Adding different value to different parts of a dataframe without loop - r

Question

I want to speed up this calculation I need to do on specific part of the dataframe, this is an example data
days <- c("01.01.2018","01.01.2018","01.01.2018",
"02.01.2018","02.01.2018","02.01.2018",
"03.01.2018","03.01.2018","03.01.2018")
time <- c("00:00:00","01:00:00","02:00:00",
"00:00:00","01:00:00","02:00:00",
"00:00:00","01:00:00","02:00:00")
a <- c(1,2,3,
1,2,3,
1,2,3)
b <- c(1,2,3,
5,6,7,
10,11,12)
results <- NA
df1 <- data.frame(days,time,a,results)
df2 <- data.frame(days,time,b)
I need to add the value from df2$b at 00:00:00 of each day to the same entire day values in df1$a and store it in results.
Right now I'm doing it like this:
ndays <- unique(df1$days)
for(i in 1:length(ndays)) {
factor <- df2[(df2$days == ndays[i] & df2$time == "00:00:00"),]$b
df1[df1$days == ndays[i],]$results <- df1[df1$days == ndays[i],]$a + factor
}
The problem is, I've got huge dataframes with lot of days and cycling them one by one is slow. Is there a fastest way to do so?
edit: This is the filled results column after the cycle
df1
days time a results
1 01.01.2018 00:00:00 1 2 # results = a + df$b # 01.01.2018 00:00:00
2 01.01.2018 01:00:00 2 3 # results = a + df$b # 01.01.2018 00:00:00
3 01.01.2018 02:00:00 3 4 # results = a + df$b # 01.01.2018 00:00:00
4 02.01.2018 00:00:00 1 6 # results = a + df$b # 02.01.2018 00:00:00
5 02.01.2018 01:00:00 2 7 # results = a + df$b # 02.01.2018 00:00:00
6 02.01.2018 02:00:00 3 8 # results = a + df$b # 02.01.2018 00:00:00
7 03.01.2018 00:00:00 1 11 # results = a + df$b # 03.01.2018 00:00:00
8 03.01.2018 01:00:00 2 12 # results = a + df$b # 03.01.2018 00:00:00
9 03.01.2018 02:00:00 3 13 # results = a + df$b # 03.01.2018 00:00:00

Answer 1

You can do this with a merge instead of a for loop which will be much faster. In the below answer I'm also using data.table, a fast version of data.frames that are very useful when working with large tables.
# install.packages("data.table") # Uncomment if necessary
library(data.table)
df1 <- data.frame(days,time,a) # You don't need to create the result column yet
df2 <- data.frame(days,time,b)
df1 <- data.table(df1)
df2 <- data.table(df2)
# Merge the two tables on the days column
df3 <- merge(df1, df2[time=="00:00:00"], by="days")
# This is your result
answer <- df3[, .(days, time=time.x, a, results=a+b)]
Output:
> answer
days time a results
1: 01.01.2018 00:00:00 1 2
2: 01.01.2018 01:00:00 2 3
3: 01.01.2018 02:00:00 3 4
4: 02.01.2018 00:00:00 1 6
5: 02.01.2018 01:00:00 2 7
6: 02.01.2018 02:00:00 3 8
7: 03.01.2018 00:00:00 1 11
8: 03.01.2018 01:00:00 2 12
9: 03.01.2018 02:00:00 3 13

Answer 2

One solution using dplyr could be as below. The approach of the solution is to:
1) First filter all time other than 00:00:00 from the df2
2) Then inner_join both df1 and df2 on days. This will enable to select value of b from df2 to every matching day in merged dataframe. Finally add a and b to find result.
df1 <- data.frame(days,time,a,results, stringsAsFactors = FALSE)
df2 <- data.frame(days,time,b, stringsAsFactors = FALSE)
library(dplyr)
df2 %>%
filter(time == "00:00:00") %>%
inner_join(df1, by="days") %>%
mutate(time = time.y, results = a+b) %>%
select( days, time, a, b, results)
#Result:
days time a b results
1 01.01.2018 00:00:00 1 1 2
2 01.01.2018 01:00:00 2 1 3
3 01.01.2018 02:00:00 3 1 4
4 02.01.2018 00:00:00 1 5 6
5 02.01.2018 01:00:00 2 5 7
6 02.01.2018 02:00:00 3 5 8
7 03.01.2018 00:00:00 1 10 11
8 03.01.2018 01:00:00 2 10 12
9 03.01.2018 02:00:00 3 10 13

Answer 3

transform(merge(df1,aggregate(b~days,df2,function(x)x[1])),results=a+b)
days time a results b
1 01.01.2018 00:00:00 1 2 1
2 01.01.2018 01:00:00 2 3 1
3 01.01.2018 02:00:00 3 4 1
4 02.01.2018 00:00:00 1 6 5
5 02.01.2018 01:00:00 2 7 5
6 02.01.2018 02:00:00 3 8 5
7 03.01.2018 00:00:00 1 11 10
8 03.01.2018 01:00:00 2 12 10
9 03.01.2018 02:00:00 3 13 10
One thing to note. This assumes time in df2 is arranged chronologically and that the first value for any given day is of time 00:00:00.