How to use if else statement in a dataframe when comparing dates? - r

Question

I have a dataframe D and I would want to calculate a daily return of "Close" only if they share the same month. So for example there would be 0 for 1995-08-01
Date Close Month
1 1995-07-27 163.32 1995-07
2 1995-07-28 161.36 1995-07
3 1995-07-30 162.91 1995-07
4 1995-08-01 162.95 1995-08
5 1995-08-02 162.69 1995-08
I am trying to use an if-else statement and looping to apply it on other dataframes.
D1 <- D[-1,]
for (i in c("Close"))
{ TT <- dim(D)[1]
if (D[1:(TT-1),"Month"] == D[2:TT,"Month"]) {
D1[,i] = round((100*(log(D[2:TT,i]/D[1:(TT-1),i]))), digits = 4)
}
else {
D1[i] = 0 }
}
I get these results but in the forth row it should be 0.0000 because the forth row is a from different month than the the third row. Moreover, I get this warning message : "Warning message: In if (D[1:(TT - 1), "Month"] == D[2:TT, "Month"]) { : the condition has length > 1 and only the first element will be used". Can you please help me out? Thank you.
Date Close Month
1 1995-07-27 0.5903 1995-07
2 1995-07-28 1.4577 1995-07
3 1995-07-30 0.9139 1995-07
4 1995-08-01 0.0006 1995-08
5 1995-08-02 0.0255 1995-08

Answer 1

Next time you should REALLY provide a reproducible example here I did it for you. My solution uses diff and ifelse as requested.
month <- c(1,1:5,5:6)
data <- (1:8)*(1:8)
df <- data.frame(cbind(month, data))
diffs <- sapply(df, diff)
diffs <- data.frame(rbind(NA, diffs))
df$result <- ifelse(diffs$month==0, diffs$data, 0)
df
month data result
1 1 1 NA
2 1 4 3
3 2 9 0
4 3 16 0
5 4 25 0
6 5 36 0
7 5 49 13
8 6 64 0

Answer 2

if() expects a single value (usually TRUE or FALSE, but can also be 0 or 1, and it can handle other single values, e.g., it treats positive values like ones). You are feeding in a vector of values. The warning message is telling you that it is ignoring all the other values of the vector except the first, which is usually a strong indication that your code is not doing what you intend it to do.
Here's one do-it-yourself approach with no loops (I'm sure some time-series package has a function to calculate returns):
# create your example dataset
D <- data.frame(
Date = (as.Date("1995-07-27") + 0:6)[-c(3,5)],
Close = 162 + c(1.32, -.64, .91, .95, .69)
)
# get lagged values as new columns
D$Close_lag <- dplyr::lag(D$Close)
D$Date_lag <- dplyr::lag(D$Date)
# calculate all returns
D$return <- D$Close / D$Close_lag - 1
# identify month switches
D$new_month <- lubridate::month(D$Date) != lubridate::month(D$Date_lag)
# replace returns with zeros when month switches
D[!is.na(D$return) & D$new_month==TRUE, "return"] <- 0
# print results
D