I have a dataset of 2015 with every day of the year. In this dataset, there are actions that happen on any given day. Some days have more actions than others, therefore some days have many more entries than others.
I am trying to create a function that will create an individual dataset per day of the year without having to code 365 of these:
df <- subset(dataset, date== "2015-01-01")
I have looked at dyplyr's group_by(), however I do not want a summary per day, it is important that I get to see the whole observation on any given day for graphing purposes.
I have a relative large xts object. It contains the daily adjusted closing prices from 2012-2021 for each company in the STOXX 600 Europe. I want to calculate the yearly volatility of the stocks for each year for each company.
Here is a link of how my dataset look like:
https://imgur.com/a/oS7ROCL
So I started by calculate the log differences by:
XTS.LOGDIFFS <- diff(log(XTS.ADJCLOSE))
The next step would be to calculate the volatility of the stocks for a specific period of time, for example from 2012-05-01 till 2012-12-31, by using the standard deviation and multiply it with the square root of 252 ( 252 are the average trading days).
So my idea is this: First I want to extract the data from my dataset for a specific period of time, in this case from 2012-05-01 till 2012-12-31.
I tried this XT1<-xts(XTS.LOGDIFFS [1:174]).
As an alternativ I have thought about this:
start_date<- as.Date("2012-05-01")
end_date<-as.Date("2012-12-31")
The next step would be to calculate the volatility for the extracted xts object.
So I tried this: vol<-sd(XTS.1, na.rm = TRUE) *sqrt (252).
But this only give me the stock volatility for all combined and not for every single one.
So I think I need a function to get the stock volatility for every company for the extracted period of time in the xts object. But I have no clue how this should look like.
I have two data frames: rainfall data collected daily and nitrate concentrations of water samples collected irregularly, approximately once a month. I would like to create a vector of values for each nitrate concentration that is the sum of the previous 5 days' rainfall. Basically, I need to match the nitrate date with the rain date, sum the previous 5 days' rainfall, then print the sum with the nitrate data.
I think I need to either make a function, a for loop, or use tapply to do this, but I don't know how. I'm not an expert at any of those, though I've used them in simple cases. I've searched for similar posts, but none get at this exactly. This one deals with summing by factor groups. This one deals with summing each possible pair of rows. This one deals with summing by aggregate.
Here are 2 example data frames:
# rainfall df
mm<- c(0,0,0,0,5, 0,0,2,0,0, 10,0,0,0,0)
date<- c(1:15)
rain <- data.frame(cbind(mm, date))
# b/c sums of rainfall depend on correct chronological order, make sure the data are in order by date.
rain[ do.call(order, list(rain$date)),]
# nitrate df
nconc <- c(15, 12, 14, 20, 8.5) # nitrate concentration
ndate<- c(6,8,11,13,14)
nitrate <- data.frame(cbind(nconc, ndate))
I would like to have a way of finding the matching rainfall date for each nitrate measurement, such as:
match(nitrate$date[i] %in% rain$date)
(Note: Will match work with as.Date dates?) And then sum the preceding 5 days' rainfall (not including the measurement date), such as:
sum(rain$mm[j-6:j-1]
And prints the sum in a new column in nitrate
print(nitrate$mm_sum[i])
To make sure it's clear what result I'm looking for, here's how to do the calculation 'by hand'. The first nitrate concentration was collected on day 6, so the sum of rainfall on days 1-5 is 5mm.
Many thanks in advance.
You were more or less there!
nitrate$prev_five_rainfall = NA
for (i in 1:length(nitrate$ndate)) {
day = nitrate$ndate[i]
nitrate$prev_five_rainfall[i] = sum(rain$mm[(day-6):(day-1)])
}
Step by step explanation:
Initialize empty result column:
nitrate$prev_five_rainfall = NA
For each line in the nitrate df: (i = 1,2,3,4,5)
for (i in 1:length(nitrate$ndate)) {
Grab the day we want final result for:
day = nitrate$ndate[i]
Take the rainfull sum and it put in in the results column
nitrate$prev_five_rainfall[i] = sum(rain$mm[(day-6):(day-1)])
Close the for loop :)
}
Disclaimer: This answer is basic in that:
It will break if nitrate's ndate < 6
It will be incorrect if some dates are missing in the rain dataframe
It will be slow on larger data
As you get more experience with R, you might use data manipulation packages like dplyr or data.table for these types of manipulations.
#nelsonauner's answer does all the heavy lifting. But one thing to note, in my actual data my dates are not numerical like they are in the example above, they are dates listed as MM/DD/YYYY with the appropriate as.Date(nitrate$date, "%m/%d/%Y").
I found that the for loop above gave me all zeros for nitrate$prev_five_rainfall and I suspected it was a problem with the dates.
So I changed my dates in both data sets to numerical using the difference in number of days between a common start date and the recorded date, so that the for loop would look for a matching number of days in each data frame rather than a date. First, make a column of the start date using rep_len() and format it:
nitrate$startdate <- rep_len("01/01/1980", nrow(nitrate))
nitrate$startdate <- as.Date(all$startdate, "%m/%d/%Y")
Then, calculate the difference using difftime():
nitrate$diffdays <- as.numeric(difftime(nitrate$date, nitrate$startdate, units="days"))
Do the same for the rain data frame. Finally, the for loop looks like this:
nitrate$prev_five_rainfall = NA
for (i in 1:length(nitrate$diffdays)) {
day = nitrate$diffdays[i]
nitrate$prev_five_rainfall[i] = sum(rain$mm[(day-5):(day-1)]) # 5 days
}
I have daily data over several years for several currencies. I would like to lag variables in the data set by exactly one month (i.e. June 15 to July 15, not necessarily 30 days). NA's are fine where that is not possible.
I have gotten by so far by writing something like this:
ddply(Data, .(Currency), function(x){ #First column is date, 2nd is Currency, rest data.
y=x[,-(3)] #this is all data I want lagged
y$date=as.Date(y$date) %m+% months(1) #This increases the dates by one month
x$date=as.Date(x$date) #what data I dont want lagged is x[, (1:3)] below
z=merge(x[,(1:3)], y, by=c("date", "Currency")) #merge by date,
#lagged stuff merges with non-lagged stuff 1 month later than original obs date
return(z)
})
I can include the data if necessary, but given that I have something that works already I dont want anyone to spend time on it.
I just want to check that I cant use the lubridate %m+% months(1) syntax within the lag function. I have tried the lag function from package "statar" that uses the along_by syntax but haven't been able to figure it out.
Thanks!
Imagine an intra-day set of data, e.g. hourly intervals. Thanks to Google and valuable Joshua's answers to other people, I managed to create new columns in the xts object carrying DAILY Open/High/Low/Close values. These are daily values applied on intra-day intervals so all rows of the same day have the same value in particular column. Since the HLC values are look-ahead biased, I want to move them to the next day. Let's focus on just one column called Prev.Day.Close.
Actual status:
My Prev.Day.Close column caries proper values for the current day. All "2010-01-01 ??:??" rows have the same value - Close of 2010-01-01 trading session. So it is not PREVIOUS day at the moment how the column name says.
What I need:
Lag the Prev.Day.Close column to the NEXT DAY OF THE SET.
I cannot lag it using lag() because it works on row (not day) basis. It must not be fixed calendar day like:
C <- ave(x$Close, .indexday(x), FUN = last)
index(C) <- index(C) + 86400
x$Prev.Day.Close <- C
Because this solution does not care about real data in the set. For example it adds new rows because the original data set has holes on weekends and holidays. Moreover, two particular days may not have the same number of intervals (rows) so the shifted data will not fit.
Desired result:
All rows of the first day in the set have NA in Prev.Day.Close because there is no previous day to get data from.
All rows of the second day have the same value in Prev.Day.Close - Any of the values I actually have in Prev.Day.Close of previous day.
The same for every next row.
If I understand correctly, here's one way to do it:
require(xts)
# sample data
dt <- .POSIXct(seq(1, 86400*4, 3600), tz="UTC")-1
x <- xts(seq_along(dt), dt)
# get the last value for each calendar day
daily.last <- apply.daily(x, last)
# merge the last value of the day with the origianl data set
y <- merge(x, daily.last)
# now lag the last value of the day and carry the NA forward
# y$daily.last <- na.locf(lag(y$daily.last))
y$daily.last <- lag(y$daily.last)
y$daily.last <- na.locf(y$daily.last)
Basically, you want to get the end of day values, merge them with the original data, then lag them. That will align the previous end of day values with the beginning of the day.