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
spector at 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
>
> [[alternative HTML version deleted]]
>
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