I have two dataframes Vobs and Vest. See the example below:
dput(head(Vobs,20))
structure(list(ID = c("LAM_1", "LAM_2", "LAM_3", "LAM_4", "LAM_5",
"LAM_6", "LAM_7", "AUR_1", "AUR_2", "AUR_3", "AUR_4", "AUR_5",
"AUR_6"), SOS = c(2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24,
26), EOS = c(3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27)), row.names = c(NA,
-13L), class = c("tbl_df", "tbl", "data.frame"))
dput(head(Vest,30))
structure(list(ID = c("LAM", "LAM", "LAM", "LAM", "LAM", "AUR",
"AUR", "AUR", "AUR", "AUR", "AUR", "P0", "P01", "P01", "P02",
"P1", "P2", "P3", "P4", "P13", "P14", "P15", "P17", "P18", "P19",
"P20", "P22", "P23", "P24"), EVI_SOS = c(2, 6, 10, 14, NA, 20,
24, 28, 32, 36, NA, 42, 42, NA, 48, 48, 52, 56, 60, 64, 68, NA,
NA, 72, NA, 78, 82, 86, 90), EVI_EOS = c(3, 7, 11, 15, NA, 21,
25, 29, 33, 37, NA, 43, 43, NA, 49, 49, 53, 57, 61, 65, 69, NA,
NA, 73, NA, 79, 83, 87, 91), NDVI_SOS = c(4, 8, 12, 16, 18, 22,
26, 30, 34, 38, 40, 44, 44, 46, 50, 50, 54, 58, 62, 66, 70, NA,
NA, 74, 76, 80, 84, 88, 92), NDVI_EOS = c(5, 9, 13, 17, 19, 23,
27, 31, 35, 39, 41, 45, 45, 47, 51, 51, 55, 59, 63, 67, 71, NA,
NA, 75, 77, 81, 85, 89, 93)), row.names = c(NA, -29L), class = c("tbl_df",
"tbl", "data.frame"))
I want to do the root mean square error (RMSE) between the two dataframes. As an example, I pretend to do the RMSE between SOS column of Vobs and EVI_SOS column of Vest concerning the LAM ID (which exists in both dataframes).
In other words, I want to subset the data for the ID of interest. In this example, I’m interested in the LAM ID, for Vest and LAM_3 to LAM_7 (that is LAM_3, LAM_4, LAM_5, LAM_6, LAM_7) for Vobs.
I have been using this code:
sqrt(mean( ( (Vobs$SOS)-(Vest$EVI_SOS) )^2 , na.rm = TRUE )) but I missed the ID subset for both columns of the two different dataframes. How can I do the subset using this code?
Any help will be much appreciated.
>Solution :
You could get the subsets of the relevant data as:
diff <- subset(Vobs, ID %in% paste0("LAM_", 3:7))$SOS -
subset(Vest, str_detect(ID, "LAM"))$EVI_SOS
sqrt(mean(diff^2, na.rm=TRUE))
#> [1] 2.44949