My question is similar to replace duplicate values with NA in time series data using dplyr but while applying to other time series which are like below :
box_num date x y
6-WQ 2018-11-18 20.2 8
6-WQ 2018-11-25 500.75 7.2
6-WQ 2018-12-2 500.75 23
25-LR 2018-11-18 374.95 4.3
25-LR 2018-11-25 0.134 9.3
25-LR 2018-12-2 0.134 4
73-IU 2018-12-2 225.54 0.7562
73-IU 2018-12-9 28 0.7562
73-IU 2018-12-16 225.54 52.8
library(dplyr)
df %>%
group_by(box_num) %>%
mutate_at(vars(x:y), funs(replace(., duplicated(.), NA)))
The above code can identify and replace with NA, but the underlying problem is I'm trying to replace all NA with a linear trend in the coming step. Since it's a time series.But when we see for box_num : 6-WQ after 20.2 we can see directly a large shift which we can say it's a imputed value so I would to replace both the imputed values as NA and the other case is like for box_num 73-IU imputed values got entered after one week so I would like to replace imputed values with NA
Expected output :
box_num date x y
6-WQ 2018-11-18 20.2 8
6-WQ 2018-11-25 NA 7.2
6-WQ 2018-12-2 NA 23
25-LR 2018-11-18 374.95 4.3
25-LR 2018-11-25 NA 9.3
25-LR 2018-12-2 NA 4
73-IU 2018-12-2 NA NA
73-IU 2018-12-9 28 NA
73-IU 2018-12-16 NA 52.8
foo = function(x){
replace(x, ave(x, x, FUN = length) > 1, NA)
}
myCols = c("x", "y")
df1[myCols] = lapply(df1[myCols], foo)
df1
# box_num date x y
#1 6-WQ 2018-11-18 20.20 8.0
#2 6-WQ 2018-11-25 NA 7.2
#3 6-WQ 2018-12-2 NA 23.0
#4 25-LR 2018-11-18 374.95 4.3
#5 25-LR 2018-11-25 NA 9.3
#6 25-LR 2018-12-2 NA 4.0
#7 73-IU 2018-12-2 NA NA
#8 73-IU 2018-12-9 28.00 NA
#9 73-IU 2018-12-16 NA 52.8
#DATA
df1 = structure(list(box_num = c("6-WQ", "6-WQ", "6-WQ", "25-LR", "25-LR",
"25-LR", "73-IU", "73-IU", "73-IU"), date = c("2018-11-18", "2018-11-25",
"2018-12-2", "2018-11-18", "2018-11-25", "2018-12-2", "2018-12-2",
"2018-12-9", "2018-12-16"), x = c(20.2, 500.75, 500.75, 374.95,
0.134, 0.134, 225.54, 28, 225.54), y = c(8, 7.2, 23, 4.3, 9.3,
4, 0.7562, 0.7562, 52.8)), class = "data.frame", row.names = c(NA,
-9L))
With tidyverse you can do:
df %>%
group_by(box_num) %>%
mutate_at(vars(x:y), funs(ifelse(. %in% subset(rle(sort(.))$values, rle(sort(.))$length > 1), NA, .)))
box_num date x y
<fct> <fct> <dbl> <dbl>
1 6-WQ 2018-11-18 20.2 8.00
2 6-WQ 2018-11-25 NA 7.20
3 6-WQ 2018-12-2 NA 23.0
4 25-LR 2018-11-18 375. 4.30
5 25-LR 2018-11-25 NA 9.30
6 25-LR 2018-12-2 NA 4.00
7 73-IU 2018-12-2 NA NA
8 73-IU 2018-12-9 28.0 NA
9 73-IU 2018-12-16 NA 52.8
First, it sorts the values in "x" and "y" and computes the run length of equal values. Second, it creates a subset for those values that have a run length > 1. Finally, it compares whether the values in "x" and "y" are in the subset, and if so, they get NA.
Related
I have a column with numeric data with NA and also ending on NA:
df <- data.frame(
Diam_av = c(12.3, 13, 15.5, NA, NA, NA, NA, 13.7, NA, NA, NA, 9.98, 4,0, 8.76, NA, NA, NA)
)
I want to interpolate the missing values. This works fine with zoo's function na.approx as long as there are positive boundary values to interpolate from but it fails if, as in my case, one of the boundary values is NA (at the end of the column Daim_av:
library(zoo)
df %>%
mutate(Diam_intpl = na.approx(Diam_av))
Error: Problem with `mutate()` input `Diam_intpl`.
x Input `Diam_intpl` can't be recycled to size 18.
ℹ Input `Diam_intpl` is `na.approx(Diam_av)`.
ℹ Input `Diam_intpl` must be size 18 or 1, not 15.
Any idea how to exclude/neutralize column-final NA values?
Add na.rm=F to remove the error message. Add rule=2 to get the value from the last non-NA value.
df %>%
mutate(Diam_intpl = na.approx(Diam_av, na.rm=F),
Diam_intpl2 = na.approx(Diam_av, na.rm=F, rule=2))
Diam_av Diam_intpl Diam_intpl2
1 12.30 12.30 12.30
2 13.00 13.00 13.00
3 15.50 15.50 15.50
4 NA 15.14 15.14
5 NA 14.78 14.78
6 NA 14.42 14.42
7 NA 14.06 14.06
8 13.70 13.70 13.70
9 NA 12.77 12.77
10 NA 11.84 11.84
11 NA 10.91 10.91
12 9.98 9.98 9.98
13 4.00 4.00 4.00
14 0.00 0.00 0.00
15 8.76 8.76 8.76
16 NA NA 8.76
17 NA NA 8.76
18 NA NA 8.76
If I understand well, you can replace NAs with imputeTS::na_interpolation(), that has many options:
library(imputeTS)
df$interpolated <- na_interpolation(df,option = 'linear')$Diam_av
Diam_av interpolated
1 12.30 12.30
2 13.00 13.00
3 15.50 15.50
4 NA 15.14
5 NA 14.78
6 NA 14.42
7 NA 14.06
8 13.70 13.70
9 NA 12.77
10 NA 11.84
11 NA 10.91
12 9.98 9.98
13 4.00 4.00
14 0.00 0.00
15 8.76 8.76
16 NA 8.76
17 NA 8.76
18 NA 8.76
I have a quarterly time series. I am trying to apply a function which is supposed calculate the year-to-year growth and year-to-year difference and multiply a variable by (-1).
I already used a similar function for calculating quarter-to-quarter changes and it worked.
I modified this function for yoy changes and it does not have any effect on my data frame. And any error popped up.
Do you have any suggestion how to modify the function or how to accomplish to apply the yoy change function on a time series?
Here is the code:
Date <- c("2004-01-01","2004-04-01", "2004-07-01","2004-10-01","2005-01-01","2005-04-01","2005-07-01","2005-10-01","2006-01-01","2006-04-01","2006-07-01","2006-10-01","2007-01-01","2007-04-01","2007-07-01","2007-10-01")
B1 <- c(3189.30,3482.05,3792.03,4128.66,4443.62,4876.54,5393.01,5885.01,6360.00,6930.00,7430.00,7901.00,8279.00,8867.00,9439.00,10101.00)
B2 <- c(7939.97,7950.58,7834.06,7746.23,7760.59,8209.00,8583.05,8930.74,9424.00,9992.00,10041.00,10900.00,11149.00,12022.00,12662.00,13470.00)
B3 <- as.numeric(c("","","","",140.20,140.30,147.30,151.20,159.60,165.60,173.20,177.30,185.30,199.30,217.10,234.90))
B4 <- as.numeric(c("","","","",-3.50,-14.60,-11.60,-10.20,-3.10,-16.00,-4.90,-17.60,-5.30,-10.90,-12.80,-8.40))
df <- data.frame(Date,B1,B2,B3,B4)
The code will produce following data frame:
Date B1 B2 B3 B4
1 2004-01-01 3189.30 7939.97 NA NA
2 2004-04-01 3482.05 7950.58 NA NA
3 2004-07-01 3792.03 7834.06 NA NA
4 2004-10-01 4128.66 7746.23 NA NA
5 2005-01-01 4443.62 7760.59 140.2 -3.5
6 2005-04-01 4876.54 8209.00 140.3 -14.6
7 2005-07-01 5393.01 8583.05 147.3 -11.6
8 2005-10-01 5885.01 8930.74 151.2 -10.2
9 2006-01-01 6360.00 9424.00 159.6 -3.1
10 2006-04-01 6930.00 9992.00 165.6 -16.0
11 2006-07-01 7430.00 10041.00 173.2 -4.9
12 2006-10-01 7901.00 10900.00 177.3 -17.6
13 2007-01-01 8279.00 11149.00 185.3 -5.3
14 2007-04-01 8867.00 12022.00 199.3 -10.9
15 2007-07-01 9439.00 12662.00 217.1 -12.8
16 2007-10-01 10101.00 13470.00 234.9 -8.4
And I want to apply following changes on the variables:
# yoy absolute difference change
abs.diff = c("B1","B2")
# yoy percentage change
percent.change = c("B3")
# make the variable negative
negative = c("B4")
This is the fuction that I am trying to use for my data frame.
transformation = function(D,abs.diff,percent.change,negative)
{
TT <- dim(D)[1]
DData <- D[-1,]
nms <- c()
for (i in c(2:dim(D)[2])) {
# yoy absolute difference change
if (names(D)[i] %in% abs.diff)
{ DData[,i] = (D[5:TT,i]-D[1:(TT-4),i])
names(DData)[i] = paste('a',names(D)[i],sep='') }
# yoy percent. change
if (names(D)[i] %in% percent.change)
{ DData[,i] = 100*(D[5:TT,i]-D[1:(TT-4),i])/D[1:(TT-4),i]
names(DData)[i] = paste('p',names(D)[i],sep='') }
#CA.deficit
if (names(D)[i] %in% negative)
{ DData[,i] = (-1)*D[1:TT,i] }
}
return(DData)
}
This is what I would like to get :
Date pB1 pB2 aB3 B4
1 2004-01-01 NA NA NA NA
2 2004-04-01 NA NA NA NA
3 2004-07-01 NA NA NA NA
4 2004-10-01 NA NA NA NA
5 2005-01-01 39.33 -2.26 NA 3.5
6 2005-04-01 40.05 3.25 NA 14.6
7 2005-07-01 42.22 9.56 NA 11.6
8 2005-10-01 42.54 15.29 11.0 10.2
9 2006-01-01 43.13 21.43 19.3 3.1
10 2006-04-01 42.11 21.72 18.3 16.0
11 2006-07-01 37.77 16.99 22.0 4.9
12 2006-10-01 34.26 22.05 17.7 17.6
13 2007-01-01 30.17 18.3 19.7 5.3
14 2007-04-01 27.95 20.32 26.1 10.9
15 2007-07-01 27.04 26.1 39.8 12.8
16 2007-10-01 27.84 23.58 49.6 8.4
Grouping by the months, i.e. 6th and 7th substring using ave and do the necessary calculations. With sapply we may loop over the columns.
f <- function(x) {
g <- substr(Date, 6, 7)
l <- length(unique(g))
o <- ave(x, g, FUN=function(x) 100/x * c(x[-1], NA) - 100)
c(rep(NA, l), head(o, -4))
}
cbind(df[1], sapply(df[-1], f))
# Date B1 B2 B3 B4
# 1 2004-01-01 NA NA NA NA
# 2 2004-04-01 NA NA NA NA
# 3 2004-07-01 NA NA NA NA
# 4 2004-10-01 NA NA NA NA
# 5 2005-01-01 39.32901 -2.259202 NA NA
# 6 2005-04-01 40.04796 3.250329 NA NA
# 7 2005-07-01 42.21960 9.560688 NA NA
# 8 2005-10-01 42.54044 15.291439 NA NA
# 9 2006-01-01 43.12655 21.434066 13.83738 -11.428571
# 10 2006-04-01 42.10895 21.720063 18.03279 9.589041
# 11 2006-07-01 37.77093 16.986386 17.58316 -57.758621
# 12 2006-10-01 34.25636 22.050356 17.26190 72.549020
# 13 2007-01-01 30.17296 18.304329 16.10276 70.967742
# 14 2007-04-01 27.95094 20.316253 20.35024 -31.875000
# 15 2007-07-01 27.03903 26.102978 25.34642 161.224490
# 16 2007-10-01 27.84458 23.577982 32.48731 -52.272727
Say I have a data.frame as follows, each month has one entry of data:
df <- read.table(text="date,gmsl
2009-01-17,58.4
2009-02-17,59.1
2009-04-16,60.9
2009-06-16,62.3
2009-09-16,64.6
2009-12-16,68.3",sep=",",header=TRUE)
## > df
## date gmsl
## 1 2009-01-17 58.4
## 2 2009-02-17 59.1
## 3 2009-04-16 60.9
## 4 2009-06-16 62.3
## 5 2009-09-16 64.6
## 6 2009-12-16 68.3
Just wondering how could I fill missing month with gmsl as NaN for date range from 2009-01 to 2009-12?
I have extracted year and month for date column by df$Month_Yr <- format(as.Date(df$date), "%Y-%m").
Here's a way to this with tidyr::complete
library(dplyr)
df %>%
mutate(date = as.Date(date),
first_date = as.Date(format(date, "%Y-%m-01"))) %>%
tidyr::complete(first_date = seq(min(first_date), max(first_date), "1 month"))
# A tibble: 12 x 3
# first_date date gmsl
# <date> <date> <dbl>
# 1 2009-01-01 2009-01-17 58.4
# 2 2009-02-01 2009-02-17 59.1
# 3 2009-03-01 NA NA
# 4 2009-04-01 2009-04-16 60.9
# 5 2009-05-01 NA NA
# 6 2009-06-01 2009-06-16 62.3
# 7 2009-07-01 NA NA
# 8 2009-08-01 NA NA
# 9 2009-09-01 2009-09-16 64.6
#10 2009-10-01 NA NA
#11 2009-11-01 NA NA
#12 2009-12-01 2009-12-16 68.3
You can then decide which column to keep, either first_date or date or combine them both.
data
df <- structure(list(date = structure(1:6, .Label = c("2009-01-17",
"2009-02-17", "2009-04-16", "2009-06-16", "2009-09-16", "2009-12-16"
), class = "factor"), gmsl = c(58.4, 59.1, 60.9, 62.3, 64.6,
68.3)), class = "data.frame", row.names = c(NA, -6L))
In base R you could match (using %in%) the substrings of a seq.Date.
dt.match <- seq.Date(ISOdate(2009, 1, 1), ISOdate(2009, 12, 1), "month")
sub <-
cbind(date=substr(dt.match, 1, 10)[!substr(dt.match, 1, 7) %in% substr(dat$date, 1, 7)],
gmsl=NA)
merge(dat, sub, all=TRUE)
# date gmsl
# 1 2009-01-17 58.4
# 2 2009-02-17 59.1
# 3 2009-03-01 <NA>
# 4 2009-04-16 60.9
# 5 2009-05-01 <NA>
# 6 2009-06-16 62.3
# 7 2009-07-01 <NA>
# 8 2009-08-01 <NA>
# 9 2009-09-16 64.6
# 10 2009-10-01 <NA>
# 11 2009-11-01 <NA>
# 12 2009-12-16 68.3
Data
dat <- structure(list(date = c("2009-01-17", "2009-02-17", "2009-04-16",
"2009-06-16", "2009-09-16", "2009-12-16"), gmsl = c(58.4, 59.1,
60.9, 62.3, 64.6, 68.3)), row.names = c(NA, -6L), class = "data.frame")
I was trying to forecast a time series problem using lm() and my data looks like below
Customer_key date sales
A35 2018-05-13 31
A35 2018-05-20 20
A35 2018-05-27 43
A35 2018-06-03 31
BH22 2018-05-13 60
BH22 2018-05-20 67
BH22 2018-05-27 78
BH22 2018-06-03 55
Converted my df to a list format by
df <- dcast(df, date ~ customer_key,value.var = c("sales"))
df <- subset(df, select = -c(dt))
demandWithKey <- as.list(df)
Trying to write a function such that applying this function across all customers
my_fun <- function(x) {
fit <- lm(ds_load ~ date, data=df) ## After changing to list ds_load and date column names
## are no longer available for formula
fit_b <- forecast(fit$fitted.values, h=20) ## forecast using lm()
return(data.frame(c(fit$fitted.values, fit_b[["mean"]])))
}
fcast <- lapply(df, my_fun)
I know the above function doesn't work, but basically I'm looking for getting both the fitted values and forecasted values for a grouped data.
But I've tried all other methods using tslm() (converting into time series data) and so on but no luck I can get the lm() work somehow on just one customer though. Also many questions/posts were on just fitting the model but I would like to forecast too at same time.
lm() is for a regression model
but here you have a time serie so for forecasting the serie you have to use one of the time serie model (ARMA ARCH GARCH...)
so you can use the function in r : auto.arima() in "forecast" package
I don't know what you're up to exactly, but you could make this less complicated.
Using by avoids the need to reshape your data, it splits your data e.g. by customer ID as in your case and applies a function on the subsets (i.e. it's a combination of split and lapply; see ?by).
Since you want to compare fitted and forecasted values somehow in your result, you probably need predict rather than $fitted.values, otherwise the values won't be of same length. Because your independent variable is a date in weekly intervals, you may use seq.Date and take the first date as a starting value; the sequence has length actual values (nrow each customer) plus h= argument of the forecast.
For demonstration purposes I add the fitted values as first column in the following.
res <- by(dat, dat$cus_key, function(x) {
H <- 20 ## globally define 'h'
fit <- lm(sales ~ date, x)
fitted <- fit$fitted.values
pred <- predict(fit, newdata=data.frame(
date=seq(x$date[1], length.out= nrow(x) + H, by="week")))
fcst <- c(fitted, forecast(fitted, h=H)$mean)
fit.na <- `length<-`(unname(fitted), length(pred)) ## for demonstration
return(cbind(fit.na, pred, fcst))
})
Result
res
# dat$cus_key: A28
# fit.na pred fcst
# 1 41.4 41.4 41.4
# 2 47.4 47.4 47.4
# 3 53.4 53.4 53.4
# 4 59.4 59.4 59.4
# 5 65.4 65.4 65.4
# 6 NA 71.4 71.4
# 7 NA 77.4 77.4
# 8 NA 83.4 83.4
# 9 NA 89.4 89.4
# 10 NA 95.4 95.4
# 11 NA 101.4 101.4
# 12 NA 107.4 107.4
# 13 NA 113.4 113.4
# 14 NA 119.4 119.4
# 15 NA 125.4 125.4
# 16 NA 131.4 131.4
# 17 NA 137.4 137.4
# 18 NA 143.4 143.4
# 19 NA 149.4 149.4
# 20 NA 155.4 155.4
# 21 NA 161.4 161.4
# 22 NA 167.4 167.4
# 23 NA 173.4 173.4
# 24 NA 179.4 179.4
# 25 NA 185.4 185.4
# ----------------------------------------------------------------
# dat$cus_key: B16
# fit.na pred fcst
# 1 49.0 49.0 49.0
# 2 47.7 47.7 47.7
# 3 46.4 46.4 46.4
# 4 45.1 45.1 45.1
# 5 43.8 43.8 43.8
# 6 NA 42.5 42.5
# 7 NA 41.2 41.2
# 8 NA 39.9 39.9
# 9 NA 38.6 38.6
# 10 NA 37.3 37.3
# 11 NA 36.0 36.0
# 12 NA 34.7 34.7
# 13 NA 33.4 33.4
# 14 NA 32.1 32.1
# 15 NA 30.8 30.8
# 16 NA 29.5 29.5
# 17 NA 28.2 28.2
# 18 NA 26.9 26.9
# 19 NA 25.6 25.6
# 20 NA 24.3 24.3
# 21 NA 23.0 23.0
# 22 NA 21.7 21.7
# 23 NA 20.4 20.4
# 24 NA 19.1 19.1
# 25 NA 17.8 17.8
# ----------------------------------------------------------------
# dat$cus_key: C12
# fit.na pred fcst
# 1 56.4 56.4 56.4
# 2 53.2 53.2 53.2
# 3 50.0 50.0 50.0
# 4 46.8 46.8 46.8
# 5 43.6 43.6 43.6
# 6 NA 40.4 40.4
# 7 NA 37.2 37.2
# 8 NA 34.0 34.0
# 9 NA 30.8 30.8
# 10 NA 27.6 27.6
# 11 NA 24.4 24.4
# 12 NA 21.2 21.2
# 13 NA 18.0 18.0
# 14 NA 14.8 14.8
# 15 NA 11.6 11.6
# 16 NA 8.4 8.4
# 17 NA 5.2 5.2
# 18 NA 2.0 2.0
# 19 NA -1.2 -1.2
# 20 NA -4.4 -4.4
# 21 NA -7.6 -7.6
# 22 NA -10.8 -10.8
# 23 NA -14.0 -14.0
# 24 NA -17.2 -17.2
# 25 NA -20.4 -20.4
As you can see, prediction and forecast yield the same values, since both methods are based on the same single explanatory variable date in this case.
Toy data:
set.seed(42)
dat <- transform(expand.grid(cus_key=paste0(LETTERS[1:3], sample(12:43, 3)),
date=seq.Date(as.Date("2018-05-13"), length.out=5, by="week")),
sales=sample(20:80, 15, replace=TRUE))
Given the following set of data:
transect <- c("B","N","C","D","H","J","E","L","I","I")
sampler <- c(rep("J",5),rep("W",5))
species <- c("ROB","HAW","HAW","ROB","PIG","HAW","PIG","PIG","HAW","HAW")
weight <- c(2.80,52.00,56.00,2.80,16.00,55.00,16.20,18.30,52.50,57.00)
wingspan <- c(13.9, 52.0, 57.0, 13.7, 11.0,52.5, 10.7, 11.1, 52.3, 55.1)
week <- c(1,2,3,4,5,6,7,8,9,9)
# Warning to R newbs: Really bad idea to use this code
ex <- as.data.frame(cbind(transect,sampler,species,weight,wingspan,week))
What I’m trying to achieve is to transpose the species and its associated information on weight and wingspan. For a better idea of the expected result please see below. My data set is about half a million lines long with approximately 200 different species so it will be a very large dataframe.
transect sampler week ROBweight HAWweight PIGweight ROBwingspan HAWwingspan PIGwingspan
1 B J 1 2.8 0.0 0.0 13.9 0.0 0.0
2 N J 2 0.0 52.0 0.0 0.0 52.0 0.0
3 C J 3 0.0 56.0 0.0 0.0 57.0 0.0
4 D J 4 2.8 0.0 0.0 13.7 0.0 0.0
5 H J 5 0.0 0.0 16.0 0.0 0.0 11.0
6 J W 6 0.0 55.0 0.0 0.0 52.5 0.0
7 E W 7 0.0 0.0 16.2 0.0 0.0 10.7
8 L W 8 0.0 0.0 18.3 0.0 0.0 11.1
9 I W 9 0.0 52.5 0.0 0.0 52.3 0.0
10 I W 9 0.0 57.0 0.0 0.0 55.1 0.0
The main problem is that you don't currently have unique "id" variables, which will create problems for the usual suspects of reshape and dcast.
Here's a solution. I've used getanID from my "splitstackshape" package, but it's pretty easy to create your own unique ID variable using many different methods.
library(splitstackshape)
library(reshape2)
idvars <- c("transect", "sampler", "week")
ex <- getanID(ex, id.vars=idvars)
From here, you have two options:
reshape from base R:
reshape(ex, direction = "wide",
idvar=c("transect", "sampler", "week", ".id"),
timevar="species")
melt and dcast from "reshape2"
First, melt your data into a "long" form.
exL <- melt(ex, id.vars=c(idvars, ".id", "species"))
Then, cast your data into a wide form.
dcast(exL, transect + sampler + week + .id ~ species + variable)
# transect sampler week .id HAW_weight HAW_wingspan PIG_weight PIG_wingspan ROB_weight ROB_wingspan
# 1 B J 1 1 NA NA NA NA 2.8 13.9
# 2 C J 3 1 56.0 57.0 NA NA NA NA
# 3 D J 4 1 NA NA NA NA 2.8 13.7
# 4 E W 7 1 NA NA 16.2 10.7 NA NA
# 5 H J 5 1 NA NA 16.0 11.0 NA NA
# 6 I W 9 1 52.5 52.3 NA NA NA NA
# 7 I W 9 2 57.0 55.1 NA NA NA NA
# 8 J W 6 1 55.0 52.5 NA NA NA NA
# 9 L W 8 1 NA NA 18.3 11.1 NA NA
# 10 N J 2 1 52.0 52.0 NA NA NA NA
A better option: "data.table"
Alternatively (and perhaps preferably), you can use the "data.table" package (at least version 1.8.11) as follows:
library(data.table)
library(reshape2) ## Also required here
packageVersion("data.table")
# [1] ‘1.8.11’
DT <- data.table(ex)
DT[, .id := sequence(.N), by = c("transect", "sampler", "week")]
DTL <- melt(DT, measure.vars=c("weight", "wingspan"))
dcast.data.table(DTL, transect + sampler + week + .id ~ species + variable)
# transect sampler week .id HAW_weight HAW_wingspan PIG_weight PIG_wingspan ROB_weight ROB_wingspan
# 1: B J 1 1 NA NA NA NA 2.8 13.9
# 2: C J 3 1 56.0 57.0 NA NA NA NA
# 3: D J 4 1 NA NA NA NA 2.8 13.7
# 4: E W 7 1 NA NA 16.2 10.7 NA NA
# 5: H J 5 1 NA NA 16.0 11.0 NA NA
# 6: I W 9 1 52.5 52.3 NA NA NA NA
# 7: I W 9 2 57.0 55.1 NA NA NA NA
# 8: J W 6 1 55.0 52.5 NA NA NA NA
# 9: L W 8 1 NA NA 18.3 11.1 NA NA
# 10: N J 2 1 52.0 52.0 NA NA NA NA
Add fill = 0 to either of the dcast versions to replace NA values with 0.