[R] assigning suitability index value

Jeff Newmiller jdnewm|| @end|ng |rom dcn@d@v|@@c@@u@
Wed Aug 11 08:16:08 CEST 2021


Piecewise linear interpolation is implemented in the ?approx function. It does not agree exactly with your Output, I don't know if there is something else you are accounting for it if your Output is in error.

temp$index <- approx( table2$temp, table2$Index, temp$temp )$y

BTW your code was usable but messed up... please set your email program to send plain text email so your formatting does not mess with your code.


On August 10, 2021 10:30:57 PM PDT, Marna Wagley <marna.wagley using gmail.com> wrote:
>Hi R Users,
>I have two tables, one is temperature data (temp) and another table is a
>suitability index. I wanted to assign the suitability index value in the
>temperature data (temp) based on Table 2 (or graph, which is a suitability
>curve), but I could not figure it out.
>Are there any suggestions for me how I can assign the suitability index
>value in table1 (temp) based on the suitability graph? I have a very big
>data set but showing only a few data to illustrate the problem.
>
>temp<-structure(list(X = c(1468285.96, 1468476.96, 1468479.96, 1468482.96,
>
>1468485.96, 1468467.96, 1468470.96, 1468473.96, 1468476.96, 1468479.96,
>
>1468482.96, 1468485.96, 1468458.96, 1468461.96, 1468464.96, 1468467.96,
>
>1468470.96, 1468473.96, 1468476.96), Y = c(415099.27, 415096.27,
>
>415096.27, 415096.27, 415096.27, 415093.27, 415093.27, 415093.27,
>
>415093.27, 415093.27, 415093.27, 415093.27, 415090.27, 415090.27,
>
>415090.27, 415090.27, 415090.27, 415090.27, 415090.27), temp = c(1.959473,
>
>15.092773, 15.128174, 14.368896, 9.892578, 15.720215, 15.767822,
>
>15.26001, 14.642334, 14.6521, 13.916016, 10.3479, 16.052246,
>
>16.094971, 15.167236, 15.455322, 15.472412, 24.741211, 14.755859
>
>)), class = "data.frame", row.names = c(NA, -19L))
>
>
>print(temp)
>
>
>table2<-structure(list(temp = c(0L, 10L, 15L, 17L, 25L, 30L), Index = c(0,
>
>0.3, 1, 1, 0.5, 0)), class = "data.frame", row.names = c(NA,
>
>-6L))
>
>print(table2)
>
>
>ggplot(data=table2, aes(x=temp, y=Index)) +
>
>  geom_path()+
>
>  geom_point()
>
>
>
># now I would like to assign the index value of table 2 into table 1
>(temp), and I was looking for the following table as an output. The index
>value in the output I put manually.
>
>
>Output<-structure(list(X = c(1468285.96, 1468476.96, 1468479.96, 1468482.96,
> 1468485.96, 1468467.96, 1468470.96, 1468473.96, 1468476.96, 1468479.96,
>
>1468482.96, 1468485.96, 1468458.96, 1468461.96, 1468464.96, 1468467.96,
>
>1468470.96, 1468473.96, 1468476.96), Y = c(415099.27, 415096.27,
>
>415096.27, 415096.27, 415096.27, 415093.27, 415093.27, 415093.27,
>
>415093.27, 415093.27, 415093.27, 415093.27, 415090.27, 415090.27,
>
>415090.27, 415090.27, 415090.27, 415090.27, 415090.27), temp = c(1.959473,
>
>0.092773, 15.128174, 14.368896, 9.892578, 15.720215, 15.767822,
>
>15.26001, 14.642334, 14.6521, 13.916016, 10.3479, 16.052246,
>
>16.094971, 15.167236, 15.455322, 15.472412, 24.741211, 14.755859
>
>), index = c(0.012, 0.001, 1, 0.9, 0.31, 1, 1, 1, 0.91, 0.921,
>
>0.824, 0.254, 1, 1, 1, 1, 1, 0.652, 0.93)), class = "data.frame", row.names
>= c(NA,
>
>-19L))
>
>
>print(Output)
>
>
>Thank you very much for your help.
>
>MW
>
>	[[alternative HTML version deleted]]
>
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-- 
Sent from my phone. Please excuse my brevity.



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