Check differences of various DATE inside one variables R - r

Question

I want to split the line when the variable contain different YEAR,
also split the col : "Price" with evenly divided by the numbers of date appear
--> count (" ; ") +1
There is a table with the variable that is not yet be splitted.
# Dataset call df
Price Date
500 2016-01-01
400 2016-01-03;2016-01-09
1000 2016-01-04;2017-09-01;2017-08-10;2018-01-01
25 2016-01-04;2017-09-01
304 2015-01-02
238 2018-01-02;2018-02-02
Desire Outlook
# Targeted df
Price Date
500 2016-01-01
400 2016-01-03;2016-01-09
250 2016-01-04
250 2017-09-01
250 2017-08-10
250 2018-01-01
12.5 2016-01-04
12.5 2017-09-01
304 2015-01-02
238 2018-01-02;2018-02-02
Once the variable contains different year is defined , below is the operation
have to do .(It is just a example .)
mutate(Price = ifelse(DIFFERENT_DATE_ROW,
as.numeric(Price) / (str_count(Date,";")+1),
as.numeric(Price)),
Date = ifelse(DIFFERENT_DATE_ROW,
strsplit(as.character(Date),";"),
Date)) %>%
unnest()
I meet some constraints that cannot use dplyr's function "if_else" because
else NO operation cannot be recognized .Only ifelse work properly.
How to find out there is differences of the year in one variables to
PROVOKE the split line & split price calculations ?
so far the operation to split the element like
unlist(lapply(unlist(strsplit(df1$noFDate[8],";")),FUN = year))
cannot solve the problem.
I am beginner of coding , please feel free to change all operation above with considering the real data have over 2 million rows and 50 cols.

Answer 1

This might not be the most efficient one but can be used to get the required answer.
#Get the row indices which we need to separate
inds <- sapply(strsplit(df$Date, ";"), function(x)
#Format the date into year and count number of unique values
#Return TRUE if number of unique values is greater than 1
length(unique(format(as.Date(x), "%Y"))) > 1
)
library(tidyverse)
library(stringr)
#Select those indices
df[inds, ] %>%
# divide the price by number of dates in that row
mutate(Price = Price / (str_count(Date,";") + 1)) %>%
# separate `;` delimited values in separate rows
separate_rows(Date, sep = ";") %>%
# bind the remaining rows as it is
bind_rows(df[!inds,])
# Price Date
#1 250.0 2016-01-04
#2 250.0 2017-09-01
#3 250.0 2017-08-10
#4 250.0 2018-01-01
#5 12.5 2016-01-04
#6 12.5 2017-09-01
#7 500.0 2016-01-01
#8 400.0 2016-01-03;2016-01-09
#9 304.0 2015-01-02
#10 238.0 2018-01-02;2018-02-02

Answer 2

A bit cumbersome but you could do:
d_new = lapply(1:nrow(dat),function(x) {
a = dat[x,]
b = unlist(strsplit(as.character(a$Date),";"))
l = length(b)
if (l==1) check = 0 else check = ifelse(var(as.numeric(strftime(b,"%Y")))==0,0,1)
if (check==0) {
a
} else {
data.frame(Date = b, Price = rep(a$Price / l,l))
}
})
do.call(rbind,d_new)