R function 或循环重复选择满足条件的行，另存为单独的 object 和重命名列标题

Question

I have 16 large datasets of landcover variables around routes. Example dataset "Trial1":

RtNo     TYPE    CA      PLAND   NP      PD      LPI     TE 
2001     cls_11     996.57  6.4297  22  0.1419  6.3055  31080
2010     cls_11     56.34   0.3654  23  0.1492  0.1669  15480
18003    cls_11     141.12  0.9899  37  0.2596  0.1503  38700
18014    cls_11     797.58  5.3499  47  0.3153  1.3969  98310
2001     cls_21     1514.97 9.7744  592 3.8195  0.8443  761670
2010     cls_21     638.55  4.1414  95  0.6161  0.7489  463260
18003    cls_21     904.68  6.3463  612 4.2931  0.8769  549780
18014    cls_21     1189.89 7.9814  759 5.0911  0.4123  769650
2001     cls_22     732.33  4.7249  653 4.2131  0.7212  377430
2010     cls_22     32.31   0.2096  168 1.0896  0.0198  31470
18003    cls_22     275.85  1.9351  781 5.4787  0.0423  237390
18014    cls_22     469.44  3.1488  104 6.7345  0.1014  377580

I want to first select rows that meet a condition, for example, all rows in column "TYPE" that is cls_21. I know the following code does this work:

Trial21 <-subset(Trial1, TYPE==" cls_21 ")

(yes the invisible space before and after the categorical variable caused me a considerable headache). And there are several other ways of doing this as shown in [https://stackoverflow.com/questions/5391124/select-rows-of-a-matrix-that-meet-a-condition]

I get the following output (sorry this one has extra columns, but shouldn't affect my question):

    RtNo    TYPE    CA     PLAND     NP  PD    LPI     TE       ED      LSI
2   18003   cls_21  904.68  6.3463  612 4.2931  0.8769  549780  38.5668 46.1194
18  18014   cls_21  1189.89 7.9814  759 5.0911  0.4123  769650  51.6255 56.2522
34  2001    cls_21  1514.97 9.7744  592 3.8195  0.8443  761670  49.1418 49.3462
50  2010    cls_21  638.55  4.1414  95  0.6161  0.7489  463260  30.0457 46.0118
62  2020    cls_21  625.5   4.1165  180 1.1846  0.5064  384840  25.3268 38.6407
85  2021    cls_21  503.55  2.7926  214 1.1868  0.1178  348330  19.3175 38.9267

I want to rename the columns in this subset so they uniquely identify the class by adding "L21" at the back of existing column names, and I can do this using

library(data.table)
setnames(Trial21, old = c('CA', 'PLAND', 'NP', 'PD', 'LPI', 'TE', 'ED', 'LSI'), 
         new = c('CAL21', 'PLANDL21', 'NPL21', 'PDL21', 'LPIL21', 'TEL21', 'EDL21', 'LSIL21'))

I want help to develop a function or a loop that automates this process so I don't have to spend days repeating the same codes for 15 different classes and 16 datasets (240 times). Also, decrease the risk of errors. I may have to do the same for additional datasets. Any help to speed the process will be greatly appreciated.

Answer 1

You could do:

a <- split(df, df$TYPE)

b <- sapply(names(a), function(x)setNames(a[[x]],
              paste0(names(a[[x]]), sub(".*_", 'L', x))), simplify = FALSE)

Answer 2

Here is a start that should work for your example:

library(dplyr)

myfilter <- function(data, number) {
  data %>%
    filter(TYPE == sprintf(" cls_%s ") %>%
    rename_with(\(x) sprintf("%s%s", x, suffix), !1:2)
}

myfilter(example_data, 21)

Given a list of numbers (here: 21 to 31) you could then automatically use them to filter a single dataframe:

multifilter <- function(data) {
  purrr::map(21:31, \(i) myfilter(data, i))
}

multifilter(example_data)

Finally, given a list of dataframes, you can automatically apply the filters to them:

purrr::map(list_of_dataframes, multifilter)

Answer 3

You can use ls to get the variable names of the datasets, and manipulate them as you wish inside a loop and with get function, then create new datasets with assign .

sets = grep("Trial", ls(), value=TRUE) #Assuming every dataset has "Trial" in the name

for(i in sets){
  classes = unique(get(i)$TYPE)
  
  for(j in classes){
    number = gsub("(.+)([0-9]{2})( )", "\\2", j)#this might be an overly complicated way of getting just the number, you can look for better options if you want
    assign(paste0("Trial", number),
           subset(Trial1, TYPE==j) %>% rename_with(function(x){paste0(x, number)}))}}