detecting measurement of specific id in column in R

Dear all,

if the attachment didn?t arrive, maybe it works now.
I would like to substract the "IN" values (= the air which goes into the
chambers) for "d13C", "ppm_CO2" and "ppm_13CO2"
from the "d13C", "ppm_CO2" and "ppm_13CO2" for every single chamber.
I need to substract the "IN" values which were measured *before* the
chamber.

So the calculation would look like df$d13C [chambers] - df$d13C [IN]
                                                       df$ppm_CO2
[chambers] - df$ppm_13CO2 [IN]
                                                       df$ppm_13CO2
[chambers] - df$ ppm_13CO2 [IN]
--> for chamber 101-111 this should be the first "IN" (No 1)
--> for chamber 1-11 this should be the second "IN" (No 12)
...and so on

I tried sth. like which(abs(date-x) == min(abs(date-x), but it just gives
me the closest "IN" in time and not the "IN" before.

I would appreciate any help!
Thank you so much, Vicci

Am Do., 22. Nov. 2018 um 17:53 Uhr schrieb Rui Barradas <
ruipbarradas at sapo.pt>:

Hello,

No attachment arrived.
R-help allows only a limited number of file types to be attached, if you
file is a text file, change its extension to .txt an try again, please.

Hope this helps,

Rui Barradas

?s 09:33 de 22/11/2018, Romy Rehschuh via R-help escreveu:

Dear all,

I hope this is the right way to ask questions.
I have a problem with R regarding the detection of the measurement of a
specific sample_id (see example file attached). I have to substract the
"IN" values (means the air which goes into the chambers) from the values

of

"d13C", "ppm_CO2" and "ppm_13CO2" for every single chamber (=sample ID).
The "IN" values have to be the ones which were measured* before *the
measurements of the single chambers in time. I measured "IN" once and

then

up to 10 chambers in a row to safe time, then "IN" again, but it can

change.

Therefore, searching for the closest "IN" does not work.

Do you have any suggestions? Would it be possible to write a loop for

this?

I would really much appreciate your help!

Best, Vicci

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