I do it regularly.
Base R:
result <- split( DF[ , 4, drop=FALSE ], DF[ , -4 ] )
Tidyverse:
library(tidyr)
result <- nest( DF, time )
filter( result, "a2"==a & "b1"==b & "c1"==c )[[ "data" ]]
On Thu, 13 Jun 2019, Bert Gunter wrote:
Why? I suspect that there is no reason that you need to do this.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Thu, Jun 13, 2019 at 1:22 PM Tina Chatterjee <tinamunim2...@gmail.com>
wrote:
Hello everyone!
I have the following dataframe(df).
a<-c("a1","a2","a2","a1","a1","a1")
b<-c("b1","b1","b1","b1","b1","b2")
c<-c("c1","c1","c1","c1","c1","c2")
time <- c(runif(6,0,1))
df<-data.frame(a,b,c,time)
df
a b c time
1 a1 b1 c1 0.28781082
2 a2 b1 c1 0.02102591
3 a2 b1 c1 0.72479220
4 a1 b1 c1 0.41947675
5 a1 b1 c1 0.58899855
6 a1 b2 c2 0.82414123
Now, I want to extract the time components corresponding
to the specific combination of the factors. Finally I have made a dataframe
(df_1) with 2 columns one with the time components and the other with the
level combinations.
df[df$a=="a1" & df$b=="b1" & df$c=="c1",]$time
df[df$a=="a2" & df$b=="b1" & df$c=="c1",]$time
df[df$a=="a1" & df$b=="b2" & df$c=="c2",]$time
val <- c(df[df$a=="a1" & df$b=="b1" & df$c=="c1",]$time,df[df$a=="a2" &
df$b=="b1" & df$c=="c1",]$time,df[df$a=="a1" & df$b=="b2" &
df$c=="c2",]$time)
name <- c(rep("a1b1c1",3),rep("a2b1c1",2),"a1b2c2")
df_1 <- data.frame(val,name)
I made it manually. In reality I have a lot of treatment combinations. So,
could you please suggest how can I do this with a loop or any control
sequence?
Thanks and regards.
Tina
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