I have 10 million+ data points which look like:
Identifier Times Data
6597104 2015-05-01 04:08:05 0.15512575543732
In order to study these I want to add a Period (1, 2,...) column so the oldest row with the 6597104 identifier is period 1 and the second oldest is period 2 etc. However the times come irregularly so I can't just make it a time series object.
Does anyone know how to do this? Thanks in advance
Let's call your data frame data
First sort it using
data <- data[sort(data$Times,decreasing=TRUE),]
Then add a new column called Period
for i in 1:nrow(data){
data$Period[i] <- paste("Period",i,sep=" ")
}
Related
I am extracting Google Trends data looking at interest_over_time and interest_by_city.
I've noticed the interest_by_city data frame doesn't contain any date information. As I am looking to monitor the changes over time this is problematic.
Is there a way to add a new variable for the date where each observation will be the date and time the data was extracted?
If what you want is to add Sys.time() to the data you just extracted then you can call df$time <- Sys.time().
There is a data frame like this:
The first two columns in the df describe the start date (month and year) and the end date (month and year). Column names describe every single month and year of a certain time period.
I need a function/loop that insterts "1" or "0" in each cell - "1" when the date from given column name is within the period described by the two first columns, and "0" if not.
I would appreciate any help.
You want to do two different things. (a) create a dummy variable and (b) see if a particular date is in an interval.
Making a dummy variable is the easiest one, in base R you can use ifelse. For example in the iris data frame:
iris$dummy <- ifelse(iris$Sepal.Width > 2.5, 1, 0)
Now working with dates is more complicated. In this answer we will use the library lubridate. First you need to convert all those dates to a format 'Month Year' to something that R can understand. For example for February you could do:
new_format_february_2016 <- interval(ymd('2016-02-01'), ymd('2016-03-01') - dseconds(1))
#[1] 2016-02-01 UTC--2016-02-29 23:59:59 UTC
This is February, the interval of time from the 1 of February to one second before the 1 of March. You can do the same with your start date column and you end date column.
To compare two intevals of time (so, to see if a particular month fall into your other intervals) you can do:
int_overlaps(new_format_february_2016, other_interval)
If this returns true, the two intervals (one particular month and another one) overlaps. This is not the same as one being inside another, but in your case it will work. Using this you can iterate over different columns and rows and build your dummy variable.
But before doing so, I would recommend to clean your data, as your current format is complicate to work with. To get all the power that vector types in R provides ideally you would want to have one row per observation and one variable per column. This does not seem to be the case with your data frame. Take a look to the chapter 'Tidy data' of 'R for Data Science' specially the spreading and gathering subsection:
Tidy data
I have uber dataset containing variables pickup point, request time, drop time, date variable without month and year.
I need code for calculating idle time and creating a new variable idle time. Calculation as follows:
If pickup points are same for consecutive rows and date is different for consecutive rows then NA value if not difference between drop time of first row and the pickup time in second row. I have done it in excel and need to do it in R
Attached is the screenshot of data in excel
Try something like this, if this is what you are looking for
for(i in 2:nrow(df)){
df$idle[1]<-NA
if(df$Pickup.point[i]!=df$Pickup.point[i-1])
df$idle[i]<-NA
else
if(df$Date[i]!=df$Date[i-1])
df$idle[i]<-NA
else
df$idle[i]<-(df$Req[i]-df$Drop[i-1])
}
Imagine an intra-day set of data, e.g. hourly intervals. Thanks to Google and valuable Joshua's answers to other people, I managed to create new columns in the xts object carrying DAILY Open/High/Low/Close values. These are daily values applied on intra-day intervals so all rows of the same day have the same value in particular column. Since the HLC values are look-ahead biased, I want to move them to the next day. Let's focus on just one column called Prev.Day.Close.
Actual status:
My Prev.Day.Close column caries proper values for the current day. All "2010-01-01 ??:??" rows have the same value - Close of 2010-01-01 trading session. So it is not PREVIOUS day at the moment how the column name says.
What I need:
Lag the Prev.Day.Close column to the NEXT DAY OF THE SET.
I cannot lag it using lag() because it works on row (not day) basis. It must not be fixed calendar day like:
C <- ave(x$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$daily.last)
Basically, you want to get the end of day values, merge them with the original data, then lag them. That will align the previous end of day values with the beginning of the day.
Consider a dataframe of the form
id start end
2009.36220 65693384 2010-03-20 2010-07-04
2010.36221 65693592 2010-01-01 2010-12-31
2010.36222 65698250 2010-01-01 2010-12-31
2010.36223 65704349 2010-01-01 2010-12-31
where I have around 20k observations per year for 15 years.
I need to combine the rows by the following rule:
if for the same id, there exists a record that ends at the last day of the year
and a record that starts at the first day of the following year
then
- create a new row with start value of the earlier row and end value of the later year
- and delete the two original rows
Given that the same id can be visible several times (since I have more than 2 years) I will then just iterate over the script several time to combine different ids that have for example 4 rows in consecutive years that satisfy the condition.
The Question
I'd know how to program this in an iterative manner, where I would go over every single row and check if there's a row with a start date next year somewhere in the whole data frame that corresponds to the end date this year - but that's extremely slow and non satisfying from an aesthetic point of view. I'm a very beginner with R, so I have no clue of where to even look to do such a thing in a more efficient manner - I'm open for any suggestion.
Warning: this kind of code with rbind() is cancerous, but this is the easiest solution I could think of. Let df be your data.
df$start = as.POSIXct(df$start)
df$end = as.POSIXct(df$end)
df2 = data.frame()
for (i in unique(df$id)){
s = subset(df, id==i)
df2 = rbind(df2, c(id, min(s$start), max(s$end)))
}