Define different timeseries for different columns - r

I have a dataframe where some of the columns are starting later than the other. Please find a reproducible example.
set.seed(354)
df <- data.frame(Product_Id = rep(1:100, each = 50),
Date = seq(from = as.Date("2014/1/1"),
to = as.Date("2018/2/1"),
by = "month"),
Sales = rnorm(100, mean = 50, sd= 20))
df <- df[-c(251:256, 301:312, 2551:2562, 2651:2662, 2751:2762), ]
library(zoo)
z <- read.zoo(df, index = "Date", split = "Product_Id", FUN = as.yearmon)
tt <- as.ts(z)
Now for this dataframe for the columns 6,7,52,54 and 56 I want to define them as timeseries starting from a different date as compared to the rest of the dataframe. Supposedly the data begins from Jan 2000, column 6 will begin from July 2000, column 7 from Jan 2001 and so on. How should I proceed to do this?
Later, I want to perform a forecast on this dataset. Any inputs on this? Should I consider each column as a seperate dataframe and do the forecasting. Or can I convert each column to a different timeseries object that starts from the first non NA value?

Now for this dataframe for the columns 6,7,52,54 and 56 I want to define them as timeseries starting from a different date as compared to the rest of the dataframe. Supposedly the data begins from Jan 2000, column 6 will begin from July 2000, column 7 from Jan 2001 and so on. How should I proceed to do this?
There, AFAIK, no way to do this in R in a time series matrix. And if each column started at a different date, then (since each column has the same number of entries), each column would also need to end at a different date. Is this really what you need? A collection of time series that all happen to be of the same length (so they can fit into a matrix), but that start and end with offsets? I struggle to understand where something like this would be useful, outside a kind of forecasting competition.
If you really need this, then I would recommend you put your time series into a list structure. Then each one can start and end at any date, and they can be the same or different lengths. Take inspiration from Mcomp::M3.
Later, I want to perform a forecast on this dataset. Any inputs on this? Should I consider each column as a seperate dataframe and do the forecasting. Or can I convert each column to a different timeseries object that starts from the first non NA value?
Since your tt is already a time series object, the simplest way would be simply to iterate over its columns:
fcst <- matrix(nrow=10,ncol=ncol(tt))
for ( ii in 1:ncol(tt) ) fcst <- forecast(ets(tt[,ii]),10)$mean
Note that most modeling functions in forecast will throw a warning and do something reasonable on encountering NA values. Here, e.g.:
1: In ets(tt[, ii]) :
Missing values encountered. Using longest contiguous portion of time series
Of course, you could do something yourself inside the loop, e.g., search for the last NA and start the time series for modeling right after that (but make sure you fail gracefully if the last entry is NA).

Related

creating inteval object in r using lubridate package [duplicate]

This question already has an answer here:
indicateing to which interval a date belongs
(1 answer)
Closed 4 years ago.
hi i have data from uber :
about pick ups in NYC .
im trying to add a column to the raw data, that indicates for each row, for
which time interval (which is represented by a single timepoint at the beginning of thetime interval) it belongs.
i want to Create a vector containing all relevant timepoints (i.e. every 15 minutes
Use int_diff function from lubridate package on this vector to create an
interval object.
Run a loop on all the time points in the raw data and for each data
point; indicate to which interval (which is represented by a single
timepoint at the beginning of the time interval) it belongs.
i tried looking for explanations how to use the int_diff function but i dont understand how my vector should look and how the syntax of int_diff works
tanks for the help :)
Is this what you have in mind?
start <- mdy_hm('4/11/2014 0:00') # start of the period
end <- mdy_hm('5/12/2015 0:00') # end
time_seq <- seq(from = start, to = end, by = '15 mins') # sequence by 15 minutes
times <- mdy_hm(c('4/11/2014 0:12', '4/11/2014 1:24')) # times to find intervals for
dat <- data.frame(times)
dat$intervals <- cut(times, breaks = time_seq) # assign each time to an interval
intervals_cols <- model.matrix(~ - + intervals, dat) # turn this into a set of columns, one for each interval, with a 1 indicating that this observation falls into the column

How to match dates in 2 data frames in R, then sum specific range of values up to that date?

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
}

List of Dataframes filter row based on Date

I am currently working with a list of dataframes.
Actually, I have about a hundred csv files representing forecasts of some kind, where the date on which the forecast was made is in the first line, the lines thereafter contain the predicted values. The data might look like this:
2010/04/15 10:12:51 #Date of the forecast
2010/05/02 2372 #Date for which the forecast was made and the value assigned
2010/05/09 2298
2009/04/15 10:09:13 #another forecast
....
2010/05/02 2298 #also predicts for 2010/05/02
As you might guess, the forecasts do predict values quite some time ahead (e.g. 5 years), which means predictions for the date 2010/05/02 were not only made on 2010/04/15 but also 2009/04/15 and so on (actually, forecasts are done weekly).
I would like to compare how the predicted value for a specified date (for example 2010/05/02) has changed over time.
Right now, I read in all .csv datas I have as a dataframe, and save each of the resulting dataframes in a list.
(Sadly, the date on which the prediction was made got lost-I hoped to be able to name the list elements with the respective date but have not yet figured out how to do this-still, I am pretty sure I'll find something somewhere, not the main problem here)
That's where the question title comes in: I would like to know how to filter a list of dataframes by row value.
So, I'd like to be able to use a function: function(2010/05/02) and get as a result the rows of each Element of the list (each dataframe in the list) where Date is 2010/05/02.
In this case I'd like to get:
2010/05/02 2372
2010/05/02 2298
I know how to do this using a for loop, but it needs endlessly much time.
I am happy for any suggestions.
(By this example you might understand why it is important to know when the prediction was made- which I would not have right now. I was thinking about adding a new row containing the date on which the prediction was made in each dataframe)
Threads visited until now include:
get column from list of dataframes R
convert a row of a data frame to a simple vector in R
How to get the name of a data.frame within a list? (which more or less adresses the name problem)
As you can see, no thread was particularly helpful.
As requested, a small reproducible example:
dateList <- as.Date(seq(0,100,5),origin="2010-01-01")
forecasts <- seq(2000,3000,50)
df1 <- data.frame(dateList,forecasts)
df2 <- data.frame(dateList-50,forecasts)
l <- list(df1,df2)
we have dates from 2010-01-01 in 5 days steps. I would for example like to know the predicted values for 2010-01-01 in both dataframes.
The first dataframe looks like this:
dateList forecasts
1 2010-01-01 2000
2 2010-01-06 2050
3 2010-01-11 2100
while the second looks like this:
10 2009-12-27 2450
11 2010-01-01 2500
12 2010-01-06 2550
I was hoping to find out for example the predicted values for 2010-01-01.
So, for example:
function(2010-01-01):
2000
2500
Couldn't wait for your example so I made a small one. Let me know if this is in the general direction of what you're after.
xy <- list(df1 = data.frame(dates = as.Date(c("2016-01-01", "2016-01-02", "2016-01-03")), value = runif(3)),
df2 = data.frame(dates = as.Date(c("2016-01-01", "2016-01-02", "2016-01-03")), value = runif(3)),
df3 = data.frame(dates = as.Date(c("2016-01-01", "2016-01-02", "2016-01-03")), value = runif(3))
)
getValueOnDate <- function(x, list.all) {
lapply(list.all, FUN = function(m) m[m$dates %in% x, ])
}
out <- getValueOnDate(as.Date("2016-01-02"), list.all = xy)
do.call("rbind", out)
dates value
df1 2016-01-02 0.7665590
df2 2016-01-02 0.9907976
df3 2016-01-02 0.4909025
You can obviously modify the function to return just the values.
You could alternatively use the following approach, given your list is called ls and the date column date in all data.frame's:
my.ls <- lapply(ls, subset, date == "2010/05/02")
df <- do.call("rbind", my.ls)

What does the ts function do in R

I have downloaded the historical prices between Jan-1-2010 and Dec-31-2014 for Twitter, Inc. (TWTR) -NYSE from YAHOO! FINANCE in a twitter.csv file.
I then loaded it into RStudio using:
x = read.csv("Z:/path/to/file/twitter.csv", header=T,stringsAsFactors=F)
Here is how table x looks like:
View(x)
Then I used ts function to get the time series of Adj.Close:
x.ts = ts(x$Adj.Close, frequency = 12, start=c(2010,1), end=c(2014,12)
x.ts
How the previous results have been obtained? They are really different from table x data. Do they need any adjustements?
Your problem is the scale in which the data are read. With frequency = 12, start=c(2010,1), end=c(2014,12) you are telling the function that you have one number per month. If you have one number per day, as it's your case, you should try with:
x.ts = ts(x$Adj.Close, frequency = 365, start=c(2010,1), end=c(2014,365)
Firstly, frequency should be set to 365 if you deal with daily data, 12 if monthly etc.
Secondly
Secondly, I think you need to arrange the data ascending chronologically before using the ts() function.
The function blindly follows exactly what you are telling it, e.g. the data from the chart starts with the first value 35.87 in 2014-12-31 but the start date in the code is 2010, January, meaning it will attribute that value to being associated with Jan-2010.
x <- x %>%
dplyr::arrange(date)
ts.x <- ts(x$Adj.Close, frequency = 365, start=min(x$date), end=max(x$date))

Compute average over sliding time interval (7 days ago/later) in R

I've seen a lot of solutions to working with groups of times or date, like aggregate to sum daily observations into weekly observations, or other solutions to compute a moving average, but I haven't found a way do what I want, which is to pluck relative dates out of data keyed by an additional variable.
I have daily sales data for a bunch of stores. So that is a data.frame with columns
store_id date sales
It's nearly complete, but there are some missing data points, and those missing data points are having a strong effect on our models (I suspect). So I used expand.grid to make sure we have a row for every store and every date, but at this point the sales data for those missing data points are NAs. I've found solutions like
dframe[is.na(dframe)] <- 0
or
dframe$sales[is.na(dframe$sales)] <- mean(dframe$sales, na.rm = TRUE)
but I'm not happy with the RHS of either of those. I want to replace missing sales data with our best estimate, and the best estimate of sales for a given store on a given date is the average of the sales 7 days prior and 7 days later. E.g. for Sunday the 8th, the average of Sunday the 1st and Sunday the 15th, because sales is significantly dependent on day of the week.
So I guess I can use
dframe$sales[is.na(dframe$sales)] <- my_func(dframe)
where my_func(dframe) replaces every stores' sales data with the average of the store's sales 7 days prior and 7 days later (ignoring for the first go around the situation where one of those data points is also missing), but I have no idea how to write my_func in an efficient way.
How do I match up the store_id and the dates 7 days prior and future without using a terribly inefficient for loop? Preferably using only base R packages.
Something like:
with(
dframe,
ave(sales, store_id, FUN=function(x) {
naw <- which(is.na(x))
x[naw] <- rowMeans(cbind(x[naw+7],x[naw-7]))
x
}
)
)

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