I am trying to take a column of my data that is in factor format and change it to time in the format
hours:minutes:seconds:milliseconds
I tried:
start.times <- as.POSIXct(as.character(start.times), format="%H:%M:%OS")
but it returned values with todays date and left out the milliseconds in them and that is not what I want.
I also tried downloading chron and running the code:
start.times <- times(start.times)
but this just returned NA's.....
Please help!
My data is all about start times and end times of dolphin vocalizations and I am trying to find the mean whistle duration and the inter whistle interval. Anyways, I don't really know how to get my data into the format I need it in. Thank you!
Assuming you have a factor that looks like:
start.time <- c("0:13:45.9", "3:09:44.9")
Then what you wrote should work if you change the last colon to a period
as.POSIXct(start.time, format ="%H:%M:%S.%OS")
Related
My data contains several measurements in one day. It is stored in CSV-file and looks like this:
enter image description here
The V1 column is factor type, so I'm adding a extra column which is date-time -type: vd$Vdate <- as_datetime(vd$V1) :
enter image description here
Then I'm trying to convert the vd-data into time series: vd.ts<- ts(vd, frequency = 365)
But then the dates are gone:
enter image description here
I just cannot get it what I am doing wrong! Could someone help me, please.
Your dates are gone because you need to build the ts dataframe from your variables (V1, ... V7) disregarding the date field and your ts command will order R to structure the dates.
Also, I noticed that you have what is seems like hourly data, so you need to provide the frequency that is appropriate to your time not 365. Considering what you posted your frequency seems to be a bit odd. I recommend finding a way to establish the frequency correctly. For example, if I have hourly data for 365 days of the year then I have a frequency of 365.25*24 (0.25 for the leap years).
So the following is just as an example, it still won't work properly with what I see (it is limited view of your dataset so I am not sure 100%)
# Build ts data (univariate)
vs.ts <- ts(vd$V1, frequency = 365, start = c(2019, 4)
# check to see if it is structured correctly
print(vd.ts, calendar = T)
Finally my time series is working properly. I used
ts <- zoo(measurements, date_times)
and I found out that the date_times was supposed to be converted with as_datetime() as otherwise they were character type. The measurements are converted into data.frame type.
I'm trying to convert a yyyy-mm-dd data in a data frame to the total number of days from some date to put in my survival function.
I've already tried as_date() and grepl(), but I can't seem to get it to work since there are either too many NA values in my data frame or I'm doing something wrong.
Ref.date <- ymd("1941-08-24")
Date.MI <- ymd("Date.MI")
Day <- as.numeric(difftime(Date.MI, Ref.date))
I expect just the total number of days since 1941-08-24.
How do I solve the problem?
difftime() gives you the option to specify the units for the resulting output. So maybe try something like this
as.numeric(difftime(as.POSIXct("1941-08-25"), as.POSIXct("1941-08-24"), units = c("days")))
The way to solve it:
as.numeric(difftime(as.POSIXct(Date.MI[[1]]), as.POSIXct("1941-08-24"), units = c("days")))
There were square brackets needed since that refers to the first column.
I have split my large data set by date like so to create a large list of several elements:
days <- split(df, df$Date)
My data has columns including time of sunrise, sunset etc. for each day. I now want to use a for loop to do further work on each day separately like this:
for(i in 1:length(days){
sunrisetime <- as.character(df$Sunrise[1])
# Further similar work (using time of sunrise & sunset for each date to split
into daytime hours and nighttime hours)
}
My question is about the df$Sunrise on the second line - I don't think this is the right code to use when trying to access the sunrise time of each day on the days list. I have tried all sorts of variations but am an R newbie so must just be hitting the wrong terms.
Thanks in advance.
sunrisetime<-rep(NA,length(days))
for(i in 1:length(days){
sunrisetime[i] <- as.character(df$Sunrise[i])
}
I've been trying to do a time series on my dataframe, and I need to strip times from my csv. This is what I've got:
campbell <-read.csv("campbell.csv")
campbell$date = strptime(campbell$date, "%m/%d")
campbell.ts <- xts(campbell[,-1],order.by=campbell[,1])
First, what I'm trying to do is just get xts to strip the dates as "xx/xx" meaning just the month and day. I have no year for my data. When I try that second line of code and call upon the date column, it converts it to "2013-xx-xx." These months and days have no year associated with them, and I can't figure out how to get rid of the 2013. (The csv file I'm calling on has the dates in the format "9/30,10/1...etc.)
Secondly, once I try and make a time series (the third line), I am unsure what the "order.by" command is calling on. What am I indexing?
Any help??
Thanks!
For strptime, you need to provide the full date, i.e. day, month and year. In case, any of these is not provided, current ones are assumed from the system's time and appended to the incomplete date. So, if you want to retain your date format as you have read it, first make a copy of that and store in a temporary variable and then use strptime over campbell$date to convert into R readable date format. Since, year is not a concern to you, you need not bother about it even though it is automatically appended by strptime.
campbell <-read.csv("campbell.csv")
date <- campbell$date
campbell$date <- strptime(campbell$date, "%m/%d")
Secondly, what you are doing by 'the third line' (xts(campbell[,-1],order.by=campbell[,1])) command is that, your are telling to order all the data of campbell except the first column (campbell[,-1]) according to the index provided by the time data in the first column of campbell (campbell[,1]). So, it would only work given the date is in the first column.
After ordering the data according to time-series, you can replace back the campbell$date column with date to get back the date format you wanted (although here, first you have to order date also like shown below)
date <- xts(date, order.by=campbell[,1]) # assuming campbell$date is campbell[,1]
campbell.ts <- xts(campbell[,-1], order.by=campbell[,1])
campbell.ts <- cbind(date, campbell.ts)
format(as.Date(campbell$dat, "%m/%d/%Y"), "%m/%d")
I want to create a single column with a sequence of date/time increasing every hour for one year or one month (for example). I was using a code like this to generate this sequence:
start.date<-"2012-01-15"
start.time<-"00:00:00"
interval<-60 # 60 minutes
increment.mins<-interval*60
x<-paste(start.date,start.time)
for(i in 1:365){
print(strptime(x, "%Y-%m-%d %H:%M:%S")+i*increment.mins)
}
However, I am not sure how to specify the range of the sequence of dates and hours. Also, I have been having problems dealing with the first hour "00:00:00"? Not sure what is the best way to specify the length of the date/time sequence for a month, year, etc? Any suggestion will be appreciated.
I would strongly recommend you to use the POSIXct datatype. This way you can use seq without any problems and use those data however you want.
start <- as.POSIXct("2012-01-15")
interval <- 60
end <- start + as.difftime(1, units="days")
seq(from=start, by=interval*60, to=end)
Now you can do whatever you want with your vector of timestamps.
Try this. mondate is very clever about advancing by a month. For example, it will advance the last day of Jan to last day of Feb whereas other date/time classes tend to overshoot into Mar. chron does not use time zones so you can't get the time zone bugs that code as you can using POSIXct. Here x is from the question.
library(chron)
library(mondate)
start.time.num <- as.numeric(as.chron(x))
# +1 means one month. Use +12 if you want one year.
end.time.num <- as.numeric(as.chron(paste(mondate(x)+1, start.time)))
# 1/24 means one hour. Change as needed.
hours <- as.chron(seq(start.time.num, end.time.num, 1/24))