I have a data frame 'test' like below,
day Rain SWC_11 SWC_12 SWC_13 SWC_14 SWC_21
01/01/2019 00:00:00 0.2 51 60 63 60 64
02/01/2019 00:00:00 0.2 51.5 60.3 63.4 60.8 64.4
03/01/2019 00:00:00 0.0 51.3 60.3 63.3 60.6 64.1
04/01/2019 00:00:00 0.4 51.5 60.3 63.4 60.8 64.4
15/01/2019 00:00:00 0.0 NA NA NA NA NA
16/01/2019 00:00:00 0.0 NA NA NA NA NA
17/01/2019 00:00:00 0.0 51.5 60.3 63.4 60.8 64.4
Now I want to count the days after each rain event, once it comes to the next rain events, it restarts again. My ideal outputs gonna be like below.
day Rain SWC_11 SWC_12 SWC_13 SWC_14 SWC_21 events
01/01/2019 00:00:00 0.2 51 60 63 60 64 1
02/01/2019 00:00:00 0.2 51.5 60.3 63.4 60.8 64.4 1
03/01/2019 00:00:00 0.0 51.3 60.3 63.3 60.6 64.1 2
04/01/2019 00:00:00 0.4 51.5 60.3 63.4 60.8 64.4 1
15/01/2019 00:00:00 0.0 NA NA NA NA NA 12
16/01/2019 00:00:00 0.0 NA NA NA NA NA 13
17/01/2019 00:00:00 0.0 51.5 60.3 63.4 60.8 64.4 14
my code is
test$day<- as.numeric(as.Date(test$day))
for(i in 1:(nrow(test)-1))
if (test$Rain[[i]] != 0){
test$event[i] <- 1
test$event[i+nrow(test)] <-test$day[i+nrow(test)]- test$day[i] +1
}else{
test$event <-0
}
but the results looks wired and the warning message is as below,
Error in `$<-.data.frame`(`*tmp*`, "event", value = c(0, 1, 0, 0, 0, 0, :
replacement has 12 rows, data has 10
Hope someone gonna help.
Instead of the rle-derivative I suggested earlier, I think cumulative-sum logic can be used here.
dat %>%
mutate(
day = as.Date(day, format="%d/%m/%Y"),
daylag = dplyr::lag(day, default = first(day) - 1)
) %>%
group_by(grp = cumsum(Rain > 0)) %>%
mutate(event = day - daylag[1]) %>%
ungroup()
# # A tibble: 7 x 11
# day Rain SWC_11 SWC_12 SWC_13 SWC_14 SWC_21 events daylag grp event
# <date> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <int> <date> <int> <drtn>
# 1 2019-01-01 0.2 51 60 63 60 64 1 2018-12-31 1 1 days
# 2 2019-01-02 0.2 51.5 60.3 63.4 60.8 64.4 1 2019-01-01 2 1 days
# 3 2019-01-03 0 51.3 60.3 63.3 60.6 64.1 2 2019-01-02 2 2 days
# 4 2019-01-04 0.4 51.5 60.3 63.4 60.8 64.4 1 2019-01-03 3 1 days
# 5 2019-01-15 0 NA NA NA NA NA 12 2019-01-04 3 12 days
# 6 2019-01-16 0 NA NA NA NA NA 13 2019-01-15 3 13 days
# 7 2019-01-17 0 51.5 60.3 63.4 60.8 64.4 14 2019-01-16 3 14 days
Data:
dat <- read.table(header = TRUE, text = "
day Rain SWC_11 SWC_12 SWC_13 SWC_14 SWC_21 events
01/01/2019 0.2 51 60 63 60 64 1
02/01/2019 0.2 51.5 60.3 63.4 60.8 64.4 1
03/01/2019 0.0 51.3 60.3 63.3 60.6 64.1 2
04/01/2019 0.4 51.5 60.3 63.4 60.8 64.4 1
15/01/2019 0.0 NA NA NA NA NA 12
16/01/2019 0.0 NA NA NA NA NA 13
17/01/2019 0.0 51.5 60.3 63.4 60.8 64.4 14")
I have a solution which works just with base R, but it is not as short as the one above.
# imagine if you have to do this manually, how would you achieve it step-by-step?
# then just use base R to realize every single step.
# create the "events" column and "day2" column
test$events <- NA
test$day2 <- as.Date(test$day,format="%d/%m/%Y")
# test if the first day, rains or not, and assign the value for event
for (i in 1:1){
if(test$Rain[[i]] !=0){
test$events[[i]] <- 1
}
else {
test$events[[i]] <- 0
}
}
# then starting from the 2nd row, go down one by one
# assign the value for "events" column based on your criteria
for (i in 2:(nrow(test))){
if(test$Rain[[i-1]] !=0 &
test$Rain[[i]] != 0){
test$events[[i]] <- 1
}
if (test$Rain[[i-1]] != 0 &
test$Rain[[i]] == 0){
test$events[[i]] <- test$events[[i-1]] + 1*(as.Date(as.character(test$day2[[i]]), format="%d/%m/%Y") -
as.Date(as.character(test$day2[[i-1]]), format="%d/%m/%Y"))
}
if (test$Rain[[i-1]] == 0 &
test$Rain[[i]] !=0){
test$events[[i]] <-1
}
if (test$Rain[[i-1]] == 0 &
test$Rain[[i]] ==0){
test$events[[i]] <- test$events[[i-1]] + + 1*(as.Date(as.character(test$day2[[i]]), format="%d/%m/%Y") -
as.Date(as.character(test$day2[[i-1]]), format="%d/%m/%Y"))
}
}
Now you will have the desired results. My solution is not very smart, but this is my thinking when the codes seem to be difficult.
Related
This question already has answers here:
Categorize numeric variable into group/ bins/ breaks
(4 answers)
Closed 1 year ago.
I am attempting to add a new column to the state sample data frame in R. I am hoping for this column to cluster the ID of states into broader categories (1-4). My code is close to what I am looking for but I am not getting it quite right.. I know I could enter each state ID line by line but is there a a quicker way? Thank you!
library(tidyverse)
#Add column to denote each state
States=state.x77
States=data.frame(States)
States <- tibble::rowid_to_column(States, "ID")
States
#Create new variable for state buckets
States <- States %>%
mutate(WAGE_BUCKET=case_when(ID <= c(1,12) ~ '1',
ID <= c(13,24) ~ '2',
ID <= c(25,37) ~ '3',
ID <= c(38,50) ~ '4',
TRUE ~ 'NA'))
View(States) #It is not grouping the states in the way I want/I am still getting some NA values but unsure why!
You can use cut or findInterval if all of your groups will be using contiguous ID values:
findInterval(States$ID, c(0, 12, 24, 37, 51))
# [1] 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4
If you want to make it a bit more verbose, you can use dplyr::between in your case_when:
States %>%
mutate(
WAGE_BUCKET = case_when(
between(ID, 1, 12) ~ "1",
between(ID, 13, 24) ~ "2",
between(ID, 25, 37) ~ "3",
between(ID, 38, 50) ~ "4",
TRUE ~ NA_character_)
)
# ID Population Income Illiteracy Life Exp Murder HS Grad Frost Area WAGE_BUCKET
# 1 1 3615 3624 2.1 69.05 15.1 41.3 20 50708 1
# 2 2 365 6315 1.5 69.31 11.3 66.7 152 566432 1
# 3 3 2212 4530 1.8 70.55 7.8 58.1 15 113417 1
# 4 4 2110 3378 1.9 70.66 10.1 39.9 65 51945 1
# 5 5 21198 5114 1.1 71.71 10.3 62.6 20 156361 1
# 6 6 2541 4884 0.7 72.06 6.8 63.9 166 103766 1
# 7 7 3100 5348 1.1 72.48 3.1 56.0 139 4862 1
# 8 8 579 4809 0.9 70.06 6.2 54.6 103 1982 1
# 9 9 8277 4815 1.3 70.66 10.7 52.6 11 54090 1
# 10 10 4931 4091 2.0 68.54 13.9 40.6 60 58073 1
# 11 11 868 4963 1.9 73.60 6.2 61.9 0 6425 1
# 12 12 813 4119 0.6 71.87 5.3 59.5 126 82677 1
# 13 13 11197 5107 0.9 70.14 10.3 52.6 127 55748 2
# 14 14 5313 4458 0.7 70.88 7.1 52.9 122 36097 2
# 15 15 2861 4628 0.5 72.56 2.3 59.0 140 55941 2
# 16 16 2280 4669 0.6 72.58 4.5 59.9 114 81787 2
# 17 17 3387 3712 1.6 70.10 10.6 38.5 95 39650 2
# 18 18 3806 3545 2.8 68.76 13.2 42.2 12 44930 2
# 19 19 1058 3694 0.7 70.39 2.7 54.7 161 30920 2
# 20 20 4122 5299 0.9 70.22 8.5 52.3 101 9891 2
# 21 21 5814 4755 1.1 71.83 3.3 58.5 103 7826 2
# 22 22 9111 4751 0.9 70.63 11.1 52.8 125 56817 2
# 23 23 3921 4675 0.6 72.96 2.3 57.6 160 79289 2
# 24 24 2341 3098 2.4 68.09 12.5 41.0 50 47296 2
# 25 25 4767 4254 0.8 70.69 9.3 48.8 108 68995 3
# 26 26 746 4347 0.6 70.56 5.0 59.2 155 145587 3
# 27 27 1544 4508 0.6 72.60 2.9 59.3 139 76483 3
# 28 28 590 5149 0.5 69.03 11.5 65.2 188 109889 3
# 29 29 812 4281 0.7 71.23 3.3 57.6 174 9027 3
# 30 30 7333 5237 1.1 70.93 5.2 52.5 115 7521 3
# 31 31 1144 3601 2.2 70.32 9.7 55.2 120 121412 3
# 32 32 18076 4903 1.4 70.55 10.9 52.7 82 47831 3
# 33 33 5441 3875 1.8 69.21 11.1 38.5 80 48798 3
# 34 34 637 5087 0.8 72.78 1.4 50.3 186 69273 3
# 35 35 10735 4561 0.8 70.82 7.4 53.2 124 40975 3
# 36 36 2715 3983 1.1 71.42 6.4 51.6 82 68782 3
# 37 37 2284 4660 0.6 72.13 4.2 60.0 44 96184 3
# 38 38 11860 4449 1.0 70.43 6.1 50.2 126 44966 4
# 39 39 931 4558 1.3 71.90 2.4 46.4 127 1049 4
# 40 40 2816 3635 2.3 67.96 11.6 37.8 65 30225 4
# 41 41 681 4167 0.5 72.08 1.7 53.3 172 75955 4
# 42 42 4173 3821 1.7 70.11 11.0 41.8 70 41328 4
# 43 43 12237 4188 2.2 70.90 12.2 47.4 35 262134 4
# 44 44 1203 4022 0.6 72.90 4.5 67.3 137 82096 4
# 45 45 472 3907 0.6 71.64 5.5 57.1 168 9267 4
# 46 46 4981 4701 1.4 70.08 9.5 47.8 85 39780 4
# 47 47 3559 4864 0.6 71.72 4.3 63.5 32 66570 4
# 48 48 1799 3617 1.4 69.48 6.7 41.6 100 24070 4
# 49 49 4589 4468 0.7 72.48 3.0 54.5 149 54464 4
# 50 50 376 4566 0.6 70.29 6.9 62.9 173 97203 4
It is a vector of length > 1. The comparison operators works on a single vector. We could use between
library(dplyr)
States <- States %>%
mutate(WAGE_BUCKET=case_when(between(ID, 1, 12) ~ '1',
between(ID, 13,24) ~ '2',
between(ID, 25,37) ~ '3',
between(ID, 38,50) ~ '4',
TRUE ~ NA_character_))
Or another option is to use & with > and <=
States %>%
mutate(WAGE_BUCKET=case_when(ID >= 1 & ID <=12 ~ '1',
ID >= 13 & ID <= 24) ~ '2',
ID >= 25 & ID <= 37 ~ '3',
ID >= 38 & ID <= 50 ~ '4',
TRUE ~ NA_character))
Or may be the OP meant to use %in%
States %>%
mutate(WAGE_BUCKET=case_when(ID %in% c(1,12) ~ '1',
ID %in% c(13,24) ~ '2',
ID %in% c(25,37) ~ '3',
ID %in% c(38,50) ~ '4',
TRUE ~ NA_character_))
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
This question already has answers here:
Transpose a data frame
(6 answers)
Closed 2 years ago.
Have searched the threads but can't understand a solution that will solve the problem with the data frame that I have.
My current data frame (df):
# A tibble: 8 x 29
`Athlete` Monday...2 Tuesday...3 Wednesday...4 Thursday...5 Friday...6 Saturday...7 Sunday...8
<chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr>
1 Date 29/06/2020 30/06/2020 43837.0 43868.0 43897.0 43928.0 43958.0
2 HR 47.0 54.0 51.0 56.0 59.0 NA NA
3 HRV 171.0 91.0 127.0 99.0 77.0 NA NA
4 Sleep Duration 9.11 7.12 8.59 7.15 8.32 NA NA
5 Sleep Efficien~ 92.0 94.0 89.0 90.0 90.0 NA NA
6 Recovery Score 98.0 66.0 96.0 72.0 46.0 NA NA
7 Life Stress NO NO NO NO NO NA NA
8 Sick NO NO NO NO NO NA NA
Have tried to use spread and pivot wider but I know there would require additional functions in order to get the desired output which beyond my level on understanding in R.
Do I need to u
Desired output:
Date HR HRV Sleep Duration Sleep Efficiency Recovery Score Life Stress Sick
29/06/2020 47.0 171.0 9.11
30/06/2020 54.0 91.0 7.12
43837.0 51.0 127.0 8.59
43868.0 56.0 99.0 7.15
43897.0 59.0 77.0 8.32
43928.0 NA NA NA
43958.0 NA NA NA
etc.
Thank you
In Base R you will do:
type.convert(setNames(data.frame(t(df[-1]), row.names = NULL), df[,1]))
Date HR HRV Sleep Duration Sleep Efficien~ Recovery Score Life Stress Sick
1 29/06/2020 47 171 9.11 92 98 NO NO
2 30/06/2020 54 91 7.12 94 66 NO NO
3 43837.0 51 127 8.59 89 96 NO NO
4 43868.0 56 99 7.15 90 72 NO NO
5 43897.0 59 77 8.32 90 46 NO NO
6 43928 NA NA NA NA NA <NA> <NA>
7 43958 NA NA NA NA NA <NA> <NA>
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))
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I have got monthly data in this format
PrecipMM Date
122.7 2004-01-01
54.2 2005-01-01
31.9 2006-01-01
100.5 2007-01-01
144.9 2008-01-01
96.4 2009-01-01
75.3 2010-01-01
94.8 2011-01-01
67.6 2012-01-01
93.0 2013-01-01
184.6 2014-01-01
101.0 2015-01-01
149.3 2016-01-01
50.2 2004-02-01
46.2 2005-02-01
57.7 2006-02-01
I want to calculate all of the difference of precipMM in same month of different years.
My dream output is like this:
PrecipMM Date PrecipMM_diff
122.7 2004-01-01 NA
54.2 2005-01-01 -68.5
31.9 2006-01-01 -22.3
100.5 2007-01-01 68.6
144.9 2008-01-01 44.4
96.4 2009-01-01 -48.5
75.3 2010-01-01 -21.2
94.8 2011-01-01 19.5
67.6 2012-01-01 -27.2
93.0 2013-01-01 25.4
184.6 2014-01-01 91.6
101.0 2015-01-01 -83.6
149.3 2016-01-01 48.3
50.2 2004-02-01 NA
46.2 2005-02-01 -4.0
57.7 2006-02-01 11.5
I think diff() can do this but I have no idea how.
I think you can do this with lag combined with group_by from dplyr. Here's how:
library(dplyr)
library(lubridate) # makes dealing with dates easier
# Load your example data
df <- structure(list(PrecipMM = c(4.4, 66.7, 48.2, 60.9, 108.1, 109.2,
101.7, 38.1, 53.8, 71.9, 75.4, 67.1, 92.7, 115.3, 68.9, 38.9),
Date = structure(5:20, .Label = c("101.7", "108.1", "109.2",
"115.3", "1766-01-01", "1766-02-01", "1766-03-01", "1766-04-01",
"1766-05-01", "1766-06-01", "1766-07-01", "1766-08-01", "1766-09-01",
"1766-10-01", "1766-11-01", "1766-12-01", "1767-01-01", "1767-02-01",
"1767-03-01", "1767-04-01", "38.1", "38.9", "4.4", "48.2",
"53.8", "60.9", "66.7", "67.1", "68.9", "71.9", "75.4", "92.7"
), class = "factor")), class = "data.frame", row.names = c(NA,
-16L), .Names = c("PrecipMM", "Date"))
results <- df %>%
mutate(years = year(Date), months = month(Date)) %>%
group_by(months) %>%
arrange(years) %>%
mutate(lagged.rain = lag(PrecipMM), rain.diff = PrecipMM - lagged.rain)
results
# Source: local data frame [16 x 6]
# Groups: months [12]
#
# PrecipMM Date years months lagged.rain rain.diff
# (dbl) (fctr) (dbl) (dbl) (dbl) (dbl)
# 1 4.4 1766-01-01 1766 1 NA NA
# 2 92.7 1767-01-01 1767 1 4.4 88.3
# 3 66.7 1766-02-01 1766 2 NA NA
# 4 115.3 1767-02-01 1767 2 66.7 48.6
# 5 48.2 1766-03-01 1766 3 NA NA
# 6 68.9 1767-03-01 1767 3 48.2 20.7
# 7 60.9 1766-04-01 1766 4 NA NA
# 8 38.9 1767-04-01 1767 4 60.9 -22.0
# 9 108.1 1766-05-01 1766 5 NA NA
# 10 109.2 1766-06-01 1766 6 NA NA
# 11 101.7 1766-07-01 1766 7 NA NA
# 12 38.1 1766-08-01 1766 8 NA NA
# 13 53.8 1766-09-01 1766 9 NA NA
# 14 71.9 1766-10-01 1766 10 NA NA
# 15 75.4 1766-11-01 1766 11 NA NA
# 16 67.1 1766-12-01 1766 12 NA NA