Turning two columns of data in R into a series of lists as long as the columns-CodePudding

I have two columns of data referring to international abbreviations for countries (their ISO 2-Code and ISO 3-Code). I am going to use this as a "dictionary" to substitute the 2Codes for the 3Codes in another table using the gsubfn() function. I had to create the dictionary list by entering manually as follows:

d_list<-list('ABW'='AW','AFG'='AF','AGO'='AO','AIA'='AI','ALA'='AX','ALB'='AL','AND'='AD',
          'ARE'='AE','ARG'='AR','ARM'='AM','ASM'='AS','ATG'='AG','AUS'='AU','AUT'='AT',
          'AZE'='AZ','BAH'='AZ','BDI'='BI','BEL'='BE','BEN'='BJ','BFA'='BF','BGD'='BD',
          'BGR'='BG','BHR'='BH','BIH'='BA','BLM'='BL','BLR'='BY','BLZ'='BZ','BMU'='BM',
          'BOL'='BO','BRA'='BR','BRB'='BB','BRN'='BN')

d_list is a list of 212 elements ... one element per "=".

Is there a way to create d_list from a table or multiple tables that have the 2Code and 3Code in columns? ... would make this process so much faster and less error prone.

CodePudding user response：

First, some test data:

df <- data.frame(code3 = c("ABW","AFG"), code2 = c("AW", "AF"))

If each of the 3-Code values have to be unique, then split it:

split(df$code2, df$code3)
#$ABW
#[1] "AW"
#
#$AFG
#[1] "AF"

If not, make a list and set the names manually:

as.list(setNames(df$code2, df$code3))
#$ABW
#[1] "AW"
#
#$AFG
#[1] "AF"

CodePudding user response：

A loop between columns and lists should be able to accomplish this. This code (or similar) could likely be written into a function for convenience.

#Written in R version 4.2.1
#create column 1
col1 = names(d_list)
col1
 [1] "ABW" "AFG" "AGO" "AIA" "ALA" "ALB" "AND" "ARE" "ARG" "ARM" "ASM"
[12] "ATG" "AUS" "AUT" "AZE" "BAH" "BDI" "BEL" "BEN" "BFA" "BGD" "BGR"
[23] "BHR" "BIH" "BLM" "BLR" "BLZ" "BMU" "BOL" "BRA" "BRB" "BRN"
 
#create column 2
col2 = NULL
for(i in 1:length(d_list)){
col2[i] = d_list[[i]]
}
col2
 [1] "AW" "AF" "AO" "AI" "AX" "AL" "AD" "AE" "AR" "AM" "AS" "AG" "AU"
[14] "AT" "AZ" "AZ" "BI" "BE" "BJ" "BF" "BD" "BG" "BH" "BA" "BL" "BY"
[27] "BZ" "BM" "BO" "BR" "BB" "BN"

#reconstruct d_list from columns, called d_list2 here
d_list2 = list()
for(i in 1:length(d_list)){
d_list2[[i]] = col2[i]
names(d_list2)[i] = col1[i]
}
all.equal(d_list,d_list2)
[1] TRUE