Follow

Keep Up to Date with the Most Important News

By pressing the Subscribe button, you confirm that you have read and are agreeing to our Privacy Policy and Terms of Use
Contact

R: Problem with specifying column using data.table v1.14.8

I have a problem with specifying column using data.table v1.14.8. Below are the examples of what doesn’t work and what works. Using the original dataset, which I want to make work, doesn’t work.

What doesn’t work (original dataset)

  vars.row <- c("category", "subcategory", "variable")
  
  nrows <- length(vars.row)
  
  
  d <-
    structure(list(category = c("", "\\\\[-6pt]Scheduling", "\\\\[-6pt]Scheduling", 
                              "\\\\[-6pt]Scheduling", "\\\\[-6pt]Scheduling", "\\\\[-6pt]Statistics", 
                              "\\\\[-6pt]Statistics", "\\\\[-6pt]Statistics", "\\\\[-6pt]Statistics", 
                              "\\\\[-6pt]Statistics", "\\\\[-6pt]Statistics"), 
                   subcategory = c("", 
                               "New Procedures", "New Procedures", "New Partners", "New Partners", 
                               "", "", "", "", "", ""), 
                   variable = c("(Intercept)", "New procedures", 
                                 "(New procedures)$^2$", "New partners", "(New partners)$^2$", 
                                 "Adj. R$^2$", "AIC", "AICc", "BIC", "Deviance", "N")
                                 ), 
              row.names = c(NA, -11L), 
              class = c("data.table", "data.frame")
              )   # .internal.selfref = <pointer: 0x1410254e0>

  for(i in 1:nrows){
    var <- vars.row[i]
    for(j in nrow(d):2){
      if(d[j, ..var] == d[j-1, ..var]) d[j, ..var] <- ''
    }
  }
  
  

What works (fake dataset)

vars.row <- c("category", "subcategory")
  
  nrows <- length(vars.row)
  
  d <-
    data.frame(
      category = c(rep('A', 5), rep('B', 5)),
      subcategory = c(rep('a', 3), rep('b', 3), rep('c', 4))
    )
  
  for(i in 1:nrows){
    var <- vars.row[i]
    for(j in nrow(d):2){
      if(d[j, ..var] == d[j-1, ..var]) d[j, ..var] <- ''
    }
  }
  

>Solution :

MEDevel.com: Open-source for Healthcare and Education

Collecting and validating open-source software for healthcare, education, enterprise, development, medical imaging, medical records, and digital pathology.

Visit Medevel

Not sure about the bug, but we can do what I think you’re trying to do with:

for (var in vars.row) d[get(var) == shift(get(var)), c(var) := ""]
d
#                 category    subcategory             variable
#                   <char>         <char>               <char>
#  1:                                              (Intercept)
#  2: \\\\[-6pt]Scheduling New Procedures       New procedures
#  3:                                     (New procedures)$^2$
#  4:                        New Partners         New partners
#  5:                                       (New partners)$^2$
#  6: \\\\[-6pt]Statistics                          Adj. R$^2$
#  7:                                                      AIC
#  8:                                                     AICc
#  9:                                                      BIC
# 10:                                                 Deviance
# 11:                                                        N

or without the for loop:

d[, c(vars.row) := lapply(.SD, function(z) { z[z == shift(z)] <- ""; z; }), 
  .SDcols = vars.row]
Add a comment

Leave a Reply

Keep Up to Date with the Most Important News

By pressing the Subscribe button, you confirm that you have read and are agreeing to our Privacy Policy and Terms of Use

Discover more from Dev solutions

Subscribe now to keep reading and get access to the full archive.

Continue reading