How do I merge two rows in time seriesr? (in R) [duplicate] - r

I have two multivariate time series x and y, both covering approximately the same range in time (one starts two years before the other, but they end on the same date). Both series have missing observations in the form of empty columns next to the date column, and also in the sense that one of the series has several dates that are not found in the other, and vice versa.
I would like to create a data frame (or similar) with a column that lists all the dates found in x OR y, without duplicate dates. For each date (row), I would like to horizontally stack the observations from x next to the observations from y, with NA's filling the missing cells. Example:
>x
"1987-01-01" 7.1 NA 3
"1987-01-02" 5.2 5 2
"1987-01-06" 2.3 NA 9
>y
"1987-01-01" 55.3 66 45
"1987-01-03" 77.3 87 34
# result I would like
"1987-01-01" 7.1 NA 3 55.3 66 45
"1987-01-02" 5.2 5 2 NA NA NA
"1987-01-03" NA NA NA 77.3 87 34
"1987-01-06" 2.3 NA 9 NA NA NA
What I have tried: with the zoo package, I've tried the merge.zoo method, but this seems to just stack the two series next to each other, with the dates (as numbers, e.g. "1987-01-02" shown as 6210) from each series appearing in two separate columns.
I've sat for hours getting almost nowhere, so all help is appreciated.
EDIT: some code included below as per suggestion from Soumendra
atcoa <- read.csv(file = "ATCOA_full_adj.csv", header = TRUE)
atcob <- read.csv(file = "ATCOB_full_adj.csv", header = TRUE)
atcoa$date <- as.Date(atcoa$date)
atcob$date <- as.Date(atcob$date)
# only number of observations and the observations themselves differ
>str(atcoa)
'data.frame': 6151 obs. of 8 variables:
$ date :Class 'Date' num [1:6151] 6210 6213 6215 6216 6217 ...
$ max : num 4.31 4.33 4.38 4.18 4.13 4.05 4.08 4.05 4.08 4.1 ...
$ min : num 4.28 4.31 4.28 4.13 4.05 3.95 3.97 3.95 4 4.02 ...
$ close : num 4.31 4.33 4.31 4.15 4.1 3.97 4 3.97 4.08 4.02 ...
$ avg : num NA NA NA NA NA NA NA NA NA NA ...
$ tot.vol : int 877733 89724 889437 1927113 3050611 846525 1782774 1497998 2504466 5636999 ...
$ turnover : num 3762300 388900 3835900 8015900 12468100 ...
$ transactions: int 12 9 24 17 31 26 34 35 37 33 ...
>atcoa[1:1, ]
date a.max a.min a.close a.avg a.tot.vol a.turnover a.transactions
1 1987-01-02 4.31 4.28 4.31 NA 877733 3762300 12
# using timeSeries package
ts.atcoa <- timeSeries::as.timeSeries(atcoa, format = "%Y-%m-%d")
ts.atcob <- timeSeries::as.timeSeries(atcob, format = "%Y-%m-%d")
>str(ts.atcoa)
Time Series:
Name: object
Data Matrix:
Dimension: 6151 7
Column Names: a.max a.min a.close a.avg a.tot.vol a.turnover a.transactions
Row Names: 1970-01-01 01:43:30 ... 1970-01-01 04:12:35
Positions:
Start: 1970-01-01 01:43:30
End: 1970-01-01 04:12:35
With:
Format: %Y-%m-%d %H:%M:%S
FinCenter: GMT
Units: a.max a.min a.close a.avg a.tot.vol a.turnover a.transactions
Title: Time Series Object
Documentation: Wed Aug 17 13:00:50 2011
>ts.atcoa[1:1, ]
GMT
a.max a.min a.close a.avg a.tot.vol a.turnover a.transactions
1970-01-01 01:43:30 4.31 4.28 4.31 NA 877733 3762300 12
# The following will create an object of class "data frame" and mode "list", which contains observations for the days mutual for the two series
>ts.atco <- timeSeries::merge(atcoa, atcob) # produces same result as base::merge, apparently
>ts.atco[1:1, ]
date a.max a.min a.close a.avg a.tot.vol a.turnover a.transactions b.max b.min b.close b.avg b.tot.vol b.turnover b.transactions
1 1989-08-25 7.92 7.77 7.79 NA 269172 2119400 19 7.69 7.56 7.64 NA 81176693 593858000 12
EDIT: problem solved by (using zoo package)
atcoa <- read.zoo(read.csv(file = "ATCOA_full_adj.csv", header = TRUE))
atcob <- read.zoo(read.csv(file = "ATCOB_full_adj.csv", header = TRUE))
names(atcoa) <- c("a.max", "a.min", "a.close",
"a.avg", "a.tot.vol", "a.turnover", "a.transactions")
names(atcob) <- c("b.max", "b.min", "b.close",
"b.avg", "b.tot.vol", "b.turnover", "b.transactions")
atco <- merge.zoo(atcoa, atcob)
Thank you all for your help.

Try this:
Lines.x <- '"1987-01-01" 7.1 NA 3
"1987-01-02" 5.2 5 2
"1987-01-06" 2.3 NA 9'
Lines.y <- '"1987-01-01" 55.3 66 45
"1987-01-03" 77.3 87 34'
library(zoo)
# in reality x might be in a file and might be read via: x <- read.zoo("x.dat")
# ditto for y. See ?read.zoo and the zoo-read vignette if you need other args too
x <- read.zoo(text = Lines.x)
y <- read.zoo(text = Lines.y)
merge(x, y)
giving:
V2.x V3.x V4.x V2.y V3.y V4.y
1987-01-01 7.1 NA 3 55.3 66 45
1987-01-02 5.2 5 2 NA NA NA
1987-01-03 NA NA NA 77.3 87 34
1987-01-06 2.3 NA 9 NA NA NA

You can create a timeSeries (timeSeries library) object from your dates, merge them (timeSeries default merge behaviour is different from zoo and xts and does exactly what you are asking for) and then make zoo/xts objects out of the result in case you don't want to stay with timeSeries.
One quick way to test is the following, assuming you have two zoo objects zz1 and zz2 -
library(timeSeries)
as.zoo(merge(as.timeSeries(zz1), as.timeSeries(zz2)))
Compare the output of the above command with
merge(zz1, zz2)
You can also cbind -
cbind(zz1, zz2)
provided there are no shared columns with same names. Even if such column are there, you can choose the columns by which you cbind, and you will get a zoo object.
cbind(zz1[, 1:2], zz2[, 2:3]) #Assuming other columns are common

here, i found a more generic aproach from stat.ethz.ch
a <- ts(1:10, start=c(2014,6), frequency=12)
b <- ts(1:12, start=c(2015,1), frequency=12)
library(zoo)
m <- merge(a = as.zoo(a), b = as.zoo(b))
to get a ts object back:
as.ts(m)

How about this:
## Generate unique sorted time values.
i <- sort(unique(c(index(x), index(y))))
## Empty data matrix.
v <- matrix(nrow=length(i), ncol=6, NA)
## Pull in data items.
v[match(index(x), i), 1:3] <- coredata(x)
v[match(index(y), i), 4:6] <- coredata(y)
## Build new zoo object.
d <- zoo(v, order.by=i)

Related

Create data frame variables based on a function with two matching variable arguments where argument order matters

Here is a toy data frame
df <- data.frame(alpha = c(rep(.005,5)),
a1 = c(1:5),
b1 = c(4:8),
c1 = c(10:14),
a2 = c(9:13),
b2 = c(3:7),
c2 = c(15:19))
Here is a nonsensical toy function that requires two variables, both of which must have the same letter prefix. The specific function calculation is not important. Rather, the issue is how to pass two or more separate named variables to the function from the data frame where the order of the arguments matters.
toy_function <- function(x,y){
z = x+y
w = x/y
v = z+w
return(v)
}
Manual calculation of new variables using the function would look like this. Not practical when you've got dozens or hundreds of variable pairs.
df2 <- df %>%
mutate(va = toy_function(a1,a2),
vb = toy_function(b1,b2),
vc = toy_function(c1,c2)
)
How can I do this across all matching pairs of variables? This problem seems similar to How to use map from purrr with dplyr::mutate to create multiple new columns based on column pairs but that example was applying a simple mathematical function (e.g., +) in which variable order does not matter. I'm having trouble figuring out how to modify it for this case.
Here is one base R approach using split.default.
cbind(df, sapply(split.default(df[-1],
sub('\\d+', '', names(df)[-1])), function(x)
toy_function(x[[1]], x[[2]])))
# alpha a1 b1 c1 a2 b2 c2 a b c
#1 0.005 1 4 10 9 3 15 10.1 8.33 25.7
#2 0.005 2 5 11 10 4 16 12.2 10.25 27.7
#3 0.005 3 6 12 11 5 17 14.3 12.20 29.7
#4 0.005 4 7 13 12 6 18 16.3 14.17 31.7
#5 0.005 5 8 14 13 7 19 18.4 16.14 33.7
We ignore the first column ([-1]) since we don't want to include that in the calculation and create a group of similarly named column and split them into lists. Using sapply we apply toy_function to each element in the list.
sub is used to remove the numbers from the names and create groups to split on.
sub('\\d+', '', names(df)[-1])
#[1] "a" "b" "c" "a" "b" "c"
If you wish to use the tidyverse approach you could do :
library(dplyr)
library(purrr)
unique_names <- unique(sub('\\d+', '', names(df)[-1]))
map_dfc(unique_names, ~df[-1] %>%
select(matches(.x)) %>%
mutate(!!paste0('v', .x) := toy_function(.[[1]], .[[2]])))
# a1 a2 va b1 b2 vb c1 c2 vc
#1 1 9 10.1 4 3 8.33 10 15 25.7
#2 2 10 12.2 5 4 10.25 11 16 27.7
#3 3 11 14.3 6 5 12.20 12 17 29.7
#4 4 12 16.3 7 6 14.17 13 18 31.7
#5 5 13 18.4 8 7 16.14 14 19 33.7
You can do something like this
First, create a dataframe with the function arguments as columns and the values to be used for each function call as rows.
vars <- letters[1:3]
args <- tibble(
arg1 = setNames(paste0(vars, 1), paste0("set_output_names_like_this_", vars)),
arg2 = paste0(vars, 2)
)
> str(args)
tibble [3 x 2] (S3: tbl_df/tbl/data.frame)
$ arg1: Named chr [1:3] "a1" "b1" "c1"
..- attr(*, "names")= chr [1:3] "set_output_names_like_this_a" "set_output_names_like_this_b" "set_output_names_like_this_c"
$ arg2: chr [1:3] "a2" "b2" "c2"
Then, use pmap_dfc
df %>% mutate(pmap_dfc(args, function(arg1, arg2, d) toy_function(d[[arg1]], d[[arg2]]), .data))
Output
alpha a1 b1 c1 a2 b2 c2 set_output_names_like_this_a set_output_names_like_this_b set_output_names_like_this_c
1 0.005 1 4 10 9 3 15 10.11111 8.333333 25.66667
2 0.005 2 5 11 10 4 16 12.20000 10.250000 27.68750
3 0.005 3 6 12 11 5 17 14.27273 12.200000 29.70588
4 0.005 4 7 13 12 6 18 16.33333 14.166667 31.72222
5 0.005 5 8 14 13 7 19 18.38462 16.142857 33.73684

How to group by and fill NA with closest not NA in R dataframe column with condition on another column

I have a data frame of blood test markers results and I want to fill in the NA's by the following criteria:
For each group of ID (TIME is in ascending order) if the marker value is NA then fill it with the closest not NA value in this group (may be past or future) but only if the time difference is less than 14.
this example of my data:
df<-data.frame(ID=c(rep(2,5),rep(4,3)), TIME =c(1,22,33,43,85,-48,1,30),
CEA = c(1.32,1.42,1.81,2.33,2.23,29.7,23.34,18.23),
CA.15.3 = c(14.62,14.59,16.8,22.34,36.33,56.02,94.09,121.5),
CA.125 = c(33.98,27.56,30.31,NA,39.57,1171.00,956.50,825.30),
CA.19.9 = c(6.18,7.11,5.72, NA, 7.38,39.30,118.20,98.26),
CA.72.4 = c(rep(NA,5),1.32, NA, NA),
NSE = c(NA, 13.21, rep(NA,6)))
ID TIME CEA CA.15.3 CA.125 CA.19.9 CA.72.4 NSE
2 1 1.32 14.62 33.98 6.18 NA NA
2 22 1.42 14.59 27.56 7.11 NA 13.21
2 33 1.81 16.80 30.31 5.72 NA NA
2 43 2.33 22.34 NA NA NA NA
2 85 2.23 36.33 39.57 7.38 NA NA
4 -48 29.70 56.02 1171.00 39.30 1.32 NA
4 1 23.34 94.09 956.50 118.20 NA NA
4 30 18.23 121.50 825.30 98.26 NA NA
ID is the patient.
The TIME is the time of the blood test.
The others are the markers.
The only way I could do it is with loops which I try to avoid as much as possible.
I expect the output to be:
ID TIME CEA CA.15.3 CA.125 CA.19.9 CA.72.4 NSE
2 1 1.32 14.62 33.98 6.18 NA NA
2 22 1.42 14.59 27.56 7.11 NA 13.21
2 33 1.81 16.80 30.31 5.72 NA 13.21
2 43 2.33 22.34 30.31 5.72 NA NA
2 85 2.23 36.33 39.57 7.38 NA NA
4 -48 29.70 56.02 1171.00 39.30 1.32 NA
4 1 23.34 94.09 956.50 118.20 NA NA
4 30 18.23 121.50 825.30 98.26 NA NA
CA.19.9 and CA.124 are filled with the previous (10 days before)
NSE filled with the previous (11 days)
CA.72.4 not filled since the time difference of 1.32 which is -48 is 49 days from the next measure.
I bet there is a much simpler, vectorized solution but the following works.
fill_NA <- function(DF){
sp <- split(df, df$ID)
sp <- lapply(sp, function(DF){
d <- diff(DF$TIME)
i_diff <- c(FALSE, d < 14)
res <- sapply(DF[-(1:2)], function(X){
inx <- i_diff & is.na(X)
if(any(inx)){
inx <- which(inx)
last_change <- -1
for(i in inx){
if(i > last_change + 1){
if(i == 1){
X[i] <- X[i + 1]
}else{
X[i] <- X[i - 1]
}
last_change <- i
}
}
}
X
})
cbind(DF[1:2], res)
})
res <- do.call(rbind, sp)
row.names(res) <- NULL
res
}
fill_NA(df)
# ID TIME CEA CA.15.3 CA.125 CA.19.9 CA.72.4 NSE
#1 2 1 1.32 14.62 33.98 6.18 NA NA
#2 2 22 1.42 14.59 27.56 7.11 NA 13.21
#3 2 33 1.81 16.80 30.31 5.72 NA 13.21
#4 2 43 2.33 22.34 30.31 5.72 NA NA
#5 2 85 2.23 36.33 39.57 7.38 NA NA
#6 4 -48 29.70 56.02 1171.00 39.30 1.32 NA
#7 4 1 23.34 94.09 956.50 118.20 NA NA
#8 4 30 18.23 121.50 825.30 98.26 NA NA
Yes, you can have a vectorized solution. first let us consider the case in which you only impute using the future value. You need to create few auxiliary variables:
a variable that tells you whether the next observation belong to the same id (so it can be used to impute),
a variable that tells you whether the next observation is less than 14 days apart from the current one.
These do not depend on the specific variable you want to impute. for each variable to be imputed you will also need a variable that tells you whether the next variable is missing.
Then you can vectorize the following logic: when the next observation has the same id, and when it is less than 14 days from the current one and it is not missing copy its value in the current one.
Things get more complicated when you need to decide whether to use the past or future value, but the logic is the same. the code is below, it is a bit long but you can simplify it, I just wanted to be clear about what it does.
Hope this helps
x <-data.frame(ID=c(rep(2,5),rep(4,3)), TIME =c(1,22,33,43,85,-48,1,30),
CEA = c(1.32,1.42,1.81,2.33,2.23,29.7,23.34,18.23),
CA.15.3 = c(14.62,14.59,16.8,22.34,36.33,56.02,94.09,121.5),
CA.125 = c(33.98,27.56,30.31,NA,39.57,1171.00,956.50,825.30),
CA.19.9 = c(6.18,7.11,5.72, NA, 7.38,39.30,118.20,98.26),
CA.72.4 = c(rep(NA,5),1.32, NA, NA),
NSE = c(NA, 13.21, rep(NA,6)))
### these are the columns we want to input
cols.to.impute <- colnames(x)[! colnames(x) %in% c("ID","TIME")]
### is the next id the same?
x$diffidf <- NA
x$diffidf[1:(nrow(x)-1)] <- diff(x$ID)
x$diffidf[x$diffidf > 0] <- NA
### is the previous id the same?
x$diffidb <- NA
x$diffidb[2:nrow(x)] <- diff(x$ID)
x$diffidb[x$diffidb > 0] <- NA
### diff in time with next observation
x$difftimef <- NA
x$difftimef[1:(nrow(x)-1)] <- diff(x$TIME)
### diff in time with previous observation
x$difftimeb <- NA
x$difftimeb[2:nrow(x)] <- diff(x$TIME)
### if next (previous) id is not the same time difference is not meaningful
x$difftimef[is.na(x$diffidf)] <- NA
x$difftimeb[is.na(x$diffidb)] <- NA
### we do not need diffid anymore (due to previous statement)
x$diffidf <- x$diffidb <- NULL
### if next (previous) point in time is more than 14 days it is not useful for imputation
x$difftimef[abs(x$difftimef) > 14] <- NA
x$difftimeb[abs(x$difftimeb) > 14] <- NA
### create variable usef that tells us whether we should attempt to use the forward observation for imputation
### it is 1 only if difftime forward is less than difftime backward
x$usef <- NA
x$usef[!is.na(x$difftimef) & x$difftimef < x$difftimeb] <- 1
x$usef[!is.na(x$difftimef) & is.na(x$difftimeb)] <- 1
x$usef[is.na(x$difftimef) & !is.na(x$difftimeb)] <- 0
if (!is.na(x$usef[nrow(x)]))
stop("\nlast observation usef is not missing\n")
### now we get into column specific operations.
for (col in cols.to.impute){
### we will store the results in x$imputed, and copy into c[,col] at the end
x$imputed <- x[,col]
### x$usef needs to be modified depending on the specific column, so we define a local version of it
x$usef.local <- x$usef
### if a variable is not missing no point in looking at usef.local, so we make it missing
x$usef.local[!is.na(x[,col])] <- NA
### when usef.local is 1 but the next observation is missing it cannot be used for imputation, so we
### make it 0. but a value of 0 does not mean we can use the previous observation because that may
### be missing too. so first we make usef 0 and next we check the previous observation and if that
### is missing too we make usef missing
x$previous.value <- c(NA,x[1:(nrow(x)-1),col])
x$next.value <- c(x[2:nrow(x),col],NA)
x$next.missing <- is.na(x$next.value)
x$previous.missing <- is.na(x$previous.value)
x$usef.local[x$next.missing & x$usef.local == 1] <- 0
x$usef.local[x$previous.missing & x$usef.local == 0] <- NA
### now we can impute properly: use next value when usef.local is 1 and previous value when usef.local is 0
tmp <- rep(FALSE,nrow(x))
tmp[x$usef.local == 1] <- TRUE
x$imputed[tmp] <- x$next.value[tmp]
tmp <- rep(FALSE,nrow(x))
tmp[x$usef.local == 0] <- TRUE
x$imputed[tmp] <- x$previous.value[tmp]
### copy to column
x[,col] <- x$imputed
}
### get rid of useless temporary stuff
x$previous.value <- x$previous.missing <- x$next.value <- x$next.missing <- x$imputed <- x$usef.local <- NULL
ID TIME CEA CA.15.3 CA.125 CA.19.9 CA.72.4 NSE difftimef difftimeb usef
1 2 1 1.32 14.62 33.98 6.18 NA NA NA NA NA
2 2 22 1.42 14.59 27.56 7.11 NA 13.21 11 NA 1
3 2 33 1.81 16.80 30.31 5.72 NA 13.21 10 11 1
4 2 43 2.33 22.34 30.31 5.72 NA NA NA 10 0
5 2 85 2.23 36.33 39.57 7.38 NA NA NA NA NA
6 4 -48 29.70 56.02 1171.00 39.30 1.32 NA NA NA NA
7 4 1 23.34 94.09 956.50 118.20 NA NA NA NA NA
8 4 30 18.23 121.50 825.30 98.26 NA NA NA NA NA
>

R generate bins from a data frame respecting blanks

I need to generate bins from a data.frame based on the values of one column. I have tried the function "cut".
For example: I want to create bins of air temperature values in the column "AirTDay" in a data frame:
AirTDay (oC)
8.16
10.88
5.28
19.82
23.62
13.14
28.84
32.21
17.44
31.21
I need the bin intervals to include all values in a range of 2 degrees centigrade from that initial value (i.e. 8-9.99, 10-11.99, 12-13.99...), to be labelled with the average value of the range (i.e. 9.5, 10.5, 12.5...), and to respect blank cells, returning "NA" in the bins column.
The output should look as:
Air_T (oC) TBins
8.16 8.5
10.88 10.5
5.28 NA
NA
19.82 20.5
23.62 24.5
13.14 14.5
NA
NA
28.84 28.5
32.21 32.5
17.44 18.5
31.21 32.5
I've gotten as far as:
setwd('C:/Users/xxx')
temp_data <- read.csv("temperature.csv", sep = ",", header = TRUE)
TAir <- temp_data$AirTDay
Tmin <- round(min(TAir, na.rm = FALSE), digits = 0) # is start at minimum value
Tmax <- round(max(TAir, na.rm = FALSE), digits = 0)
int <- 2 # bin ranges 2 degrees
mean_int <- int/2
int_range <- seq(Tmin, Tmax + int, int) # generate bin sequence
bin_label <- seq(Tmin + mean_int, Tmax + mean_int, int) # generate labels
temp_data$TBins <- cut(TAir, breaks = int_range, ordered_result = FALSE, labels = bin_label)
The output table looks correct, but for some reason it shows a sequential additional column, shifts column names, and collapse all values eliminating blank cells. Something like this:
Air_T (oC) TBins
1 8.16 8.5
2 10.88 10.5
3 5.28 NA
4 19.82 20.5
5 23.62 24.5
6 13.14 14.5
7 28.84 28.5
8 32.21 32.5
9 17.44 18.5
10 31.21 32.5
Any ideas on where am I failing and how to solve it?
v<-ceiling(max(dat$V1,na.rm=T))
breaks<-seq(8,v,2)
labels=seq(8.5,length.out=length(s)-1,by=2)
transform(dat,Tbins=cut(V1,breaks,labels))
V1 Tbins
1 8.16 8.5
2 10.88 10.5
3 5.28 <NA>
4 NA <NA>
5 19.82 18.5
6 23.62 22.5
7 13.14 12.5
8 NA <NA>
9 NA <NA>
10 28.84 28.5
11 32.21 <NA>
12 17.44 16.5
13 31.21 30.5
This result follows the logic given: we have
paste(seq(8,v,2),seq(9.99,v,by=2),sep="-")
[1] "8-9.99" "10-11.99" "12-13.99" "14-15.99" "16-17.99" "18-19.99" "20-21.99"
[8] "22-23.99" "24-25.99" "26-27.99" "28-29.99" "30-31.99"
From this we can tell that 19.82 will lie between 18 and 20 thus given the value 18.5, similar to 10.88 being between 10-11.99 thus assigned the value 10.5

Reshape to wide based on standard monthly periods

I am working with a large dataset in long format (after using melt) that I would now like to switch to a wide format before running a calculation on each row of the df
This long-form dataset ("dflong") has 7 columns: name, returnmonth, startmonth, descriptive1, descriptive2, descriptive3, and return.
I use the long format because I have multiple returnmonths for each name/startmonth pair. Specifically, I have the next 12 months of returns for each name and startmonth, and the startmonths span 10 years. This makes the df large: it has 1.2 million rows and those 7 columns.
What I would like next is to standardize the next 12 months for each name/startmonth and have a wide format where the right-most 12 columns are the returns in month startmonth+1, startmonth+2, etc regardless of what that actual month is. But I cannot just use the formula below because the returnmonths themselves differ, which I imagine would create a very, very wide df, which I do not want. (at this point, it's just giving me an error when I try to run it).
dfwide=reshape(dflong,idvar=c("name","startmonth","descriptive1","descriptive2","descriptive3"),timevar="returnmonth",direction="wide")
Is there a combination of reshape and some other tool that would allow me to both convert from long to wide but also not take into account the particular month that the return is coming from. This would form a long but manageable n rows*17 column dataframe to work with.
Appreciate your help very much.
This old question hasn't got an answer so far.
The OP has requested to reshape from long to wide format where the returns values should be arranged in 12 columns representing the 12 months after each startmonth.
This can be accomplished using the dcast() and rowid() functions from the data.table package.
library(data.table)
dcast(setDT(DT), name + startmonth + d1 + d2 + d3 ~ sprintf("M_%02i", rowid(name, startmonth)),
value.var = "return")
name startmonth d1 d2 d3 M_01 M_02 M_03 M_04 M_05 M_06 M_07 M_08 M_09 M_10 M_11 M_12
1: 0001 2000-01-01 A B C -56.0 -23.0 155.9 7.1 12.9 171.5 46.1 -126.5 -68.7 -44.6 122.4 36.0
2: 0001 2000-02-01 A B C 40.1 11.1 -55.6 178.7 49.8 -196.7 70.1 -47.3 -106.8 -21.8 -102.6 -72.9
3: 0001 2000-03-01 A B C -62.5 -168.7 83.8 15.3 -113.8 125.4 42.6 -29.5 89.5 87.8 82.2 68.9
4: 0001 2000-04-01 A B C 55.4 -6.2 -30.6 -38.0 -69.5 -20.8 -126.5 216.9 120.8 -112.3 -40.3 -46.7
5: 0001 2000-05-01 A B C 78.0 -8.3 25.3 -2.9 -4.3 136.9 -22.6 151.6 -154.9 58.5 12.4 21.6
---
95996: 0800 2009-08-01 A B C 136.5 79.6 -140.4 -216.5 35.3 31.4 -57.0 78.1 -53.3 -18.0 60.9 100.7
95997: 0800 2009-09-01 A B C 33.8 -12.5 -17.2 90.8 -24.7 -19.3 101.4 -39.1 139.4 15.5 -33.5 17.7
95998: 0800 2009-10-01 A B C 56.8 -121.2 242.4 -1.0 -9.9 78.2 34.6 31.8 -51.7 -11.1 45.0 89.1
95999: 0800 2009-11-01 A B C 158.8 -101.9 -271.8 -21.1 -24.6 108.0 185.3 82.3 -54.9 116.4 149.2 -30.4
96000: 0800 2009-12-01 A B C 34.9 -34.0 186.4 -3.7 -26.8 97.0 8.7 84.5 -35.5 -106.2 -165.0 188.0
Data
The OP hasn't provided a reproducible example, so we make up our own dummy data:
library(data.table)
library(lubridate)
# 10 years of startmonths times 12 returnmonths times 800 names
nn <- 800L
ns <- 10L * 12L
nr <- 12L
# cross join to create all combinations
DT <- CJ(name = sprintf("%04i", seq_len(nn)),
startmonth = seq(as.Date("2000-01-01"), length.out = ns, by = "month"),
returnmonth = seq_len(nr)
)
# compute returnmonth as sequence of 12 months after each startmonth
DT[, returnmonth := startmonth + months(returnmonth)][]
# append other data columns
DT[, paste0("d", 1:3) := .("A", "B", "C")][]
set.seed(123L)
DT[, return := round(100 * rnorm(nrow(DT)), 1L)][]
str(DT)
Classes ‘data.table’ and 'data.frame': 1152000 obs. of 7 variables:
$ name : chr "0001" "0001" "0001" "0001" ...
$ startmonth : Date, format: "2000-01-01" "2000-01-01" "2000-01-01" "2000-01-01" ...
$ returnmonth: Date, format: "2000-02-01" "2000-03-01" "2000-04-01" "2000-05-01" ...
$ d1 : chr "A" "A" "A" "A" ...
$ d2 : chr "B" "B" "B" "B" ...
$ d3 : chr "C" "C" "C" "C" ...
$ return : num -56 -23 155.9 7.1 12.9 ...
- attr(*, ".internal.selfref")=<externalptr>

R: merge two irregular time series

I have two multivariate time series x and y, both covering approximately the same range in time (one starts two years before the other, but they end on the same date). Both series have missing observations in the form of empty columns next to the date column, and also in the sense that one of the series has several dates that are not found in the other, and vice versa.
I would like to create a data frame (or similar) with a column that lists all the dates found in x OR y, without duplicate dates. For each date (row), I would like to horizontally stack the observations from x next to the observations from y, with NA's filling the missing cells. Example:
>x
"1987-01-01" 7.1 NA 3
"1987-01-02" 5.2 5 2
"1987-01-06" 2.3 NA 9
>y
"1987-01-01" 55.3 66 45
"1987-01-03" 77.3 87 34
# result I would like
"1987-01-01" 7.1 NA 3 55.3 66 45
"1987-01-02" 5.2 5 2 NA NA NA
"1987-01-03" NA NA NA 77.3 87 34
"1987-01-06" 2.3 NA 9 NA NA NA
What I have tried: with the zoo package, I've tried the merge.zoo method, but this seems to just stack the two series next to each other, with the dates (as numbers, e.g. "1987-01-02" shown as 6210) from each series appearing in two separate columns.
I've sat for hours getting almost nowhere, so all help is appreciated.
EDIT: some code included below as per suggestion from Soumendra
atcoa <- read.csv(file = "ATCOA_full_adj.csv", header = TRUE)
atcob <- read.csv(file = "ATCOB_full_adj.csv", header = TRUE)
atcoa$date <- as.Date(atcoa$date)
atcob$date <- as.Date(atcob$date)
# only number of observations and the observations themselves differ
>str(atcoa)
'data.frame': 6151 obs. of 8 variables:
$ date :Class 'Date' num [1:6151] 6210 6213 6215 6216 6217 ...
$ max : num 4.31 4.33 4.38 4.18 4.13 4.05 4.08 4.05 4.08 4.1 ...
$ min : num 4.28 4.31 4.28 4.13 4.05 3.95 3.97 3.95 4 4.02 ...
$ close : num 4.31 4.33 4.31 4.15 4.1 3.97 4 3.97 4.08 4.02 ...
$ avg : num NA NA NA NA NA NA NA NA NA NA ...
$ tot.vol : int 877733 89724 889437 1927113 3050611 846525 1782774 1497998 2504466 5636999 ...
$ turnover : num 3762300 388900 3835900 8015900 12468100 ...
$ transactions: int 12 9 24 17 31 26 34 35 37 33 ...
>atcoa[1:1, ]
date a.max a.min a.close a.avg a.tot.vol a.turnover a.transactions
1 1987-01-02 4.31 4.28 4.31 NA 877733 3762300 12
# using timeSeries package
ts.atcoa <- timeSeries::as.timeSeries(atcoa, format = "%Y-%m-%d")
ts.atcob <- timeSeries::as.timeSeries(atcob, format = "%Y-%m-%d")
>str(ts.atcoa)
Time Series:
Name: object
Data Matrix:
Dimension: 6151 7
Column Names: a.max a.min a.close a.avg a.tot.vol a.turnover a.transactions
Row Names: 1970-01-01 01:43:30 ... 1970-01-01 04:12:35
Positions:
Start: 1970-01-01 01:43:30
End: 1970-01-01 04:12:35
With:
Format: %Y-%m-%d %H:%M:%S
FinCenter: GMT
Units: a.max a.min a.close a.avg a.tot.vol a.turnover a.transactions
Title: Time Series Object
Documentation: Wed Aug 17 13:00:50 2011
>ts.atcoa[1:1, ]
GMT
a.max a.min a.close a.avg a.tot.vol a.turnover a.transactions
1970-01-01 01:43:30 4.31 4.28 4.31 NA 877733 3762300 12
# The following will create an object of class "data frame" and mode "list", which contains observations for the days mutual for the two series
>ts.atco <- timeSeries::merge(atcoa, atcob) # produces same result as base::merge, apparently
>ts.atco[1:1, ]
date a.max a.min a.close a.avg a.tot.vol a.turnover a.transactions b.max b.min b.close b.avg b.tot.vol b.turnover b.transactions
1 1989-08-25 7.92 7.77 7.79 NA 269172 2119400 19 7.69 7.56 7.64 NA 81176693 593858000 12
EDIT: problem solved by (using zoo package)
atcoa <- read.zoo(read.csv(file = "ATCOA_full_adj.csv", header = TRUE))
atcob <- read.zoo(read.csv(file = "ATCOB_full_adj.csv", header = TRUE))
names(atcoa) <- c("a.max", "a.min", "a.close",
"a.avg", "a.tot.vol", "a.turnover", "a.transactions")
names(atcob) <- c("b.max", "b.min", "b.close",
"b.avg", "b.tot.vol", "b.turnover", "b.transactions")
atco <- merge.zoo(atcoa, atcob)
Thank you all for your help.
Try this:
Lines.x <- '"1987-01-01" 7.1 NA 3
"1987-01-02" 5.2 5 2
"1987-01-06" 2.3 NA 9'
Lines.y <- '"1987-01-01" 55.3 66 45
"1987-01-03" 77.3 87 34'
library(zoo)
# in reality x might be in a file and might be read via: x <- read.zoo("x.dat")
# ditto for y. See ?read.zoo and the zoo-read vignette if you need other args too
x <- read.zoo(text = Lines.x)
y <- read.zoo(text = Lines.y)
merge(x, y)
giving:
V2.x V3.x V4.x V2.y V3.y V4.y
1987-01-01 7.1 NA 3 55.3 66 45
1987-01-02 5.2 5 2 NA NA NA
1987-01-03 NA NA NA 77.3 87 34
1987-01-06 2.3 NA 9 NA NA NA
You can create a timeSeries (timeSeries library) object from your dates, merge them (timeSeries default merge behaviour is different from zoo and xts and does exactly what you are asking for) and then make zoo/xts objects out of the result in case you don't want to stay with timeSeries.
One quick way to test is the following, assuming you have two zoo objects zz1 and zz2 -
library(timeSeries)
as.zoo(merge(as.timeSeries(zz1), as.timeSeries(zz2)))
Compare the output of the above command with
merge(zz1, zz2)
You can also cbind -
cbind(zz1, zz2)
provided there are no shared columns with same names. Even if such column are there, you can choose the columns by which you cbind, and you will get a zoo object.
cbind(zz1[, 1:2], zz2[, 2:3]) #Assuming other columns are common
here, i found a more generic aproach from stat.ethz.ch
a <- ts(1:10, start=c(2014,6), frequency=12)
b <- ts(1:12, start=c(2015,1), frequency=12)
library(zoo)
m <- merge(a = as.zoo(a), b = as.zoo(b))
to get a ts object back:
as.ts(m)
How about this:
## Generate unique sorted time values.
i <- sort(unique(c(index(x), index(y))))
## Empty data matrix.
v <- matrix(nrow=length(i), ncol=6, NA)
## Pull in data items.
v[match(index(x), i), 1:3] <- coredata(x)
v[match(index(y), i), 4:6] <- coredata(y)
## Build new zoo object.
d <- zoo(v, order.by=i)

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