Create new data frame based on the match of values between a list and a column of data frame

Question

I am trying to create a new data frame using a list of values that match the values of a column in the old data frame. Also for the new data frame I want to preserve the order from the list of values used for match. Here is an example of what I want to achieve:

#A list of values used for matching
time.new <- c(2, 3, 4, 3, 4, 5, 4, 5, 6)
#The old data frame which I would match on the column of **time.old**
old <- data.frame(time.old=1:10, y=rnorm(10))
   time.old        y
          1  0.20320
          2 -0.74696
          3 -0.73716
          4 -0.61959
          5  1.12733
          6  2.58322
          7 -0.08138
          8 -0.10436
          9 -0.13081
         10 -1.20050
#Here is the expected new data frame
       time        y
          2 -0.74696
          3 -0.73716
          4 -0.61959
          3 -0.73716
          4 -0.61959
          5  1.12733
          4 -0.61959
          5  1.12733
          6  2.58322

Answer 1

Try the left_join from dplyr . First, convert time.new to a column of a data frame:

library(tidyverse)
time.new <- c(2, 3, 4, 3, 4, 5, 4, 5, 6)
#The old data frame which I would match on the column of **time.old**
old <- data.frame(time.old=1:10, y=rnorm(10))

time.new <- data.frame(time=time.new) 
new_dataframe <- left_join(time.new, old, by=c("time"="time.old"))

In base R use merge:

merge(x = time.new, y = old, by.x = "time", by.y="time.old", all.x = TRUE)

If you want to preserve the order of time.new you need to add an auxiliary row number column to your data, merge, order on row number and delete the id column:

time.new <- c(2, 3, 4, 3, 4, 5, 4, 5, 6)
old <- data.frame(time.old=1:10, y=rnorm(10))
    
time.new <- data.frame(id = 1:length(time.new), time=time.new)

new_dataframe <- merge(x = time.new, y = old, by.x = "time", by.y="time.old", all.x = TRUE)
new_dataframe <- new_dataframe[order(new_dataframe$id), ]
new_dataframe$id <- NULL