring NestkastNummer=",
ind2, "\n", " EBScore=", XOind1, "\n", " EBScore Neighbour element=",
XOind2,
"\n", " avg=", mean(abs(XOind1 - XOind2)), sep = " ", "\n")
mean(abs(XOind1 - XOind2)
, "50,000-99,999.99", "1,00,000-1,49,999.99",
"1,50,000-1,79,999.99", "above 1,80,000"
# INCOME INCOME_RANGE
#1 20100 below 25,000
#2 26800 25,000-49,999.99
#3 5 50,000-99,999.99
#4 18 above 1,80,000
A.K.
Hi Arun,
Thanks
tly with the
wavelet package for this specific example. I am trying to identify the packages.
Many thanks again
Barnaby
On Monday, May 26, 2014 1:22 PM, arun wrote:
Hi,
I couldn't reproduce the error.
str(wt.aligned)
Formal class 'dwt' [package "wavelets"] with 11 slot
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