I'm struggeling to get a good performing script for this problem: I have a table with a score, x, y. I want to sort the table by score and than build groups based on the x value. Each group should have an equal sum (not counts) of x. x is a metric number in the dataset and resembles the historic turnover of a customer.
score x y
0.436024136 3 435
0.282303336 46 56
0.532358015 24 34
0.644236597 0 2
0.99623626 0 4
0.557673456 56 46
0.08898779 0 7
0.702941303 453 2
0.415717835 23 1
0.017497461 234 3
0.426239166 23 59
0.638896238 234 86
0.629610596 26 68
0.073107526 0 35
0.85741877 0 977
0.468612039 0 324
0.740704267 23 56
0.720147257 0 68
0.965212467 23 0
a good way to do so is adding a group variable to the data.frame with cumsum! Now you can easily sum the groups with e. g. subset.
data.frame$group <-cumsum(as.numeric(data.frame$x)) %/% (ceiling(sum(data.frame$x) / 3)) + 1
remarks:
in big data.frames cumsum(as.numeric()) works reliably
%/% is a division where you get an integer back
the '+1' just let your groups start with 1 instead of 0
thank you #Ronak Shah!
I would like to produce nested tables for a multilevel factorial experiment. I have 10 paints examined for time to reach an end point under 4 levels of humidity, 3 temperatures and 2 wind speeds. Of course I have searched on line but without success.
Some sample code can be generated using:
## Made Up Data # NB the data is continuous whereas observations were made 40/168 so data is censored.
time3 <- 4*seq(1:24) # Dependent: times in hrs, runif is not really representative but will do
wind <- c(1,2) # Independent: factor draught on or off
RH <- c(0,35,75,95) # Independent: value for RH but can be processes as a factor
temp <- c(5,11,20) # Independent: value for temperature but can be processed as a factor
paint <- c("paintA", "paintB", "paintC") # Independent: Experimental material
# Combine into dataframe
dfa <- data.frame(rep(temp,8))
dfa$RH <- rep(RH,6)
dfa$wind <- rep(wind,12)
dfa$time3 <- time3
dfa$paint <- rep(paint[1],24)
# Replicate for different paints
dfb <- dfa
dfb$paint <- paint[2]
dfc <- dfa
dfc$paint <- paint[3]
dfx <- do.call("rbind", list(dfa,dfb,dfc))
# Rename first col
colnames(dfx)[1] <- "temp"
# Prepare xtab tables
tx <- xtabs(dfx$time3 ~ dfx$wind + dfx$RH + dfx$temp + dfx$paint)
tx
And the target I hope to obtain would be like this xtab example
This
tx <- xtabs(dfx$time3 ~ dfx$wind + dfx$RH + dfx$temp)
does not work well enough. I would also like to write to C:\file.csv for printing and reporting etc. Please advise on how to achieve the desired output.
You can paste the two variables you want to nest together. Since the items will be ordered lexicographically, you will need to zero-pad the temp variable, to get numerical ordering.
xtabs(time3~wind+paste(sprintf("%02d",temp),RH,sep=":")+paint,dfx)
, , paint = paintA
paste(sprintf("%02d", temp), RH, sep = ":")
wind 05:0 05:35 05:75 05:95 11:0 11:35 11:75 11:95 20:0 20:35 20:75 20:95
1 56 0 104 0 88 0 136 0 120 0 72 0
2 0 128 0 80 0 64 0 112 0 96 0 144
, , paint = paintB
paste(sprintf("%02d", temp), RH, sep = ":")
wind 05:0 05:35 05:75 05:95 11:0 11:35 11:75 11:95 20:0 20:35 20:75 20:95
1 56 0 104 0 88 0 136 0 120 0 72 0
2 0 128 0 80 0 64 0 112 0 96 0 144
, , paint = paintC
paste(sprintf("%02d", temp), RH, sep = ":")
wind 05:0 05:35 05:75 05:95 11:0 11:35 11:75 11:95 20:0 20:35 20:75 20:95
1 56 0 104 0 88 0 136 0 120 0 72 0
2 0 128 0 80 0 64 0 112 0 96 0 144
I want col c phys_pos to be the value in col a position plus the accumulative value of col b length. In excel the calculation is: =A2+SUM($B$2:B2), but excel can't handle such a lot of data. Thanks all.
The data I would like:
position length phys_pos
12 45 57
97 0 142
135 0 180
498 0 543
512 0 557
16 67 128
76 0 188
89 0 201
101 0 213
152 0 264
3 103 218
19 0 234
76 0 291
88 0 303
Look into dplyr https://cran.rstudio.com/web/packages/dplyr/vignettes/introduction.html
install.packages("dplyr")
library(dplyr)
df <- df %>% mutate(phys_pos=cumsum(length)+position)
I am assuming your data.frame is named df
Or with base R
df$phys_pos <- cumsum(df$length) + df$position
Assuming your data is stored in a dataframe called "dat":
acc <- 0
for(i in 1:nrow(dat)){
acc <- acc + dat[i,"length"]
dat[i,"phys_pos"] <- dat[i,"position"]+acc
}
This is simple stuff. If you would do some tutorials you could learn to do it on your own pretty fast.
The libraries used are: library(survival)
library(splines)
library(boot)
library(frailtypack) and the function used is in the library frailty pack.
In my data I have two recurrent events(delta.stable and delta.unstable) and one terminal event (delta.censor). There are some time-varying explanatory variables, like unemployment rate(u.rate) (is quarterly) that's why my dataset has been splitted by quarters.
Here there is a link to the subsample used in the code just below, just in case it may be helpful to see the mistake. https://www.dropbox.com/s/spfywobydr94bml/cr_05_males_services.rda
The problem is that it takes a lot of time running until the warning message appear.
Main variables of the Survival function are:
I have two recurrent events:
delta.unstable (unst.): takes value one when the individual find an unstable job.
delta.stable (stable): takes value one when the individual find a stable job.
And one terminal event
delta.censor (d.censor): takes value one when the individual has death, retired or emigrated.
row id contadorbis unst. stable d.censor .t0 .t
1 78 1 0 1 0 0 88
2 101 2 0 1 0 0 46
3 155 3 0 1 0 0 27
4 170 4 0 0 0 0 61
5 170 4 1 0 0 61 86
6 213 5 0 0 0 0 92
7 213 5 0 0 0 92 182
8 213 5 0 0 0 182 273
9 213 5 0 0 0 273 365
10 213 5 1 0 0 365 394
11 334 6 0 1 0 0 6
12 334 7 1 0 0 0 38
13 369 8 0 0 0 0 27
14 369 8 0 0 0 27 119
15 369 8 0 0 0 119 209
16 369 8 0 0 0 209 300
17 369 8 0 0 0 300 392
When I apply multivePenal I obtain the following message:
Error en aggregate.data.frame(as.data.frame(x), ...) :
arguments must have same length
Además: Mensajes de aviso perdidos
In Surv(.t0, .t, delta.stable) : Stop time must be > start time, NA created
#### multivePenal function
fit.joint.05_malesP<multivePenal(Surv(.t0,.t,delta.stable)~cluster(contadorbis)+terminal(as.factor(delta.censor))+event2(delta.unstable),formula.terminalEvent=~1, formula2=~as.factor(h.skill),data=cr_05_males_serv,Frailty=TRUE,recurrentAG=TRUE,cross.validation=F,n.knots=c(7,7,7), kappa=c(1,1,1), maxit=1000, hazard="Splines")
I have checked if Surv(.t0,.t,delta.stable) contains NA, and there are no NA's.
In addition, when I apply for the same data the function frailtyPenal for both possible combinations, the function run well and I get results. I take one week looking at this and I do not find the key. I would appreciate some of light to this problem.
#delta unstable+death
enter code here
fit.joint.05_males<-frailtyPenal(Surv(.t0,.t,delta.unstable)~cluster(id)+u.rate+as.factor(h.skill)+as.factor(m.skill)+as.factor(non.manual)+as.factor(municipio)+as.factor(spanish.speakers)+ as.factor(no.spanish.speaker)+as.factor(Aged.16.19)+as.factor(Aged.20.24)+as.factor(Aged.25.29)+as.factor(Aged.30.34)+as.factor(Aged.35.39)+ as.factor(Aged.40.44)+as.factor(Aged.45.51)+as.factor(older61)+ as.factor(responsabilities)+
terminal(delta.censor),formula.terminalEvent=~u.rate+as.factor(h.skill)+as.factor(m.skill)+as.factor(municipio)+as.factor(spanish.speakers)+as.factor(no.spanish.speaker)+as.factor(Aged.16.19)+as.factor(Aged.20.24)+as.factor(Aged.25.29)+as.factor(Aged.30.34)+as.factor(Aged.35.39)+as.factor(Aged.40.44)+as.factor(Aged.45.51)+as.factor(older61)+ as.factor(responsabilities),data=cr_05_males_services,n.knots=12,kappa1=1000,kappa2=1000,maxit=1000, Frailty=TRUE,joint=TRUE, recurrentAG=TRUE)
###Be patient. The program is computing ...
###The program took 2259.42 seconds
#delta stable+death
fit.joint.05_males<frailtyPenal(Surv(.t0,.t,delta.stable)~cluster(id)+u.rate+as.factor(h.skill)+as.factor(m.skill)+as.factor(non.manual)+as.factor(municipio)+as.factor(spanish.speakers)+as.factor(no.spanish.speaker)+as.factor(Aged.16.19)+as.factor(Aged.20.24)+as.factor(Aged.25.29)+as.factor(Aged.30.34)+as.factor(Aged.35.39)+as.factor(Aged.40.44)+as.factor(Aged.45.51)+as.factor(older61)+as.factor(responsabilities)+terminal(delta.censor),formula.terminalEvent=~u.rate+as.factor(h.skill)+as.factor(m.skill)+as.factor(municipio)+as.factor(spanish.speakers)+as.factor(no.spanish.speaker)+as.factor(Aged.16.19)+as.factor(Aged.20.24)+as.factor(Aged.25.29)+as.factor(Aged.30.34)+as.factor(Aged.35.39)+as.factor(Aged.40.44)+as.factor(Aged.45.51)+as.factor(older61)+as.factor(responsabilities),data=cr_05_males_services,n.knots=12,kappa1=1000,kappa2=1000,maxit=1000, Frailty=TRUE,joint=TRUE, recurrentAG=TRUE)
###The program took 3167.15 seconds
Because you neither provide information about the packages used, nor the data necessary to run multivepenal or frailtyPenal, I can only help you with the Surv part (because I happened to have that package loaded).
The Surv warning message you provided (In Surv(.t0, .t, delta.stable) : Stop time must be > start time, NA created) suggests that something is strange with your variables .t0 (the time argument in Surv, refered to as 'start time' in the warning), and/or .t (time2 argument, 'Stop time' in the warning). I check this possibility with a simple example
# read the data you feed `Surv` with
df <- read.table(text = "row id contadorbis unst. stable d.censor .t0 .t
1 78 1 0 1 0 0 88
2 101 2 0 1 0 0 46
3 155 3 0 1 0 0 27
4 170 4 0 0 0 0 61
5 170 4 1 0 0 61 86
6 213 5 0 0 0 0 92
7 213 5 0 0 0 92 182
8 213 5 0 0 0 182 273
9 213 5 0 0 0 273 365
10 213 5 1 0 0 365 394
11 334 6 0 1 0 0 6
12 334 7 1 0 0 0 38
13 369 8 0 0 0 0 27
14 369 8 0 0 0 27 119
15 369 8 0 0 0 119 209
16 369 8 0 0 0 209 300
17 369 8 0 0 0 300 392", header = TRUE)
# create survival object
mysurv <- with(df, Surv(time = .t0, time2 = .t, event = stable))
mysurv
# create a new data set where one .t for some reason is less than .to
# on row five .t0 is 61, so I set .t to 60
df2 <- df
df2$.t[df2$.t == 86] <- 60
# create survival object using new data which contains at least one Stop time that is less than Start time
mysurv2 <- with(df2, Surv(time = .t0, time2 = .t, event = stable))
# Warning message:
# In Surv(time = .t0, time2 = .t, event = stable) :
# Stop time must be > start time, NA created
# i.e. the same warning message as you got
# check the survival object
mysurv2
# as you can see, the fifth interval contains NA
# I would recommend you check .t0 and .t in your data set carefully
# one way to examine rows where Stop time (.t) is less than start time (.t0) is:
df2[which(df2$.t0 > df2$.t), ]
I am not familiar with multivepenal but it seems that it does not accept a survival object which contains intervals with NA, whereas might frailtyPenal might do so.
The authors of the package have told me that the function is not finished yet, so perhaps that is the reason that it is not working well.
I encountered the same error and arrived at this solution.
frailtyPenal() will not accept data.frames of different length. The data.frame used in Surv and data.frame named in data= in frailtyPenal must be the same length. I used a Cox regression to identify the incomplete cases, reset the survival object to exclude the missing cases and, finally, run frailtyPenal:
library(survival)
library(frailtypack)
data(readmission)
#Reproduce the error
#change the first start time to NA
readmission[1,3] <- NA
#create a survival object with one missing time
surv.obj1 <- with(readmission, Surv(t.start, t.stop, event))
#observe the error
frailtyPenal(surv.obj1 ~ cluster(id) + dukes,
data=readmission,
cross.validation=FALSE,
n.knots=10,
kappa=1,
hazard="Splines")
#repair by resetting the surv object to omit the missing value(s)
#identify NAs using a Cox model
cox.na <- coxph(surv.obj1 ~ dukes, data = readmission)
#remove the NA cases from the original set to create complete cases
readmission2 <- readmission[-cox.na$na.action,]
#reset the survival object using the complete cases
surv.obj2 <- with(readmission2, Surv(t.start, t.stop, event))
#run frailtyPenal using the complete cases dataset and the complete cases Surv object
frailtyPenal(surv.obj2 ~ cluster(id) + dukes,
data = readmission2,
cross.validation = FALSE,
n.knots = 10,
kappa = 1,
hazard = "Splines")
My array looks like this:
Slide Index A B C DoseGroup
482 778 l 0 0 2 13Gy_p_75_42wk
483 778 r 0 0 2 13Gy_p_75_42wk
484 779 l 0 0 2 13Gy_p_75_42wk
485 779 r 0 0 2 13Gy_p_75_42wk
486 4700 l 2 2 2 14.25Gy_C_50pl_42wk
487 4700 r 0 0 1 14.25Gy_C_50pl_42wk
488 4701 l 0 0 1 14.25Gy_C_50pl_42wk
I would like to use the DoseGroup column's entries to be able to select the respective entries in the other columns. I would like to be able to tell R, e.g., "Do a wilcox.test between the 13Gy_p_75_42wk and the 14.25Gy_C_50pl_42wk datasets using column C."
How can I do this with R? Is there some kind of way to select all columns having the entry 14.25Gy_C_50pl_42wk?
I modified your data to add a third level in DoseGroup to make it more realistic.
txt <- "Slide Index A B C DoseGroup
778 l 0 0 2 13Gy_p_75_42wk
778 r 0 0 2 13Gy_p_75_42wk
779 l 0 0 2 13Gy_p_75_42wk
779 r 0 0 2 13Gy_p_75_42wk
4700 l 2 2 2 14.25Gy_C_50pl_42wk
4700 r 0 0 1 14.25Gy_C_50pl_42wk
4701 l 0 0 1 14.25Gy_C_50pl_42wk
4702 l 0 0 10 15Gy_C_50pl_42wk"
dat <- read.table(text = txt, header = TRUE)
wilcox.test(C ~ DoseGroup, data = dat,
subset = DoseGroup %in% c("13Gy_p_75_42wk", "14.25Gy_C_50pl_42wk"))
## Wilcoxon rank sum test with continuity correction
## data: C by DoseGroup
## W = 10, p-value = 0.1175
## alternative hypothesis: true location shift is not equal to 0
To select data, you can use one of these two command.
dat[dat$DoseGroup == "14.25Gy_C_50pl_42wk", ]
subset(dat, DoseGroup == "14.25Gy_C_50pl_42wk")
Those commands are basics in R and if you read any introduction to R, you'll be able to do same.
So I urge you to do so, I you want to really enjoy R.