R previous week total [duplicate] - r

This question already has an answer here:
Aggregate by week in R
(1 answer)
Closed 7 years ago.
I have Date (column B) and Total (column A) variables - how can I create a new variable in R that sums the previous seven days' worth of Totals?
In Excel, I have the following formula:
=SUMIFS($A:$A,$B:$B, ">="&$B20-7,$B:$B,"<"&$B20)
and I just don't know how to convert this to work in R. Suggestions?

This will do it too, advanced, but short - essentially a one-liner.
# Initialze some data
date <- seq(as.Date("2001-01-01"),as.Date("2001-01-31"),"days")
tot <- trunc(rnorm(31,100,20))
df <- data.frame(date,tot)
# Now compute week sum by summing a subsetted df for each date
df$wktot <- sapply(df$date,function(x)sum(df[difftime(df$date,x,,"days") %in% 0:-6,]$tot))
Changed the variable names to match the posed problem.
It also handles the data in any order and multiple entries per day.
Edited to add comments and make it fit in a window.

If there is one total per day, this function may help:
rollSums <- function(totals, roll) {
res <- c()
for(i in 1:(length(totals)-roll)) {
res <- c(res, sum(totals[0:(roll-1)+i]))
}
res
}
df1
Total Date
1 3 2015-01-01
2 8 2015-01-01
3 4 2015-01-02
4 7 2015-01-03
5 6 2015-01-04
6 1 2015-01-04
7 10 2015-01-05
8 9 2015-01-06
9 2 2015-01-07
10 5 2015-01-08
rollSums(df1$Total, 3)
[1] 15 19 17 14 17 20 21
rollSums(df1$Total, 4)
[1] 22 25 18 24 26 22
It will take two arguments, the vector with the totals and how many days you'd like in each sum.
Data
dput(df1)
structure(list(Total = c(3L, 8L, 4L, 7L, 6L, 1L, 10L, 9L, 2L,
5L), Date = structure(c(16436, 16436, 16437, 16438, 16439, 16439,
16440, 16441, 16442, 16443), class = "Date")), .Names = c("Total",
"Date"), row.names = c(NA, -10L), class = "data.frame")
Update
In case you run into a situation with multiple values on the same day, here's a solution. Surprisingly, #MikeWise has a one-liner that can do all of this. See other answer.
grouped.roll <- function(DF, Values, Group, roll) {
totals <- eval(substitute(with(DF, tapply(Values, Group, sum))))
newsums <- rollSums(totals, roll)
data.frame(Group=names(totals), Sums=c(rep(NA, roll), newsums))
}
It uses the rollSums that I used earlier. It will spit out NAs until the desired day grouping begins. That may be the only advantage over the other answer. But they could easily edit that in, I'm sure. Just providing more options for reference.
grouped.roll(df1, Total, Date, 3)
Group Sums
1 2015-01-01 NA
2 2015-01-02 NA
3 2015-01-03 NA
4 2015-01-04 22
5 2015-01-05 18
6 2015-01-06 24
7 2015-01-07 26
8 2015-01-08 21

Related

For-loop, calculate predicted movement, delete row if outside range

I have a data-set containing vehicle movement, where vehicle time and position is logged with GPS. The issue is that there exist duplicate values (spoofing) for several of the vehicles, and I am not able to identify the true vehicle, outside the guess that the first time the vehicle is introduced it is the true vehicle. My intent is to create a fore-loop that calculates predicted movement from one position to the next , and if the next value is outside this value the row will be deleted. This will also remove outliers, where one position is extremely off for some reason.
Dataset
Vehicle ID Time Lat Long Max Speed (kts)
1 01.01.2013 12:00:00 9.535 18.536 20
1 01.01.2013 12:10:00 9.539 18.539 20
1 01.01.2013 12:20:00 65.535 35.545 20
1 01.01.2013 12:30:00 65.835 35.545 20
1 01.01.2013 12:40:00 9.541 18.542 20
1 01.01.2013 12:50:00 66.135 35.536 20
1 01.01.2013 13:00:00 9.543 18.545 20
2 05.01.2013 17:00:00 13.535 15.536 30
The idea is to run a loop that takes the position from row 1 if the Vehicle ID = Vehicle Id + 1, and calculates the maximum distance possibly traveled by calculating the time between time and time for next row (time + 1) and multiple this by the maximum speed. Then calculate a maximum and minimum latitude and longitude for where the vehicle can be theoretically be at (time + 1) and if the position is outside these maximum values, the row will be deleted and the loop will run the same statement on next row.
Something like this:
if vehicle ID = vehicle ID[n+1],
then (create latmax and latmin) ( time[n+1] - time ) * maximum speed +- latitude &
then (create lonmax and lonmin) ( time[n+1] - time ) * maximum speed +- longitude
then if lat[n+1] > latmax & lat[n+1] < latmin & lon[n+1] > lonmax & lon[n+1] < lonmax (deleterow) if not, do the same at next line
This should result in deleting row 3, 4 and 6 in my sample. For row 8 there is a new vehicle and a large deviation in position is allowed.
This method does not take account in the fact that the earth is circular and distance between latitudes decrease the closer we get to the north or south pole. The best solution would take this into account by solving for this mathematically in the formula, or using distm or similar to calculate the true distance. Implementing cosinus in the formula is properly the easiest method. However the deviation between the outliers and the true position is normalt so large, that the curvature of the earth does not matter in practice for this data-set.
Ok you have 2 problems here, you have a research problem where you need to define the appropriate distance to compare the spoofing pairs in function of coordinates, but foremost you need to define the spoofing pairs and the id of their previous known non-spoofing coordinates. The first problem is a research problem, and I won't go deep into it, but perhaps an internet search on how to calculate distances based on coordinates would help you. A solution to the second problem, the coding part, is proposed below, assuming you want to choose the minimal distance to the latest known non-spoofing position.
First you can take my same example by running this:
dput(df)
structure(list(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L), structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L), .Label = c("01.01.2013", "05.01.2013"
), class = "factor"), structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L,
8L, 3L), .Label = c("12:00:00", "12:10:00", "12:20:00", "12:30:00",
"12:40:00", "12:50:00", "13:00:00", "17:00:00"), class = "factor"),
c(9.535, 9.635, 65.535, 65.835, 9.935, 66.135, 10.235, 13.535,
40.535), c(18.536, 18.636, 35.536, 35.536, 18.936, 35.536,
19.236, 15.536, 40.545), c(20L, 20L, 20L, 20L, 20L, 20L,
20L, 30L, 20L)), .Names = c("Vehicle ID", "date", "Time",
"Lat", "Long", "Max Speed (kts)"), class = "data.frame", row.names = c(NA,
-9L))
My method is to use a series of apply functions. I am also interested if someone knows a more elegant way of doing it other than explicit loops, which might perhaps do the job in fewer steps, but I tend to avoid those.
spoofingtestdb <- df[,1:3]
df$spoofing <- duplicated(spoofingtestdb)|duplicated(spoofingtestdb, fromLast = T)
df$datetime <- dmy_hms(paste0(df$date,"-", df$Time))
df$candidatespreviousposition <- apply(df, 1, function(x) which(df$`Vehicle ID`== x["Vehicle ID"] & !df$spoofing & (as_datetime(df$datetime) < as_datetime(x["datetime"])) ) )
df$latestpreviousposition <- NA
for(i in 1: nrow(df)){
if(length(df$candidatespreviousposition[[i]]>0)) df$latestpreviousposition[[i]] <- df$candidatespreviousposition[[i]][which.max(df$datetime[df$candidatespreviousposition[[i]]])]
}
df$spoofingkey <- paste0(df$`Vehicle ID`, df$datetime)
df$spoofingid <- ifelse(df$spoofing, apply(df, 1, function(x) which(df$spoofingkey==x["spoofingkey"])), NA)
df$lat1 <- apply(df, 1, function(x) df$Lat[x[["spoofingid"]][1]][which(!is.na(df$Lat[x[["spoofingid"]][1]]))] )
df$long1 <- apply(df, 1, function(x) df$Long[x[["spoofingid"]][1]][which(!is.na(df$Long[x[["spoofingid"]][1]]))] )
df$latinit <- apply(df, 1, function(x) df$Lat[x["latestpreviousposition"]])
df$latinit <- ifelse(df$spoofing, df$Lat[df$latestpreviousposition], NA)
df$longinit <- ifelse(df$spoofing, df$Long[df$latestpreviousposition], NA)
getdistance <- function(latinit, longinit, lat, long) {
distance1 <- abs(lat-latinit)+abs(long-longinit)
}
df$distance <- ifelse(df$spoofing, getdistance(df$latinit, df$longinit, df$Lat, df$Long), NA )
df$spoofingnumber <- apply(df, 1, function(x) paste0(x["spoofingid"], collapse=""))
#apply(df, 1, function(x) which(df$spoofingnumber==x["spoofingnumber"]))
df$ismindistance <- apply(df, 1, function(x) x["distance"] == min(df$distance[which(df$spoofingnumber==x["spoofingnumber"])]))
df$tokeep <- ifelse(is.na(df$ismindistance)|df$ismindistance, T, F)
result <- df[df$tokeep,]
result
Here just using a basic distance calculation function. The result is below, as you can see the second row has been deleted in my example, only the minimum distance was kept with respect to the previous known position.
Vehicle ID date Time Lat Long Max Speed (kts) spoofing datetime candidatespreviousposition
1 1 01.01.2013 12:00:00 9.535 18.536 20 FALSE 2013-01-01 12:00:00
2 1 01.01.2013 12:10:00 9.635 18.636 20 FALSE 2013-01-01 12:10:00 1
4 1 01.01.2013 12:30:00 65.835 35.536 20 FALSE 2013-01-01 12:30:00 1, 2
5 1 01.01.2013 12:40:00 9.935 18.936 20 FALSE 2013-01-01 12:40:00 1, 2, 4
6 1 01.01.2013 12:50:00 66.135 35.536 20 FALSE 2013-01-01 12:50:00 1, 2, 4, 5
7 1 01.01.2013 13:00:00 10.235 19.236 20 FALSE 2013-01-01 13:00:00 1, 2, 4, 5, 6
8 2 05.01.2013 17:00:00 13.535 15.536 30 FALSE 2013-01-05 17:00:00
9 1 01.01.2013 12:20:00 40.535 40.545 20 TRUE 2013-01-01 12:20:00 1, 2
latestpreviousposition spoofingkey spoofingid lat1 long1 latinit longinit distance spoofingnumber ismindistance tokeep
1 NA 12013-01-01 12:00:00 NA NA NA NA NA NA TRUE
2 1 12013-01-01 12:10:00 NA NA NA NA NA NA TRUE
4 2 12013-01-01 12:30:00 NA NA NA NA NA NA TRUE
5 4 12013-01-01 12:40:00 NA NA NA NA NA NA TRUE
6 5 12013-01-01 12:50:00 NA NA NA NA NA NA TRUE
7 6 12013-01-01 13:00:00 NA NA NA NA NA NA TRUE
8 NA 22013-01-05 17:00:00 NA NA NA NA NA NA TRUE
9 2 12013-01-01 12:20:00 3, 9 65.535 35.536 9.635 18.636 52.809 c(3, 9) TRUE TRUE
After you choose which distance function is appropriate for you, you can just replace the getdistance() function above.

R - Having some trouble with plotting using time data

I am currently trying to write a forecasting algorithm in R, but I'm having an issue extracting my time data from a txt file.
I currently have a test text file with the following data
x
1 2010-01-01
2 2010-07-02
3 2010-08-03
4 2011-02-04
5 2011-11-05
6 2011-12-06
7 2012-06-07
8 2012-08-30
9 2013-04-16
10 2013-03-18
11 2014-02-22
12 2014-01-27
13 2015-12-15
14 2015-09-28
15 2016-05-04
16 2017-11-07
17 2017-09-22
18 2017-04-04
When I extract it and try to plot it with the following code:
library(forecast)
library(ggplot2)
Quantity <- c(read.table("....Path..../Quantity.txt"))
Time <- c(read.table("....Path..../Time.txt"))
x <- ts(as.Date(unlist(Time)))
y <- unlist(Quantity)
plot(x,y)
The resulting graph displays all the points on the graph correctly, except for the labels for time (which are 14500, 16000, and 17500). The labels should show the dates from the file, but the way I see it, its probably treating the data as a maths sum (and does a calculation resulting in those values) and not dates.
I also have an issue that the time data is not being plotted in chronological order, but instead in the order from the files.
Here's the data from the other file just for reference:
x
1 5
2 3
3 8
4 4
5 0
6 5
7 2
8 7
9 4
10 2
11 6
12 8
13 4
14 7
15 8
16 9
17 4
18 6
How can I correct these 2 issues?
Thanks in advance.
Here is one of the many possible solutions.
I hope it can help you.
# A dataset with date and x values
# Important: the format of date is "character"
df <- structure(list(date = c("2010-01-01", "2010-07-02", "2010-08-03",
"2011-02-04", "2011-11-05", "2011-12-06", "2012-06-07", "2012-08-30",
"2013-04-16", "2013-03-18", "2014-02-22", "2014-01-27", "2015-12-15",
"2015-09-28", "2016-05-04", "2017-11-07", "2017-09-22", "2017-04-04"
), x = c(5L, 3L, 8L, 4L, 0L, 5L, 2L, 7L, 4L, 2L, 6L, 8L, 4L,
7L, 8L, 9L, 4L, 6L)), .Names = c("date", "x"), row.names = c(NA,
-18L), class = "data.frame")
str(df)
# Create a x vector with dates as rownames
x <- as.matrix(df$x)
rownames(x) <- df$date
# Convert in a xts object
library(xts)
x <- as.xts(x)
# Plot the xts object
plot(x, grid.col="white")
To answer your ggplot question, using the data frame that Marco provided above, you would simply use:
ggplot(df, aes(x = date, y = x)) + geom_line(group = 1)
Since you have only one group or one set of points, you must use the group = 1 arg in geom_line.
One of the things I will point out is that your time series data has irregular periods, and you will have to make sure that you account for that in your time series object. Most time series packages have their own specialized functions for handling the data and plotting.

Date Interval removal

I have data that looks like
ID CLM_ID Date1 Date2
1 718182 1/1/2014 1/17/2014
1 718184 1/2/2014 1/8/2014
1 885236 1/15/2014 1/17/2014
1 885362 3/20/2014 3/21/2014
2 589963 3/18/2015 3/22/2015
2 589999 2/27/2015 5/9/2015
2 594226 4/11/2015 4/17/2015
2 689959 5/10/2015 6/10/2015
3 656696 5/1/2016 5/5/2016
3 669625 5/6/2016 5/22/2016
4 777777 2/21/2015 3/4/2015
4 778952 2/1/2015 2/28/2015
4 778965 3/1/2015 3/22/2015
I am working on two different problems with this. The first one was answered in a previous post about how to roll dates up (Date roll-up in R) and the second now is that I have intervals that are within intervals and I am trying to get rid of them. So the final outcome should look like
ID CLM_ID Date1 Date2
1 718182 1/1/2014 1/17/2014
1 885362 3/20/2014 3/21/2014
2 589999 2/27/2015 5/9/2015
3 656696 5/1/2016 5/22/2016
4 778952 2/1/2015 3/22/2015
Now I know I will have to create the extended intervals via the date rollup first, but then how do I get rid of these sub-intervals (a term I am making up for intervals within intervals)? I am also looking for a solution that is efficient since I actually have 75,000 records to go through (i.e. I am trying to avoid iterative solutions).
Using non-equi joins from the current development version of data.table, v1.9.7,
require(data.table) # v1.9.7+
dt[dt, .(CLM_IDs = CLM_IDs[.N==1L]), on=.(ID, Date1<=Date1, Date2>=Date2), by=.EACHI]
# ID Date1 Date2 CLM_ID
# 1: 1 2014-01-01 2014-01-17 718182
# 2: 1 2014-03-20 2014-03-21 885362
# 3: 2 2015-02-27 2015-05-09 589999
# 4: 2 2015-05-10 2015-06-10 689959
# 5: 3 2016-05-01 2016-05-05 656696
# 6: 3 2016-05-06 2016-05-22 669625
# 7: 4 2015-02-21 2015-03-04 777777
# 8: 4 2015-02-01 2015-02-28 778952
# 9: 4 2015-03-01 2015-03-22 778965
What this does is, for each row in dt (the one inside of square bracket), it looks up which rows match in dt (on the outside) based on the condition provided to the on argument.
The matching row indices are returned iff the only match is a self-match (since the condition includes equality as well). This is done by CLM_IDs[.N == 1L], where .N holds the number of observations for each group.
"I am also looking for a solution that is efficient ... (i.e. I am trying to avoid iterative solutions)."
"Your assumptions are your windows on the world. Scrub them off every once in a while, or the light won't come in." - Isaac Asimov
Below is a super fast base R iterative solution. It returns the correct results for very large data frames virtually instantly. (it also "rolls-up" the data, so there is no need to carry out two algorithms):
MakeDFSubInt <- function(df, includeCost = FALSE) {
## Sorting the data frame to allow for fast
## creation of the "Contained" logical vector below
tempDF <- df[order(df$ID, df$Date1, df$Date2), ]
UniIDs <- unique(tempDF$ID)
Len <- length(UniIDs)
## Determine starting (i.e. "s") and ending (i.e. "e")
## points of the respective groups of IDs
e <- which(diff(tempDF$ID)==1)
s <- c(1L, e + 1L)
dfLen <- nrow(tempDF)
e <- c(e, dfLen)
## Converting dates to integers so that comparison
## will be faster. Internally dates are stored as
## integers, so this isn't a problem
dte1 <- as.integer(tempDF$Date1)
dte2 <- as.integer(tempDF$Date2)
## Building logical vector in order to quickly create sub-intervals
Contained <- rep(FALSE, dfLen)
BegTime <- Sys.time() ## Included to measure time of for loop execution
for (j in 1:Len) {
Compare <- ifelse(dte2[s[j]] >= (dte1[s[j]+1L]+1L), max(dte2[s[j]], dte2[s[j]+1L]), dte2[s[j]+1L])
for (x in (s[j]+1L):e[j]) {
if (!Contained[x-1L]) {
Contained[x] <- dte2[x-1L] >= (dte1[x]-1L)
} else {
Contained[x] <- Compare >= (dte1[x]-1L)
}
## could use ifelse, but this construct is faster
if (Contained[x]) {
Compare <- max(Compare, dte2[x])
} else {
Compare <- dte2[x]
}
}
}
EndTime <- Sys.time()
TotTime <- EndTime - BegTime
if (printTime) {print(paste(c("for loop execution time was: ", format(TotTime)), collapse = ""))}
## identify sub-intervals
nGrps <- which(!Contained)
## Create New fields for our new DF
ID <- tempDF$ID[nGrps]
CLM_ID <- tempDF$CLM_ID[nGrps]
Date1 <- tempDF$Date1[nGrps]
nGrps <- c(nGrps, dfLen+1L)
## as.Date is converting numbers to dates.
## N.B. This only works if origin is supplied
Date2 <- as.Date(vapply(1L:(length(nGrps) - 1L), function(x) {
max(dte2[nGrps[x]:(nGrps[x+1L]-1L)])}, 1L), origin = "1970-01-01")
## in a related question the OP had, "Cost" was
## included to show how the algorithm would handle
## generic summary information
if (includeCost) {
myCost <- tempDF$Cost
Cost <- vapply(1L:(length(nGrps) - 1L), function(x) sum(myCost[nGrps[x]:(nGrps[x+1L]-1L)]), 100.01)
NewDf <- data.frame(ID,CLM_ID,Date1,Date2,Cost)
} else {
NewDf <- data.frame(ID,CLM_ID,Date1,Date2)
}
NewDf
}
For the example given in the question, we have:
ID <- c(rep(1,4),rep(2,4),rep(3,2),rep(4,3))
CLM_ID <- c(718182, 718184, 885236, 885362, 589963, 589999, 594226, 689959, 656696, 669625, 777777, 778952, 778965)
Date1 <- c("1/1/2014","1/2/2014","1/15/2014","3/20/2014","3/18/2015","2/27/2015","4/11/2015","5/10/2015","5/1/2016","5/6/2016","2/21/2015","2/1/2015","3/1/2015")
Date2 <- c("1/17/2014","1/8/2014","1/17/2014","3/21/2014","3/22/2015","5/9/2015","4/17/2015","6/10/2015","5/5/2016","5/22/2016","3/4/2015","2/28/2015","3/22/2015")
myDF <- data.frame(ID, CLM_ID, Date1, Date2)
myDF$Date1 <- as.Date(myDF$Date1, format = "%m/%d/%Y")
myDF$Date2 <- as.Date(myDF$Date2, format = "%m/%d/%Y")
MakeDFSubInt(myDF)
ID CLM_ID Date1 Date2
1 1 718182 2014-01-01 2014-01-17
2 1 885362 2014-03-20 2014-03-21
3 2 589999 2015-02-27 2015-06-10
4 3 656696 2016-05-01 2016-05-22
5 4 778952 2015-02-01 2015-03-22
From a similar question the OP posted, we can add a Cost field, to show how we would proceed with calculations for this setup.
set.seed(7777)
myDF$Cost <- round(rnorm(13, 450, sd = 100),2)
MakeDFSubInt(myDF, includeCost = TRUE)
ID CLM_ID Date1 Date2 Cost
1 1 718182 2014-01-01 2014-01-17 1164.66
2 1 885362 2014-03-20 2014-03-21 568.16
3 2 589999 2015-02-27 2015-06-10 2019.16
4 3 656696 2016-05-01 2016-05-22 990.14
5 4 778952 2015-02-01 2015-03-22 1578.68
This algorithm scales very well. For data frames the size the OP is looking for, returning the requested DF returns almost instantaneously and for very large data frames, it returns in just seconds.
First we build a function that will generate a random data frame with n rows.
MakeRandomDF <- function(n) {
set.seed(109)
CLM_Size <- ifelse(n < 10^6, 10^6, 10^(ceiling(log10(n))))
numYears <- trunc((6/425000)*n + 5)
StrtYear <- ifelse(numYears > 16, 2000, 2016 - numYears)
numYears <- ifelse(numYears > 16, 16, numYears)
IDs <- sort(sample(trunc(n/100), n, replace = TRUE))
CLM_IDs <- sample(CLM_Size, n)
StrtDate <- as.Date(paste(c(as.character(StrtYear),"-01-01"), collapse = ""))
myPossibleDates <- StrtDate+(0:(numYears*365)) ## "numYears" years of data
Date1 <- sample(myPossibleDates, n, replace = TRUE)
Date2 <- Date1 + sample(1:100, n, replace = TRUE)
Cost <- round(rnorm(n, 850, 100), 2)
tempDF <- data.frame(IDs,CLM_IDs,Date1,Date2,Cost)
tempDF$Date1 <- as.Date(tempDF$Date1, format = "%m/%d/%Y")
tempDF$Date2 <- as.Date(tempDF$Date2, format = "%m/%d/%Y")
tempDF
}
For moderate size DFs (i.e. 75,000 rows)
TestDF <- MakeRandomDF(75000)
system.time(test1 <- MakeDFSubInt(TestDF, includeCost = TRUE, printTime = TRUE))
[1] "for loop execution time was: 0.06500006 secs"
user system elapsed
0.14 0.00 0.14
nrow(test1)
[1] 7618
head(test1)
ID CLM_ID Date1 Date2 Cost
1 1 116944 2010-01-29 2010-01-30 799.90 ## The range of dates for
2 1 515993 2010-02-15 2011-10-12 20836.83 ## each row are disjoint
3 1 408037 2011-12-13 2013-07-21 28149.26 ## as requested by the OP
4 1 20591 2013-07-25 2014-03-11 10449.51
5 1 338609 2014-04-24 2014-07-31 4219.48
6 1 628983 2014-08-03 2014-09-11 2170.93
For very large DFs (i.e. > 500,000 rows)
TestDF2 <- MakeRandomDF(500000)
system.time(test2 <- MakeDFSubInt(TestDF2, includeCost = TRUE, printTime = TRUE))
[1] "for loop execution time was: 0.3679998 secs"
user system elapsed
1.19 0.03 1.21
nrow(test2)
[1] 154839
head(test2)
ID CLM_ID Date1 Date2 Cost
1 1 71251 2004-04-19 2004-06-29 2715.69 ## The range of dates for
2 1 601676 2004-07-05 2004-09-23 2675.04 ## each row are disjoint
3 1 794409 2004-12-28 2005-04-05 1760.63 ## as requested by the OP
4 1 424671 2005-06-03 2005-08-20 1973.67
5 1 390353 2005-09-16 2005-11-06 785.81
6 1 496611 2005-11-21 2005-11-24 904.09
system.time(test3 <- MakeDFSubInt(TestDF3, includeCost = TRUE, printTime = TRUE))
[1] "for loop execution time was: 0.6930001 secs"
user system elapsed
2.68 0.08 2.79 ## 1 million rows in under 3 seconds!!!
nrow(test3)
[1] 413668
Explanation
The main part of the algorithm is generating the Contained logical vector that is used to determine the sub-intervals of continuous dates. Generation of this vector relies on the fact that the data frame is sorted, first by ID, second by Date1, and finally by Date2. We begin by locating the starting and ending rows of each group of IDs. For example, with the example provided by the OP we have:
myDF
ID CLM_ID Date1 Date2
1 1 718182 2014-01-01 2014-01-17 ## <- 1 s[1]
2 1 718184 2014-01-02 2014-01-08
3 1 885236 2014-01-15 2014-01-17
4 1 885362 2014-03-20 2014-03-21 ## <- 4 e[1]
5 2 589963 2015-03-18 2015-03-22 ## <- 5 s[2]
6 2 589999 2015-02-27 2015-05-09
7 2 594226 2015-04-11 2015-04-17
8 2 689959 2015-05-10 2015-06-10 ## <- 8 e[2]
9 3 656696 2016-05-01 2016-05-05 ## <- 9 s[3]
10 3 669625 2016-05-06 2016-05-22 ## <- 10 e[3]
11 4 777777 2015-02-21 2015-03-04 ## <- 11 s[4]
12 4 778952 2015-02-01 2015-02-28
13 4 778965 2015-03-01 2015-03-22 ## <- 13 e[4]
Below is the code that generates s and e.
## Determine starting (i.e. "s") and ending (i.e. "e")
## points of the respective groups of IDs
e <- which(diff(tempDF$ID)==1)
s <- c(1L, e + 1L)
dfLen <- nrow(tempDF)
e <- c(e, dfLen)
s
1 5 9 11
e
4 8 10 13
Now, we loop over each group and begin populating the logical vector Contained. If the date range for a particular row overlaps (or is a continuance of) the date range above it, we set that particular index of Contained to TRUE. This is why the first row in each group is set to FALSE since there is nothing above to compare it to. As we are doing this, we are updating the largest date to compare against moving forward, hence the Compare variable. It should be noted that it isn't necessarily true that Date2[n] < Date2[n+1L], this is why Compare <- max(Compare, dte2[x]) for a succession of TRUEs. The result for our example is give below.
ID CLM_ID Date1 Date2 Contained
1 1 718182 2014-01-01 2014-01-17 FALSE
2 1 718184 2014-01-02 2014-01-08 TRUE ## These two rows are contained
3 1 885236 2014-01-15 2014-01-17 TRUE ## in the date range 1/1 - 1/17
4 1 885362 2014-03-20 2014-03-21 FALSE ## This row isn't
6 2 589999 2015-02-27 2015-05-09 FALSE
5 2 589963 2015-03-18 2015-03-22 TRUE
7 2 594226 2015-04-11 2015-04-17 TRUE
8 2 689959 2015-05-10 2015-06-10 TRUE ## N.B. 5/10 is a continuance of 5/09
9 3 656696 2016-05-01 2016-05-05 FALSE
10 3 669625 2016-05-06 2016-05-22 TRUE
12 4 778952 2015-02-01 2015-02-28 FALSE
11 4 777777 2015-02-21 2015-03-04 TRUE
13 4 778965 2015-03-01 2015-03-22 TRUE
Now we can easily identify the "starting" rows by identifying all rows with a corresponding FALSE. After this, finding summary information is a breeze by simply calculating whatever you are interested in (e.g. max(Date2), sum(Cost)) over each succession of TRUEs and Voila!!
Here is a not-so-pretty solution comparing each row with the dates of all other rows. I corrected the one year 3015 to 2015. The results are different from what you are expecting, though. Either I misunderstood your question, or you misread the data.
Data:
dta <- structure(list(ID = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 4L, 4L, 4L),
CLM_ID = c(718182L, 718184L, 885236L, 885362L, 589963L, 589999L, 594226L, 689959L, 656696L, 669625L, 777777L, 778952L, 778965L),
Date1 = structure(c(1L, 3L, 2L, 9L, 8L, 6L, 10L, 12L, 11L, 13L, 5L, 4L, 7L), .Label = c("1/1/2014", "1/15/2014", "1/2/2014", "2/1/2015", "2/21/2015", "2/27/2015", "3/1/2015", "3/18/2015", "3/20/2014", "4/11/2015", "5/1/2016", "5/10/2015", "5/6/2016"), class = "factor"),
Date2 = structure(c(1L, 2L, 1L, 4L, 5L, 10L, 7L, 11L, 9L, 8L, 6L, 3L, 5L), .Label = c("1/17/2014", "1/8/2014", "2/28/2015", "3/21/2014", "3/22/2015", "3/4/2015", "4/17/2015", "5/22/2016", "5/5/2016", "5/9/2015", "6/10/2015"), class = "factor")),
.Names = c("ID", "CLM_ID", "Date1", "Date2"), class = "data.frame",
row.names = c(NA, -13L))
Code:
dta$Date1 <- as.Date(dta$Date1, format = "%m/%d/%Y")
dta$Date2 <- as.Date(dta$Date2, format = "%m/%d/%Y")
# Boolean vector to memorize results
keep <- logical(length = nrow(dta))
for(i in 1:nrow(dta)) {
match <- dta[dta$Date1 <= dta$Date1[i] & dta$Date2 >= dta$Date2[i], ]
if(nrow(match) == 1) keep[i] <- TRUE
}
# Result
dta[keep, ]

How to change the date format in my data frame using R

I wanted to plot a graph of n(y-axis) versus date(x-axis) in R, but due to the format of the date displayed in my data, the order of the date wasn't in the correct ascending order. How can I solve this? Appreciate for the help.
hybrid <- readWorksheetFromFile(excel.file, sheet="ResultSet", header=TRUE)
wb <- loadWorkbook(excel.file)
setMissingValue(wb,value=c("NA"))
hybrid1 <- readWorksheet(wb, sheet="ResultSet", header=TRUE)
I used the dplyr function. Suppose each Pub.Number have a unique code & I replaced it with one. Then, I count the number of it for a certain date.
hybrid <- mutate(hybrid1, n=sum(Publication.Number=1))
p1 <- select(hybrid1, Publication.Date, n)
pt <- count(p1, Publication.Date, wt=n)
The output look like this:
pt
Source: local data frame [627 x 2]
Publication.Date n
(chr) (dbl)
1 01.01.2013 1
2 01.01.2014 8
3 01.01.2015 10
4 01.02.2012 3
5 01.03.2012 16
6 01.04.2015 2
7 01.05.2012 1
8 01.05.2013 7
9 01.05.2014 23
10 01.06.2011 1
.. ... ...
Then, I plotted it but R recognized Pub.Date as character
qplot(x=Publication.Date, y=n, data=pt, geom="point")
x <- hybrid1[,2]
class(x)
[1] "character"
The graph I've plotted is a mess because of the wrong order of the date
I tried using the as.Date function but it seems that it's not complete (I'm using R version 3.2.2)
> pt[,1] <- as.Date(pt[,1], format='%d.%m.%Y’)
+
First convert 'Publication.Date’ to Date format, then order:
using your data:
data <- read.table(pipe('pbpaste'),sep='',header=T,stringsAsFactors = F)
data <- data[,-1]
names(data) <- c('Pub.Date', 'n’)
Pub.Date n
1 01.01.2014 8
2 01.01.2015 10
3 01.02.2012 3
4 01.03.2012 16
5 01.04.2015 2
6 01.05.2012 1
7 01.05.2013 7
8 01.05.2014 23
9 01.06.2011 1
convert ‘Pub.Date’ to date format:
data[,1] <- as.Date(data[,1],format='%d.%m.%Y’)
and order:
data[order(data$"Pub.Date",data$n), ]
Pub.Date n
9 2011-06-01 1
3 2012-02-01 3
4 2012-03-01 16
6 2012-05-01 1
7 2013-05-01 7
1 2014-01-01 8
8 2014-05-01 23
2 2015-01-01 10
5 2015-04-01 2
In the usual course of data input with R, values like " 01.01.2013" will become factor variables. Since they are not in one of the two "stadard Date formats: YYYY/MM/DD or YYYY-MM-DD, they cannot be input directly as "Date"s with "colClasses" unless you build an "as.DT" method. You will need to make sure they are character vectors either by using stringsAsFactors=FALSE in a read function or by coercing to character with as.character after they are input. That header you have displayed makes me think this data has been manipualtes dsomehow, perhaps with functions in the dplyr package?
res <- structure(list(Publication.Date = structure(1:10, .Label = c("01.01.2013",
"01.01.2014", "01.01.2015", "01.02.2012", "01.03.2012", "01.04.2015",
"01.05.2012", "01.05.2013", "01.05.2014", "01.06.2011"), class = "factor"),
n = c(1L, 8L, 10L, 3L, 16L, 2L, 1L, 7L, 23L, 1L)), .Names = c("Publication.Date",
"n"), class = "data.frame", row.names = c("1", "2", "3", "4",
"5", "6", "7", "8", "9", "10"))
> res
Publication.Date n
1 01.01.2013 1
2 01.01.2014 8
3 01.01.2015 10
4 01.02.2012 3
5 01.03.2012 16
6 01.04.2015 2
7 01.05.2012 1
8 01.05.2013 7
9 01.05.2014 23
10 01.06.2011 1
> res$Publication.Date <- as.Date( as.character(res$Publication.Date), format="%m.%d.%Y")
Then you can plot:
png(); qplot(x=Publication.Date, y=n, data=res, geom="point"); dev.off()

Aggregate (count) occurences of values over arbitrary timeframe

I have a CSV file with timestamps and certain event-types which happened at this time.
What I want is count the number of occurences of certain event-types in 6-minutes intervals.
The input-data looks like:
date,type
"Sep 22, 2011 12:54:53.081240000","2"
"Sep 22, 2011 12:54:53.083493000","2"
"Sep 22, 2011 12:54:53.084025000","2"
"Sep 22, 2011 12:54:53.086493000","2"
I load and cure the data with this piece of code:
> raw_data <- read.csv('input.csv')
> cured_dates <- c(strptime(raw_data$date, '%b %d, %Y %H:%M:%S', tz="CEST"))
> cured_data <- data.frame(cured_dates, c(raw_data$type))
> colnames(cured_data) <- c('date', 'type')
After curing the data looks like this:
> head(cured_data)
date type
1 2011-09-22 14:54:53 2
2 2011-09-22 14:54:53 2
3 2011-09-22 14:54:53 2
4 2011-09-22 14:54:53 2
5 2011-09-22 14:54:53 1
6 2011-09-22 14:54:53 1
I read a lot of samples for xts and zoo, but somehow I can't get a hang on it.
The output data should look something like:
date type count
2011-09-22 14:54:00 CEST 1 11
2011-09-22 14:54:00 CEST 2 19
2011-09-22 15:00:00 CEST 1 9
2011-09-22 15:00:00 CEST 2 12
2011-09-22 15:06:00 CEST 1 23
2011-09-22 15:06:00 CEST 2 18
Zoo's aggregate function looks promising, I found this code-snippet:
# aggregate POSIXct seconds data every 10 minutes
tt <- seq(10, 2000, 10)
x <- zoo(tt, structure(tt, class = c("POSIXt", "POSIXct")))
aggregate(x, time(x) - as.numeric(time(x)) %% 600, mean)
Now I'm just wondering how I could apply this on my use case.
Naive as I am I tried:
> zoo_data <- zoo(cured_data$type, structure(cured_data$time, class = c("POSIXt", "POSIXct")))
> aggr_data = aggregate(zoo_data$type, time(zoo_data$time), - as.numeric(time(zoo_data$time)) %% 360, count)
Error in `$.zoo`(zoo_data, type) : not possible for univariate zoo series
I must admit that I'm not really confident in R, but I try. :-)
I'm kinda lost. Could anyone point me into the right direction?
Thanks a lot!
Cheers, Alex.
Here the output of dput for a small subset of my data. The data itself is something around 80 million rows.
structure(list(date = structure(c(1316697885, 1316697885, 1316697885,
1316697885, 1316697885, 1316697885, 1316697885, 1316697885, 1316697885,
1316697885, 1316697885, 1316697885, 1316697885, 1316697885, 1316697885,
1316697885, 1316697885, 1316697885, 1316697885, 1316697885, 1316697885,
1316697885, 1316697885), class = c("POSIXct", "POSIXt"), tzone = ""),
type = c(2L, 2L, 2L, 2L, 1L, 1L, 1L, 2L, 1L, 2L, 1L, 2L,
1L, 2L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 2L)), .Names = c("date",
"type"), row.names = c(NA, -23L), class = "data.frame")
We can read it using read.csv, convert the first column to a date time binned into 6 minute intervals and add a dummy column of 1's. Then re-read it using read.zoo splitting on the type and aggregating on the dummy column:
# test data
Lines <- 'date,type
"Sep 22, 2011 12:54:53.081240000","2"
"Sep 22, 2011 12:54:53.083493000","2"
"Sep 22, 2011 12:54:53.084025000","2"
"Sep 22, 2011 12:54:53.086493000","2"
"Sep 22, 2011 12:54:53.081240000","3"
"Sep 22, 2011 12:54:53.083493000","3"
"Sep 22, 2011 12:54:53.084025000","3"
"Sep 22, 2011 12:54:53.086493000","4"'
library(zoo)
library(chron)
# convert to chron and bin into 6 minute bins using trunc
# Also add a dummy column of 1's
# and remove any leading space (removing space not needed if there is none)
DF <- read.csv(textConnection(Lines), as.is = TRUE)
fmt <- '%b %d, %Y %H:%M:%S'
DF <- transform(DF, dummy = 1,
date = trunc(as.chron(sub("^ *", "", date), format = fmt), "00:06:00"))
# split and aggregate
z <- read.zoo(DF, split = 2, aggregate = length)
With the above test data the solution looks like this:
> z
2 3 4
(09/22/11 12:54:00) 4 3 1
Note that the above has been done in wide form since that form constitutes a time series whereas the long form does not. There is one column for each type. In our test data we had types 2, 3 and 4 so there are three columns.
(We have used chron here since its trunc method fits well with binning into 6 minute groups. chron does not support time zones which can be an advantage since you can't make one of the many possible time zone errors but if you want POSIXct anyways convert it at the end, e.g. time(z) <- as.POSIXct(paste(as.Date.dates(time(z)), times(time(z)) %% 1)) . This expression is shown in a table in one of the R News 4/1 articles except we used as.Date.dates instead of just as.Date to work around a bug that seems to have been introduced since then. We could also use time(z) <- as.POSIXct(time(z)) but that would result in a different time zone.)
EDIT:
The original solution binned into dates but I noticed afterwards that you wish to bin into 6 minute periods so the solution was revised.
EDIT:
Revised based on comment.
You are almost all the way there. All you need to do now is create a zoo-isch version of that data and map it to the aggregate.zoo code. Since you want to categorize by both time and by type your second argument to aggregate.zoo must be a bit more complex and you want counts rather than means so your should use length(). I do not think that count is a base R or zoo function and the only count function I see in my workspace comes from pkg:plyr so I don't know how well it would play with aggregate.zoo. length works as most people expect for vectors but is often surprises people when working with data.frames. If you do not get what you want with length, then you should see if NROW works instead (and with your data layout they both succeed): With the new data object it is necessary to put the type argument first. AND it surns out the aggregate/zoo only handles single category classifiers so you need to put in the as.vector to remove it zoo-ness:
with(cured_data,
aggregate(as.vector(x), list(type = type,
interval=as.factor(time(x) - as.numeric(time(x)) %% 360)),
FUN=NROW)
)
# interval x
#1 2011-09-22 09:24:00 12
#2 2011-09-22 09:24:00 11
This is an example modified from where you got the code (an example on SO by WizaRd Dirk):
Aggregate (count) occurences of values over arbitrary timeframe
tt <- seq(10, 2000, 10)
x <- zoo(tt, structure(tt, class = c("POSIXt", "POSIXct")))
aggregate(as.vector(x), by=list(cat=as.factor(x),
tms = as.factor(index(x) - as.numeric(index(x)) %% 600)), length)
cat tms x
1 1 1969-12-31 19:00:00 26
2 2 1969-12-31 19:00:00 22
3 3 1969-12-31 19:00:00 11
4 1 1969-12-31 19:10:00 17
5 2 1969-12-31 19:10:00 28
6 3 1969-12-31 19:10:00 15
7 1 1969-12-31 19:20:00 17
8 2 1969-12-31 19:20:00 16
9 3 1969-12-31 19:20:00 27
10 1 1969-12-31 19:30:00 8
11 2 1969-12-31 19:30:00 4
12 3 1969-12-31 19:30:00 9

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