Whatever makes you happy.
> df1 <-
+ structure(list(x = structure(c(1L, 2L, 2L, 3L), .Label = c("AA",
+ "BB", "CC"), class = "factor"), y = 1:4), .Names = c("x", "y"
+ ), row.names = c(NA, -4L), class = "data.frame")
> mult <- c("AA"=2,"BB"=5,"CC"=1,"DD"=2)
> df1$y * mult[df1$x]
AA BB BB CC
2 10 15 4
> df1$y * mult[as.character(df1$x)]
AA BB BB CC
2 10 15 4
>
>
> df2 <- data.frame(x = c("AA","AA","BB","BB","BB","CC","DD","DD"), y = 1:8)
> df2$y * mult[df2$x]
AA AA BB BB BB CC DD DD
2 4 15 20 25 6 14 16
> df2$y * mult[as.character(df2$x)]
AA AA BB BB BB CC DD DD
2 4 15 20 25 6 14 16
On Fri, Feb 24, 2012 at 12:52 PM, Arnaud Gaboury
<arnaud.gabo...@a2ct2.com> wrote:
> In fact I need to use William tip: Use mult[as.character(df$x)] instead of
> mult[df$x].
>
>
> Let's try again with a shorter df as example:
>
> The rule: if AA, then multiply y by 2, if BB multiply y by 5, if CC do
> nothing, if DD multiply by 2.
>
>
> Let's say day 1 I have df1:
>
> df1 <-
> structure(list(x = structure(c(1L, 2L, 2L, 3L), .Label = c("AA",
> "BB", "CC"), class = "factor"), y = 1:4), .Names = c("x", "y"
> ), row.names = c(NA, -4L), class = "data.frame")
>
>> df1
> x y
> 1 AA 1
> 2 BB 2
> 3 BB 3
> 4 CC 4
>
>>mult <- c("AA"=2,"BB"=5,"CC"=1,"DD"=2)
>>df1$y <- df1$y * mult[as.character(df1$x)]
>> df1
> x y
> 1 AA 2
> 2 BB 10
> 3 BB 15
> 4 CC 4
>
> WORKING
>
> Now day 2 with df2:
>
>>df2 <- data.frame(x = c("AA","AA","BB","BB","BB","CC","DD","DD"), y = 1:8)
>>df2$y <- df2$y * mult[as.character(df2$x)]
>> df2
> x y
> 1 AA 2
> 2 AA 4
> 3 BB 15
> 4 BB 20
> 5 BB 25
> 6 CC 6
> 7 DD 14
> 8 DD 16
>
> WORKING
>
>
> Ty both of you and have a good weekend.
>
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