I am using diff() to find the difference in variables down a column. However, I would also like to display the dates the difference is found over.
For example:
Dates <- c("2017-06-07","2017-06-10","2017-06-15","2017-07-07","2017-07-12","2017-07-18")
Variable<-c(5,6,7,8,9,3)
dd<-diff(Dates)
dv<-diff(Variable)
I'd like to find a way to add columns for the start and end date for each interval, so "06-07" as the start and "06-10" as the end date for the difference between the first 2 variables. Any ideas?
The OP has requested to add columns for the start and end date for each interval.
This can be accomplished by using the head() and tail() functions:
# data provided by OP
Dates <- c("2017-06-07","2017-06-10","2017-06-15","2017-07-07","2017-07-12","2017-07-18")
Variable<-c(5,6,7,8,9,3)
start <- head(Dates, -1) # take all Dates except the last one
end <- tail(Dates, -1L) # take all Dates except the first one
dd <- diff(as.Date(Dates)) # coersion to class Date required for date arthmetic
dv <- diff(Variable)
# create data.frame of intervals
intervals <- data.frame(start, end, dd, dv)
intervals
start end dd dv
1 2017-06-07 2017-06-10 3 days 1
2 2017-06-10 2017-06-15 5 days 1
3 2017-06-15 2017-07-07 22 days 1
4 2017-07-07 2017-07-12 5 days 1
5 2017-07-12 2017-07-18 6 days -6
Note that intervals has 5 rows while the vector of breakpoints Dates it was constructed from has a length of 6.
Are you after the difference in dates?
diff(as.Date(as.character(Dates,format="%Y-%M-%D")))
Related
I have a dataframe that looks like this:
V1 V2 V3 Month_nr Date
1 2 3 1 2017-01-01
3 5 6 1 2017-01-02
6 8 9 2 2017-02-01
6 8 9 8 2017-08-01
and I want to take all variables from the data set that have Month=1 (January) and date from 2017-01-01 til 2017-01-31 (so end of January), which means that I want to take the dates as well. I would create a column with days but I have multiple observations for one day and this would be even more confusing. I tried it with this:
df<- filter(df,df$Month_nr == 1, df$Date > 2017-01-01 && df$Date < 2017-01-31)
but it did not work. I would appreciate so much your help! I am desperate at this point. My dataset has measurements for an entire year (from 1 to 12) and hence I filter for months.
The problem is that you didn't put quotation marks around 2017-01-01. Directly putting 2017-01-01 will compute the subtraction and return a number, and then you're comparing a string to a number. You can compare string to string; with string, "2" is still greater than "1", so it would work for comparing dates as strings. BTW, you don't need to write df$ when using filter; you can directly write the column names without quoting when using the tidyverse.
Why do you need to have the month as well as dates in the filter? Just the filter on the dates would work fine. However, you will have to convert the date column into a date object. You can do that as follows:
df$Date_nr <- as.Date(df$Date_nr, format = "%Y-%m-%d")
df_new <- subset(df, Date_nr >= "2017-01-01" & Date_nr <= "2017-01-31")
I have the following problem:
Suppose we have:
Idx ID StartTime EndTime
1: 1 2014-01-01 02:20:00 2014-01-01 03:42:00
2: 1 2014-01-01 14:51:00 2014-01-01 16:44:00
note: Idx is not given, but I'm simply adding it to the table view.
Now we see that person with ID=1 is using the computer from 2:20 to 3:42. Now what I would like to do is to convert this interval into a set of variables representing hour and weekday and the duration in those periods.
Idx ID Monday-0:00 Monday-1:00 ... Wednesday-2:00 Wednesday-3:00
1: 1 40 42
For the second row we would have
Idx ID Monday-0:00 Monday-1:00 ... Wednesday-14:00 Wednesday-15:00 Wednesday-16:00
2: 1 9 60 44
Now the problem is of course that it can span over multiple hours as you can see from the second row.
I would like to do this per row and I was wondering if this is possible without too much computational effort and using data.table?
PS: it is also possible that the interval spans over the day.
library(data.table)
library(lubridate)
#produce sample data
DT<-data.table(idx=1:100,ID=rep(1:20,5), StartTime=runif(100,60*60,60*60*365)+ymd('2014-01-01'))
DT[,EndTime:=StartTime+runif(1,60,60*60*8)]
#make fake start and end dates with same day of week and time but all within a single calendar week
DT[,fakestart:=as.numeric(difftime(StartTime,ymd('1970-01-01'),units="days"))%%7*60*60*24+ymd('1970-01-01')]
DT[,fakeend:=as.numeric(difftime(EndTime,ymd('1970-01-01'),units="days"))%%7*60*60*24+ymd('1970-01-01')]
setkey(DT,fakestart,fakeend)
#check that weekdays line up
nrow(DT[weekdays(EndTime)==weekdays(fakeend)])
nrow(DT[weekdays(StartTime)==weekdays(fakestart)])
#both are 100 so we're good.
#check that fakeend > fakestart
DT[fakeend<fakestart]
#uh-oh some ends are earlier than starts, let's add 7 days to those ends
DT[fakeend<fakestart,fakeend:=fakeend+days(7)]
#make data.table with all possible labels
DTin<-data.table(start=seq(from=ymd('1970-01-01'),to=DT[,floor_date(max(fakeend),"hour")],by=as.difftime(hours(1))))
DTin[,end:=start+hours(1)]
DTin[,label:=paste0(format(start,format="%A-%H:00"),' ',format(end,format="%A-%H:00"))]
#set key and use new foverlaps feature of data.table which merges by interval
setkey(DT,fakestart,fakeend)
setkey(DTin,start,end)
DTout<-foverlaps(DT,DTin,type="any")
#compute duration in each interval
DTout[,dur:=60-pmax(0,difftime(fakestart,start,unit="mins"))-pmax(0,difftime(end,fakeend,unit="mins"))]
#cast all the rows up to columns for final result
castout<-dcast.data.table(DTout,idx+ID~label,value.var="dur",fill=0)
Right now I have two dataframes. One contains over 11 million rows of a start date, end date, and other variables. The second dataframe contains daily values for heating degree days (basically a temperature measure).
set.seed(1)
library(lubridate)
date.range <- ymd(paste(2008,3,1:31,sep="-"))
daily <- data.frame(date=date.range,value=runif(31,min=0,max=45))
intervals <- data.frame(start=daily$date[1:5],end=daily$date[c(6,9,15,24,31)])
In reality my daily dataframe has every day for 9 years and my intervals dataframe has entries that span over arbitrary dates in this time period. What I wanted to do was to add a column to my intervals dataframe called nhdd that summed over the values in daily corresponding to that time interval (end exclusive).
For example, in this case the first entry of this new column would be
sum(daily$value[1:5])
and the second would be
sum(daily$value[2:8]) and so on.
I tried using the following code
intervals <- mutate(intervals,nhdd=sum(filter(daily,date>=start&date<end)$value))
This is not working and I think it might have something to do with not referencing the columns correctly but I'm not sure where to go.
I'd really like to use dplyr to solve this and not a loop because 11 million rows will take long enough using dplyr. I tried using more of lubridate but dplyr doesn't seem to support the Period class.
Edit: I'm actually using dates from as.Date now instead of lubridatebut the basic question of how to refer to a different dataframe from within mutate still stands
eps <- .Machine$double.eps
library(dplyr)
intervals %>%
rowwise() %>%
mutate(nhdd = sum(daily$value[between(daily$date, start, end - eps )]))
# start end nhdd
#1 2008-03-01 2008-03-06 144.8444
#2 2008-03-02 2008-03-09 233.4530
#3 2008-03-03 2008-03-15 319.5452
#4 2008-03-04 2008-03-24 531.7620
#5 2008-03-05 2008-03-31 614.2481
In case if you find dplyr solution bit slow (basically due torowwise), you might want to use data.table for pure speed
library(data.table)
setkey(setDT(intervals), start, end)
setDT(daily)[, date1 := date]
foverlaps(daily, by.x = c("date", "date1"), intervals)[, sum(value), by=c("start", "end")]
# start end V1
#1: 2008-03-01 2008-03-06 144.8444
#2: 2008-03-02 2008-03-09 233.4530
#3: 2008-03-03 2008-03-15 319.5452
#4: 2008-03-04 2008-03-24 531.7620
#5: 2008-03-05 2008-03-31 614.2481
I have an example dataframe:
a <- c(1:6)
b <- c("05/12/2012 05:00","05/12/2012 06:00","06/12/2012 05:00",
"06/12/2012 06:00", "07/12/2012 09:00","07/12/2012 07:00")
c <-c("0","0","0","1","1","1")
df1 <- data.frame(a,b,c,stringsAsFactors = FALSE)
Firstly, I want to make sure R recognises the date and time format, so I used:
df1$b <- strptime(df1$b, "%d/%m/%Y %H:%M")
However this can't be right as R always aborts my session as soon as I try to view the new dataframe.
Assuming that this gets resolves, I want to get a subset of the data according to whichever day in the dataframe contains the most data in 'C' that is not a zero. In the above example I should be left with the two data points on 7th Dec 2012.
I also have an additional, related question.
If I want to be left with a subset of the data with the most non zero values between a certain time period in the day (say between 07:00 and 08:00), how would I go about doing this?
Any help on the above problems would be greatly appreciated.
Well, the good news is that I have an answer for you, and the bad news is that you have more questions to ask yourself. First the bad news: you need to consider how you want to treat multiple days that have the same number of non-zero values for 'c'. I'm not going to address that in this answer.
Now the good news: this is really simple.
Step 1: First, let's reformat your data frame. Since we're changing data types on a couple of the variables (b to datetime and c to numeric), we need to create a new data frame or recalibrate the old one. I prefer to preserve the original and create a new one, like so:
a <- df1$a
b <- strptime(df1$b, "%d/%m/%Y %H:%M")
c <- as.numeric(df1$c)
hour <- as.numeric(format(b, "%H"))
date <- format(b, "%x")
df2 <- data.frame(a, b, c, hour, date)
# a b c hour date
# 1 1 2012-12-05 05:00:00 0 5 12/5/2012
# 2 2 2012-12-05 06:00:00 0 6 12/5/2012
# 3 3 2012-12-06 05:00:00 0 5 12/6/2012
# 4 4 2012-12-06 06:00:00 1 6 12/6/2012
# 5 5 2012-12-07 09:00:00 1 9 12/7/2012
# 6 6 2012-12-07 07:00:00 1 7 12/7/2012
Notice that I also added 'hour' and 'date' variables. This is to make our data easily sortable by those fields for our later aggregation function.
Step 2: Now, let's calculate how many non-zero values there are for each day between the hours of 06:00 and 08:00. Since we're using the 'hour' values, this means the values of '6' and '7' (represents 06:00 - 07:59).
library(plyr)
df2 <- ddply(df2[df2$hour %in% 6:7,], .(date), mutate, non_zero=sum(c))
# a b c hour date non_zero
# 1 2 2012-12-05 06:00:00 0 6 12/5/2012 0
# 2 4 2012-12-06 06:00:00 1 6 12/6/2012 1
# 3 6 2012-12-07 07:00:00 1 7 12/7/2012 1
The 'plyr' package is wonderful for things like this. The 'ddply' package specifically takes data frames as both input and output (hence the "dd"), and the 'mutate' function allows us to preserve all the data while adding additional columns. In this case, we're wanting a sum of 'c' for each day in .(date). Subsetting our data by the hours is taken care of in the data argument df2[df2$hour %in% 6:7,], which says to show us the rows where the hour value is in the set {6,7}.
Step 3: The final step is just to subset the data by the max number of non-zero values. We can drop the extra columns we used and go back to our original three.
subset_df <- df2[df2$non_zero==max(df2$non_zero),1:3]
# a b c
# 2 4 2012-12-06 06:00:00 1
# 3 6 2012-12-07 07:00:00 1
Good luck!
Update: At the OP's request, I am writing a new 'ddply' function that will also include a time column for plotting.
df2 <- ddply(df2[df2$hour %in% 6:7,], .(date), mutate, non_zero=sum(c), plot_time=as.numeric(format(b, "%H")) + as.numeric(format(b, "%M")) / 60)
subset_df <- df2[df2$non_zero==max(df2$non_zero),c("a","b","c","plot_time")]
We need to collapse the time down into one continuous variable, so I chose hours. Leaving any data in a time format will require us to fiddle with stuff later, and using a string format (like "hh:mm") will limit the types of functions you can use on it. Continuous numbers are the most flexible, so here we get the number of hours as.numeric(format(b, "%H")) and add it to the number of minutes divided by 60 as.numeric(format(b, "%M")) / 60 to convert the minutes into units of hours. Also, since we're dealing with more columns, I've switched the final subset statement to name the columns we want, rather than referring to the numbers. Once I'm dealing with columns that aren't in continuous order, I find that using names is easier to debug.
Agreeing with Jack. Sounds like a corrupted installation of R. First thing to try would be to delete the .Rdata file that holds the results of the prior session. They are hidden in both Mac and Windows so unless you "reveal" the 'dotfiles'(system files), the OS file manager (Finder.app and Windows Explorer) will not show them. How you find and delete that file is OS-specific task. It's going to be in your working directory and you will need to do the deletion outside of R since once R is started it will have locked access to it. It's also possible to get a corrupt .history file but in my experience that is not usually the source of the problem.
If that is not successful, you may need to reinstall R.
I have an example dataframe:
a <- c(1:6)
b <- c("05/12/2012 05:00","05/12/2012 06:00","06/12/2012 05:00",
"06/12/2012 06:00", "07/12/2012 09:00","07/12/2012 07:00")
c <-c("0","0","0","1","1","1")
df1 <- data.frame(a,b,c,stringsAsFactors = FALSE)
Firstly, I want to make sure R recognises the date and time format, so I used:
df1$b <- strptime(df1$b, "%d/%m/%Y %H:%M")
However this can't be right as R always aborts my session as soon as I try to view the new dataframe.
Assuming that this gets resolves, I want to get a subset of the data according to whichever day in the dataframe contains the most data in 'C' that is not a zero. In the above example I should be left with the two data points on 7th Dec 2012.
I also have an additional, related question.
If I want to be left with a subset of the data with the most non zero values between a certain time period in the day (say between 07:00 and 08:00), how would I go about doing this?
Any help on the above problems would be greatly appreciated.
Well, the good news is that I have an answer for you, and the bad news is that you have more questions to ask yourself. First the bad news: you need to consider how you want to treat multiple days that have the same number of non-zero values for 'c'. I'm not going to address that in this answer.
Now the good news: this is really simple.
Step 1: First, let's reformat your data frame. Since we're changing data types on a couple of the variables (b to datetime and c to numeric), we need to create a new data frame or recalibrate the old one. I prefer to preserve the original and create a new one, like so:
a <- df1$a
b <- strptime(df1$b, "%d/%m/%Y %H:%M")
c <- as.numeric(df1$c)
hour <- as.numeric(format(b, "%H"))
date <- format(b, "%x")
df2 <- data.frame(a, b, c, hour, date)
# a b c hour date
# 1 1 2012-12-05 05:00:00 0 5 12/5/2012
# 2 2 2012-12-05 06:00:00 0 6 12/5/2012
# 3 3 2012-12-06 05:00:00 0 5 12/6/2012
# 4 4 2012-12-06 06:00:00 1 6 12/6/2012
# 5 5 2012-12-07 09:00:00 1 9 12/7/2012
# 6 6 2012-12-07 07:00:00 1 7 12/7/2012
Notice that I also added 'hour' and 'date' variables. This is to make our data easily sortable by those fields for our later aggregation function.
Step 2: Now, let's calculate how many non-zero values there are for each day between the hours of 06:00 and 08:00. Since we're using the 'hour' values, this means the values of '6' and '7' (represents 06:00 - 07:59).
library(plyr)
df2 <- ddply(df2[df2$hour %in% 6:7,], .(date), mutate, non_zero=sum(c))
# a b c hour date non_zero
# 1 2 2012-12-05 06:00:00 0 6 12/5/2012 0
# 2 4 2012-12-06 06:00:00 1 6 12/6/2012 1
# 3 6 2012-12-07 07:00:00 1 7 12/7/2012 1
The 'plyr' package is wonderful for things like this. The 'ddply' package specifically takes data frames as both input and output (hence the "dd"), and the 'mutate' function allows us to preserve all the data while adding additional columns. In this case, we're wanting a sum of 'c' for each day in .(date). Subsetting our data by the hours is taken care of in the data argument df2[df2$hour %in% 6:7,], which says to show us the rows where the hour value is in the set {6,7}.
Step 3: The final step is just to subset the data by the max number of non-zero values. We can drop the extra columns we used and go back to our original three.
subset_df <- df2[df2$non_zero==max(df2$non_zero),1:3]
# a b c
# 2 4 2012-12-06 06:00:00 1
# 3 6 2012-12-07 07:00:00 1
Good luck!
Update: At the OP's request, I am writing a new 'ddply' function that will also include a time column for plotting.
df2 <- ddply(df2[df2$hour %in% 6:7,], .(date), mutate, non_zero=sum(c), plot_time=as.numeric(format(b, "%H")) + as.numeric(format(b, "%M")) / 60)
subset_df <- df2[df2$non_zero==max(df2$non_zero),c("a","b","c","plot_time")]
We need to collapse the time down into one continuous variable, so I chose hours. Leaving any data in a time format will require us to fiddle with stuff later, and using a string format (like "hh:mm") will limit the types of functions you can use on it. Continuous numbers are the most flexible, so here we get the number of hours as.numeric(format(b, "%H")) and add it to the number of minutes divided by 60 as.numeric(format(b, "%M")) / 60 to convert the minutes into units of hours. Also, since we're dealing with more columns, I've switched the final subset statement to name the columns we want, rather than referring to the numbers. Once I'm dealing with columns that aren't in continuous order, I find that using names is easier to debug.
Agreeing with Jack. Sounds like a corrupted installation of R. First thing to try would be to delete the .Rdata file that holds the results of the prior session. They are hidden in both Mac and Windows so unless you "reveal" the 'dotfiles'(system files), the OS file manager (Finder.app and Windows Explorer) will not show them. How you find and delete that file is OS-specific task. It's going to be in your working directory and you will need to do the deletion outside of R since once R is started it will have locked access to it. It's also possible to get a corrupt .history file but in my experience that is not usually the source of the problem.
If that is not successful, you may need to reinstall R.