I have data showing when an animal came to a survey station. example csv file here The first few lines of data look like this:
Site_ID DateTime HourOfDay MinTemp LunarPhase Habitat
F1 6/12/2013 14:01:00 14 -1 0 river
F1 6/12/2013 14:23:00 14 -1 0 river
F2 6/13/2013 1:21:00 1 3 1 upland
F2 6/14/2013 1:33:00 1 4 2 upland
F3 6/14/2013 1:48:00 1 4 2 river
F3 6/15/2013 11:08:00 11 0 0 river
I would like to perform a circular-linear regression in R to determine peak activity times. The dependent variable could be DateTime or HourOfDay, whichever is easier. I would like to incorporate the covariates Site_ID (random effect), plus MinTemp, LunarPhase, and Habitat into a mixed-effects model.
I have tried using the lm.circular function of program circular, and have the following code:
data<-read.csv("StackOverflowExampleData.csv")
data$DateTime<-as.POSIXct(as.character(data$DateTime), format = "%m/%d/%Y %H:%M:%S")
data$LunarPhase<-as.factor(data$LunarPhase)
str(data)
library(circular)
y<-data$DateTime
y<-circular(y, units ="hours",template = "clock24",rotation = "clock")
x<-data[,c(1,4,5,6)]
lm.circular(y=y, x=x, init=c(1,1,1,1), type='c-l', verbose=TRUE)
I keep getting the error:
Error in Ops.POSIXt(x, 12) : '/' not defined for "POSIXt" objects
Apparently this is a known bug, but I was confused by this threat about it and could not determine an appropriate work-around. Suggestions?
Also, my ultimate goal with this data was to run a circular-linear version of a glm, and then test several models against one another using AIC or some other information theoretics method. The model I'm seeking would be a circular-linear version of something like this:
glmer(HourOfDay~MinTemp+LunarPhase+Habitat+(1|Site_ID),family=binomial,data=data)
Perhaps this is an inappropriate application of the circular package. If so, I'm open to other suggestions of models and/or graphics that would investigate peak activity using the data and covariates.
Note: I did search for related discussions and found this somewhat relevant thread, but it was never answered, did not request a solution in R, and was of a different scope.
The specific problem is caused by conversion.circular. There, a POSIXlt object is divided by 12. This is an operation that has a non-defined outcome:
> as.POSIXlt('2005-07-16') / 2
Error in Ops.POSIXt(as.POSIXlt("2005-07-16"), 2) :
'/' not defined for "POSIXt" objects
So, it seems that you cannot use data of this class as input for the circular package. I could not find any mention of POSIXlt data in the examples. Maybe you need to specify the timestamps simply as a number, not as a POSIXlt object.
Related
I am stuck with a real problem.
My dataset comes from a survey and to make it usable to find statistics about the whole French population, I must weight it with weights.
For this purpose, I used the survey package, but the syntax is not really easy to use with R.
Is there a way to use ggplot while having weights?
To explain it a bit better, here is my dataset:
head(df)
Id Weight Var1
1 30 0
2 12.4 0
3 68.2 1
So my individual 1 accounts for 30 people in the French population.
I create a df_weighted dataset using the survey package.
How can I use ggplot now? df_weighted is a list!
I did something like this to try to escape the list problem but I did not work at all...
df_weighted_ggplot$var1 <- svytable(~var1, df_weighted)
df_weighted_ggplot$var_fill <- svytable(~var_fill, df_weighted)
ggplot(df_weighted_ggplot, aes(fill = var_fill , x =var1)) + geom_bar(position = "fill")
I received this predictable error:
Erreur : `data` must be a data frame, or other object coercible by `fortify()`, not a list
Do you know any other package which should help me? But I read many forums and it seems to be the most helpful...
I am trying to fit values in my algorithm so that I could predict a next month's number. I am getting a No data for variable errror when clearly I've defined what the objects are that I am putting into the equation.
I've tried to place them in vectors so that it could use one vector as a training data set to predict the new values. Current script has worked for me for a different dataset but for some reason isn't working here.
The data is small so I was wondering if that has anything to do with it. The data is:
Month io obs Units Sold
12 in 1 114
1 in 2 29
2 in 3 105
3 in 4 30
4 in 5
I'm trying to predict Units Sold with the code below
matt<-TEST1
isdf<-matt[matt$month<=3,]
isdf<-na.omit(isdf)
osdf<-matt[matt$Units.Sold==4,]
lmfit<-lm(Units.Sold~obs+Month,data=isdf,na.action=na.omit)
predict(lmFit,osdf[1,1])
I am expecting to be able to place lmfit in predict and get an output.
I have been writing a code to ease performing a discriminant analysis using the lda function. But actually I have a step which I cannot solve. And it is when I have to introduce the name of the categorical column in the code. Imagine we have the next table (called smoke), in which the column Factor represents the groups (in our cases, smoker and nsmok).
smoke
Factor Lung Heart Blood
1 smoker 7 22 15
2 smoker 8 21 12
3 nsmok 22 9 5
This is the code I have been preparing. Please, look at the XXXX's in the code (it appears twice). I want them to write automatically the name of the categorical column, instead of writing directly it twice.
lda=lda(XXXX~.,data=Smoke)
plot(lda)
lda
lda$counts
lda$svd
lda.p=predict(lda)
Tabla=table(Smoke$XXXX,lda.p$class)
Tabla
diag(prop.table(Tabla, 1))
sum(diag(prop.table(Tabla)))
I thought that writing...
colnames(Table)[1]
... would solve it. But actually there still exist some errors when running the code.
Otherwise, I though that introducing directly the name in this way:
Column_Factor-> Factor
and writing Column_Factor in the two places in the code would solve it. But it isn't.
Any ideas?
You could do something like this:
library(MASS)
#gets the column name of the factor, maybe check if there is only one factor column first
Column_Factor <- names(Smoke)[sapply(Smoke, class)=="factor"]
#creates the formula by pasting the name and the RHS
lda <- lda(as.formula(paste(Column_Factor,"~.",sep="")),data=Smoke)
plot(lda)
lda
lda$counts
lda$svd
lda.p=predict(lda)
#selects the column using the variable
Tabla=table(Smoke[,Column_Factor],lda.p$class)
Tabla
diag(prop.table(Tabla, 1))
sum(diag(prop.table(Tabla)))
I'm attempting to model customer lifetimes on subscriptions. As the data is censored I'll be using R's survival package to create a survival curve.
The original subscriptions dataset looks like this..
id start_date end_date
1 2013-06-01 2013-08-25
2 2013-06-01 NA
3 2013-08-01 2013-09-12
Which I manipulate to look like this..
id tenure_in_months status(1=cancelled, 0=active)
1 2 1
2 ? 0
3 1 1
..in order to feed the survival model:
obj <- with(subscriptions, Surv(time=tenure_in_months, event=status, type="right"))
fit <- survfit(obj~1, data=subscriptions)
plot(fit)
What shall I put in the tenure_in_months variable for the consored cases i.e. the cases where the subscription is still active today - should it be the tenure up until today or should it be NA?
First I shall say I disagree with the previous answer. For a subscription still active today, it should not be considered as tenure up until today, nor NA. What do we know exactly about those subscriptions? We know they tenured up until today, that is equivalent to say tenure_in_months for those observations, although we don't know exactly how long they are, they are longer than their tenure duration up to today.
This is a situation known as right-censor in survival analysis. See: http://en.wikipedia.org/wiki/Censoring_%28statistics%29
So your data would need to translate from
id start_date end_date
1 2013-06-01 2013-08-25
2 2013-06-01 NA
3 2013-08-01 2013-09-12
to:
id t1 t2 status(3=interval_censored)
1 2 2 3
2 3 NA 3
3 1 1 3
Then you will need to change your R surv object, from:
Surv(time=tenure_in_months, event=status, type="right")
to:
Surv(t1, t2, event=status, type="interval2")
See http://stat.ethz.ch/R-manual/R-devel/library/survival/html/Surv.html for more syntax details. A very good summary of computational details can be found: http://support.sas.com/documentation/cdl/en/statug/63033/HTML/default/viewer.htm#statug_lifereg_sect018.htm
Interval censored data can be represented in two ways. For the first use type = interval and the codes shown above. In that usage the value of the time2 argument is ignored unless event=3. The second approach is to think of each observation as a time interval with (-infinity, t) for left censored, (t, infinity) for right censored, (t,t) for exact and (t1, t2) for an interval. This is the approach used for type = interval2, with NA taking the place of infinity. It has proven to be the more useful.
If a missing end date means that the subscription is still active, then you need to take the time until the current date as censor date.
NA wont work with the survival object. I think those cases will be omitted. That is not what you want! Because these cases contain important information about the survival.
SQL code to get the time till event (use in SELECT part of query)
DATEDIFF(M,start_date,ISNULL(end_date,GETDATE()) AS tenure_in_months
BTW:
I would use difference in days, for my analysis. Does not make sense to round off the time to months.
You need to know the date the data was collected. The tenure_in_months for id 2 should then be this date minus 2013-06-01.
Otherwise I believe your encoding of the data is correct. the status of 0 for id 2 indicates it's right-censored (meaning we have a lower bound on it's lifetime, but not an upper bound).
I am trying to plot histograms with long term (several years) mean precipitation (pp) for each day of the month from a series of files. Each file has data collected from a different place (and has a different code). Each of my files looks like this:
X code year month day pp
1 2867 1945 1 1 0.0
2 2867 1945 1 2 0.0
...
And I am using the following code:
files <- list.files(pattern=".csv")
par(mfrow=c(4,6))
for (i in 1:24) {
obs <- read.table(files[i],sep=",", header=TRUE)
media.dia <- ddply(obs, .(day), summarise, daily.mean<-mean(pp))
codigo <- unique(obs$code)
hist(daily.mean, main=c("hist per day of month", codigo))
}
I get 24 histograms with 24 different codes in the title, but instead of 24 DIFFERENT histograms from 24 different locations, I get the same histogram 24 times (with 24 different titles). Can anybody tell me why? Thanks!
There are at least two errors I can see in your code.
There is an error in your ddply statement.
You are passing the wrong variable to hist, thus plotting something that may or may not exist depending on previous session actions.
The problem in your ddply statement is that you are doing an invalid assign (using <- ). Fix this by using =:
media.dia<- ddply(obs, .(day),summarise, daily.mean = mean(pp))
Then edit your hist statement:
hist(media.dia$daily.mean,main=c("hist per day of month",codigo))
I suspect the problem is that you are not passing the correct parameter to hist. The reason that your code actually produces a plot at all, is because in some previous step in your session you must have created a variable called daily.mean (as Brandon points out in the comment.)
I think the daily.mean calculated in the ddply function is assigned in a separate environment, and does not exist in an environment hist can see.
Try daily.mean<<-mean(pp)