[R] converting numeric into character strings
Hi, Im trying to put some numbers into a dataframe , I have a list of numbers (change points in a time series) like such [1] 2 11 12 20 21 98 99 but I want R to recognise this as just a character string so it will put it in one row and column, ideally I want them seperated by commas so I would have for example Person Change points (seconds) A 2,11,12,20,21,98,99 B4,5,89 etc. Is there any way I can get this I've tried this: for example if the command to get the list of numbers was b<-which(a!=s), then i have tried as.character(b) but I just end up with [1] "2" "11" "12" "20" "21" "98" "99" which is not what I want as this is more than one string and is not seperated by commas, I also tried paste(b,sep=",") but I end up with the same thing. Sorry it's a bit confusing to read but any help would be great! Melissa -- View this message in context: http://www.nabble.com/converting-numeric-into-character-strings-tp23518762p23518762.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] turning list into vector/dataframe
Hi,
I have used this command :
resamples<-lapply(1:1000,function(i) sample(lambs,replace=F))
resamples2<-lapply(resamples,Cusum)
to get a list of 1000 samples of my data. The function Cumsum is defined as
follows:
Cusum<-function(x){
SUM<-cumsum(x)-(1:length(x))*mean(x)
min<-min(cumsum(x)-(1:length(x))*mean(x))
max<-max(cumsum(x)-(1:length(x))*mean(x))
diff<-max-min
ans<-c(min,max,diff)
ans
}
where lambs is a vector of temperatures.
An example of part of my list is:
[[998]]
[1] -5.233176 6.903034 12.136210
[[999]]
[1] -9.296690 1.516233 10.812922
[[1000]]
[1] -1.502066e+01 -4.547474e-13 1.502066e+01
Now I want to convert this list into a dataframe so for example 1000 rows
with col names Min, Max and Diff. My supervisor said I first had to turn
this into a vector but I don't seem to be able to do that!
Any ideas on how to turn this list into a dataframe would be really
appreciated :) Thanks in advance
Melissa
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[R] Sequences
Hi,
I am trying to make a sequence and am using a for loop for this. I want to
start off with an initial value ie S[0]=0 then use the loop to create other
values. This is what I have so far but I just keep getting error messages.
#To calculate the culmulative sums:
s<-rep(0,207)#as this is the length of the
vector I know I will have
s<-as.vector(s)
s[0]<-0
for (i in 1:length(lambs))# where lambs is a vector of
length 207 consisting of temperature
values
{
s[i]<-s[i-1]-mean(lambs)
}
I continually get the error message:
Error in s[i] <- s[i - 1] - mean(lambs) : replacement has length zero
When I merely use s[i]<-i-mean(lambs) it works so there is obviously
something wrong with the s[i-1] but i cant see what. All I want is for each
S[i] to be the previous value for S - the mean!
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Re: [R] for loop for extracting linear model info
Uwe Ligges-3 wrote:
>
>
>
> Melissa2k9 wrote:
>> Hi,
>>
>> I have written a for loop as such:
>>
>> model<-lm(Normalised~Frame,data=All,subset=((Subject==1)&(Filmclip=="Strand")))
>> summary(model)
>>
>> ###
>> #To extract just the Adjusted R squared
>> ###
>>
>> rsq<-summary(model)[[9]]
>>
>> ###
>> #To extract just the slope
>> ###
>>
>> slope<-summary(model)[[4]][[2]]
>>
>> ###
>> #To extract only the p value from the t test for slope
>> ###
>>
>> pvalue<-summary(model)[[4]][[8]]
>>
>>
>> data<-data.frame(slope,pvalue,rsq)
>>
>>
>>
>> ###
>> #To extract this info for all films
>>
>>
>> for (i in c(1:8,10:20,22:29))
>> {
>> model_1<-lm(Normalised~Frame,data=All,subset=((Subject==i)&(Filmclip=="Strand")))
>> summary(model_1)
>> slope<-summary(model_1)[[4]][[2]]
>> pvalue<-summary(model_1)[[4]][[8]]
>> rsq<-summary(model_1)[[9]]
>> data2<-data.frame(slope,pvalue,rsq)
>> data2<-rbind(data,data2)
>> }
>>
>> I want this to run for all i but so far I am only getting two entries in
>> my
>> data frame, one for the first subject, and another.
>
>
> You are overwriting the old data2 with a new one that consists of those
> two in each iteration of the loop ...
>
>
> Uwe Ligges
>
>
>> Does anyone know where I am going wrong in my code so I can have this
>> data
>> for all subjects 1-8,10-20, and 22-29.
>>
>> Thanks
>
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
Hey,
Thanks, but do you have any advice on how to stop overwriting the old data2
and just keep adding a new row for each different subject?
Thanks :)
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[R] for loop for extracting linear model info
Hi,
I have written a for loop as such:
model<-lm(Normalised~Frame,data=All,subset=((Subject==1)&(Filmclip=="Strand")))
summary(model)
###
#To extract just the Adjusted R squared
###
rsq<-summary(model)[[9]]
###
#To extract just the slope
###
slope<-summary(model)[[4]][[2]]
###
#To extract only the p value from the t test for slope
###
pvalue<-summary(model)[[4]][[8]]
data<-data.frame(slope,pvalue,rsq)
###
#To extract this info for all films
for (i in c(1:8,10:20,22:29))
{
model_1<-lm(Normalised~Frame,data=All,subset=((Subject==i)&(Filmclip=="Strand")))
summary(model_1)
slope<-summary(model_1)[[4]][[2]]
pvalue<-summary(model_1)[[4]][[8]]
rsq<-summary(model_1)[[9]]
data2<-data.frame(slope,pvalue,rsq)
data2<-rbind(data,data2)
}
I want this to run for all i but so far I am only getting two entries in my
data frame, one for the first subject, and another.
Does anyone know where I am going wrong in my code so I can have this data
for all subjects 1-8,10-20, and 22-29.
Thanks
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[R] Linear model, finding the slope
Hi for some data I working on I am merely plotting time against temperature for a variable named filmclip. So for example, I have volunteers who watched various film clips and have used infared camera to monitor the temperature on their face at every second of the clip. The variable names I have used are Normalised ( for the temperature) and Frame (for the time in seconds). So I have fitted a linear model model<-lm(Normalised~Frame,data=All,subset=((Subject==1)&(Filmclip=="Whatever") and coef(model) gives me an intercept value and a value for the slope. Now what I want to do is find out if the slope is significant or not. So far I just have values such as 0.02211 for example and have no idea if this is to be interpreted as significant or not. Sorry if I haven't been clear but any advice on how to find out what values are significant would be greatly appreciated. -- View this message in context: http://www.nabble.com/Linear-model%2C-finding-the-slope-tp22865254p22865254.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] the by function
Hey, I have a dataframe with subjects who have watched films. The variables of interest are Pixel number and Temperature on the face. For each subject there are 8 films, for each film for each subject I need to measure the mean number of pixels then merge this vector with the data frame. I have mean.pixels<-as.data.frame(by(Final[,5],Final[,1:2],mean)[1:13,]) where column 5 is the pixels variable and 1 and 2 are subject and film number respectively. Now I can do this, but the average value of the pixel is usually around 7000. For some reason the machine has only calculated a certain second of the film as having say 500 pixels. I need to use the by function to calculate the mean pixels but I only want it to consider those values of the pixel variable that are above 1000. Does anyone know how I can modify the command I already have to make sure this happens? Sorry it's a bit confusing but I find it hard to explain. -- View this message in context: http://www.nabble.com/the-by-function-tp22723918p22723918.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
