conditional filling of data.frame - improve code

Heello,

I hadn't posted an answer because my mapply is more complicated that 
the original and much more complicated than Jeff's merge but here it 
is. But if there's a problem with the output of merge, maybe the 
mapply can be of use, only the column expressly named is created.
The result is equal to the original.
I have changed the name exp to exp1.

mydata <- data.frame(sample = c("sample2-2", "sample2-3", "sample1-1", 
"sample1-1", "sample1-1", "sample2-1"))
exp1 <- list(ex1 = c("sample1-1", "sample1-2"), ex2 = c("sample2-1", 
"sample2-2" , "sample2-3"))

for(i in names(exp1)) {
? mydata[mydata[["sample"]] %in% exp1[[i]], "experiment"] <- i
}

# must create the new column beforehand
mydata[["experiment2"]] <- NA_character_
mapply(\(value, name, s){
? i <- which(s %in% value)
? mydata[["experiment2"]][i] <<- name
}, exp1, names(exp1), MoreArgs = list(s = mydata$sample))

mydata
#???? sample experiment experiment2
#1 sample2-2??????? ex2???????? ex2
#2 sample2-3??????? ex2???????? ex2
#3 sample1-1??????? ex1???????? ex1
#4 sample1-1??????? ex1???????? ex1
#5 sample1-1??????? ex1???????? ex1
#6 sample2-1??????? ex2???????? ex2


Hope this helps,

Rui Barradas

?s 08:48 de 11/03/2022, Ivan Calandra escreveu:

In my first trials, I made a typo, which resulted in more columns 
than needed in the output of merge, which is why I needed more 
formatting. But now, it is indeed done all in one line and it is, as 
I said already, nicer anyway!

-- 
Dr. Ivan Calandra
Imaging lab
RGZM - MONREPOS Archaeological Research Centre
Schloss Monrepos
56567 Neuwied, Germany
+49 (0) 2631 9772-243
https://www.researchgate.net/profile/Ivan_Calandra

Le 11/03/2022 ? 08:47, Jeff Newmiller a ?crit?:

What a strange objection. You wouldn't keep the inline definition of 
expts in working code... that would be in a reference data file, and 
the merge is one line.

On March 10, 2022 11:24:27 PM PST, Ivan Calandra 
<ivan.calandra at rgzm.de> wrote:

Thank you Jeff and Tim for your ideas. Indeed merge/join is 
probably the
nicest way. Still, the code becomes much longer because I need more
formatting of the input and output objects than with my ugly for 
loop :)

Cheers,
Ivan

-- 
Dr. Ivan Calandra
Imaging lab
RGZM - MONREPOS Archaeological Research Centre
Schloss Monrepos
56567 Neuwied, Germany
+49 (0) 2631 9772-243
https://www.researchgate.net/profile/Ivan_Calandra

Le 10/03/2022 ? 18:58, Ebert,Timothy Aaron a ?crit?:

You could try some of the "join" commands from dplyr.
https://dplyr.tidyverse.org/reference/mutate-joins.html
https://statisticsglobe.com/r-dplyr-join-inner-left-right-full-semi-anti 



Regards,
Tim
-----Original Message-----
From: R-help <r-help-bounces at r-project.org> On Behalf Of Jeff 
Newmiller
Sent: Thursday, March 10, 2022 11:25 AM
To: r-help at r-project.org; Ivan Calandra <ivan.calandra at rgzm.de>; 
R-help <r-help at r-project.org>
Subject: Re: [R] conditional filling of data.frame - improve code

[External Email]

Use merge.

expts <- read.csv( text =
"expt,sample
ex1,sample1-1
ex1,sample1-2
ex2,sample2-1
ex2,sample2-2
ex2,sample2-3
", header=TRUE, as.is=TRUE )

mydata <- data.frame(sample = c("sample2-2", "sample2-3", 
"sample1-1", "sample1-1", "sample1-1", "sample2-1"))

merge( mydata, expts, by="sample", all.x=TRUE )


On March 10, 2022 7:50:23 AM PST, Ivan Calandra 
<ivan.calandra at rgzm.de> wrote:

Dear useRs,

I would like to improve my ugly (though working) code, but I think I
need a completely different approach and I just can't think out 
of my box!

I have some external information about which sample(s) belong to 
which
experiment. I need to get that manually into R (either typing 
directly
in a script or read a CSV file, but that makes no difference):
exp <- list(ex1 = c("sample1-1", "sample1-2"), ex2 = c("sample2-1",
"sample2-2" , "sample2-3"))

Then I have my data, only with the sample IDs:
mydata <- data.frame(sample = c("sample2-2", "sample2-3", 
"sample1-1",
"sample1-1", "sample1-1", "sample2-1"))

Now I want to add a column to mydata with the experiment ID. The 
best I
could find is that:
for (i in names(exp)) mydata[mydata[["sample"]] %in% exp[[i]],
"experiment"] <- i

In this example, the experiment ID could be extracted from the 
sample
IDs, but this is not the case with my real data so it really is a
matter of matching. Of course I also have other columns with my 
real data.

I'm pretty sure the last line (with the loop) can be improved in 
terms
of readability (speed is not an issue here). I have close to no
constraints on 'exp' (here I chose a list, but anything could 
do), the
only thing that cannot change is the format of 'mydata'.

Thank you in advance!
Ivan

-- 
Sent from my phone. Please excuse my brevity.