Dan Dan - 22 days ago 5
R Question

converting daily Total Returns and Prices to annual values in R

I've data of Total Returns and Stock Prices on a daily basis for some banks, from 1997 to 2015, such that:

DATE Bank1_TotalReturn Bank1_Price Bank2_TR Bank2_P ... and so on for all other banks
01/01/1997 103.13 10.43 NA NA
02/01/1997 104.66 11.12 153.89 23.08
03/01/1997 ... ... ... ...
...and so on
for all other
days until
31/12/2015


Using R, I need to convert them in "medium annual values", so that I obtain a single annual medium value both for Total returns and Prices, for each bank, obviously, in order to make a Panel dataset.

nb: data contains lots of missing values, code must consider that issue! :)

Answer

You can use format to extract the year from DATE (once converted to a Date class) to convert the date to the year of the date. Then using dplyr:

library(dplyr)
res <- df %>% group_by(Year=format(as.Date(DATE,format="%d/%m/%Y"),"%Y")) %>% 
              summarise_at(vars(-DATE), median, na.rm=TRUE)

We first group_by the converted Year, and then we use summarise_at to summarize the median for each column except DATE. Note that we pass in the argument na.rm=TRUE to median to ignore NA's.

Or using aggregate:

df$Year <- format(as.Date(df$DATE,format="%d/%m/%Y"),"%Y")
df <- df[,!(names(df)=="DATE")]
res <- aggregate(. ~ Year, data=df, FUN=median, na.rm=TRUE, na.action=NULL)

Here, we preprocess df to change the DATE column to Year and we use the formula . ~ Year to specify aggregating all columns grouped by Year. Note that in aggregate, we specify na.action=NULL since the default na.action will remove rows that have anyNA. Instead, we passna.rm=TRUEto the functionmedianto handleNA`'s.

To illustrate, I augmented your posted data:

df <- structure(list(DATE = c("01/01/1997", "02/01/1997", "03/01/1997", 
"04/01/1997", "01/01/1998", "02/01/1998", "03/01/1998", "04/01/1998"
), Bank1_TotalReturn = c(103.13, 104.66, 105.23, NA, 113.13, 
114.66, 115.23, NA), Bank1_Price = c(10.43, 11.12, 12.15, NA, 
11.43, 12.12, NA, 13.15), Bank2_TR = c(NA, 153.89, 145.89, 136.89, 
140.92, 153.89, 145.89, 146.89), Bank2_P = c(NA, 23.08, NA, NA, 
20.9, 23.08, 25.73, 25.98)), .Names = c("DATE", "Bank1_TotalReturn", 
"Bank1_Price", "Bank2_TR", "Bank2_P"), class = "data.frame", row.names = c(NA, 
-8L))
##        DATE Bank1_TotalReturn Bank1_Price Bank2_TR Bank2_P
##1 01/01/1997            103.13       10.43       NA      NA
##2 02/01/1997            104.66       11.12   153.89   23.08
##3 03/01/1997            105.23       12.15   145.89      NA
##4 04/01/1997                NA          NA   136.89      NA
##5 01/01/1998            113.13       11.43   140.92   20.90
##6 02/01/1998            114.66       12.12   153.89   23.08
##7 03/01/1998            115.23          NA   145.89   25.73
##8 04/01/1998                NA       13.15   146.89   25.98

With this data using either approaches, the result is:

print(res)
##   Year Bank1_TotalReturn Bank1_Price Bank2_TR Bank2_P
##1  1997            104.66       11.12   145.89  23.080
##2  1998            114.66       12.12   146.39  24.405

If the intent is to compute the mean instead of the median, then just substitute mean for median. For the aggregate solution:

res <- aggregate(. ~ Year, data=df, FUN=mean, na.rm=TRUE, na.action=NULL)
print(res)
##  Year Bank1_TotalReturn Bank1_Price Bank2_TR Bank2_P
##1 1997            104.34    11.23333 145.5567 23.0800
##2 1998            114.34    12.23333 146.8975 23.9225