Luis -
I *think* that
mapply(list,LHS,RHS,SIMPLIFY=FALSE)
will give you what you want, but without a reproducible
example it's hard to tell.
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
On Fri, 11 Jun 2010, Luis Felipe Parra wrote:
Hello, I have two lists with the same number of elements
tail(LHS)
[[1]]
[1] "antecedentes.factor_riesgo=17" "antecedentes.estado=1"
"antecedentes.medio=4" "tarjetas_flagrancia.adquiriente2="
[[2]]
[1] "antecedentes.riesgo=1" "antecedentes.estado=1"
"antecedentes.medio=4" "tarjetas_flagrancia.adquiriente2="
[[3]]
[1] "resultado_investig.pto_comp=N" "antecedentes.riesgo=1"
"antecedentes.factor_riesgo=17" "antecedentes.medio=4"
[[4]]
[1] "resultado_investig.pto_comp=N" "antecedentes.riesgo=1"
"antecedentes.factor_riesgo=17" "tarjetas_flagrancia.adquiriente2="
[[5]]
[1] "resultado_investig.pto_comp=N" "antecedentes.factor_riesgo=17"
"antecedentes.medio=4" "tarjetas_flagrancia.adquiriente2="
[[6]]
[1] "resultado_investig.pto_comp=N" "antecedentes.riesgo=1"
"antecedentes.medio=4" "tarjetas_flagrancia.adquiriente2="
tail(RHS)
[[1]]
[1] "antecedentes.riesgo=1"
[[2]]
[1] "antecedentes.factor_riesgo=17"
[[3]]
[1] "tarjetas_flagrancia.adquiriente2="
[[4]]
[1] "antecedentes.medio=4"
[[5]]
[1] "antecedentes.riesgo=1"
[[6]]
[1] "antecedentes.factor_riesgo=17"
I would like to create a new list from this two which would have in every
list entry one entry with the corresponding elements from LHS and another
entry with the corresponding element from RHS (LHS doesn't always have three
4 elements per entry). Do you know how can I do this? Thank you
Felipe Parra
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