how to count strings across various columns-CodePudding

I have a data like this

    df<- structure(list(Accession = c("Washington", "DC", "NY", "TM", 
"VA", "UTSAW", "UTDFS", "FLOR", "HYTAS", "HUTT", "ITA", "BELI"
), FirstCon = c("UNS", "UNS", "UNS", "UNS", "UNS", "", "UNS", 
"UNS", "UNS", "UNS", "UNS", "UNS"), SeconCon = c("", "", "", 
"UNS", "", "", "", "", "", "UTN", "UTN", "UTN"), Upcond = c("", 
"", "", "UNS", "", "UNS", "", "", "", "", "", "UTBTS")), class = "data.frame", row.names = c(NA, 
-12L))

it looks like the following

   Accession FirstCon SeconCon Upcond
1  Washington      UNS                
2          DC      UNS                
3          NY      UNS                
4          TM      UNS      UNS    UNS
5          VA      UNS                
6       UTSAW                      UNS
7       UTDFS      UNS                
8        FLOR      UNS                
9       HYTAS      UNS                
10       HUTT      UNS      UTN       
11        ITA      UNS      UTN       
12       BELI      UNS      UTN  UTBTS

I want to have an output like this

FirstCon SeconCon Upcond   FirstConSeconCon  FirstConUpcond  SeconConUpcond
 11        4        3           4                  2              2
FirstConSeconConUpcond
2

basically it shows how many of the strings are repeated in each column and a combination of all

for instance

Accession FirstCon SeconCon 
    4          TM      **UNS      UNS**    
    10       HUTT      **UNS      UTN**       
    11        ITA      **UNS      UTN**       
    12       BELI      **UNS      UTN**

FirstConSeconCon is 4 because FirstCon has the four UNS and the SeconCon also has the string on the same row (they dont need to be similar but as long as they have string on the same corresponding row it counts)

CodePudding user response：

As i understand here is a solution :

df<- structure(list(Accession = c("Washington", "DC", "NY", "TM", "VA", "UTSAW",
                                  "UTDFS", "FLOR", "HYTAS", "HUTT", "ITA", "BELI"),
                    FirstCon = c("UNS", "UNS", "UNS", "UNS", "UNS", "", "UNS","UNS",
                                 "UNS", "UNS", "UNS", "UNS"),
                    SeconCon = c("", "", "", "UNS", "", "", "", "", "", "UTN",
                                 "UTN", "UTN"),
                    Upcond = c("","", "", "UNS", "", "UNS", "", "", "", "", "", "UTBTS")),
               class = "data.frame", row.names = c(NA,-12L))
# this function returns the existing of strings in the rows
occurence <- function(df){
  oc <- 0L
  for(i in 1:nrow(df)){
    if(all(nchar(df[i , ]) > 0)){
      oc <- oc   1L
    }
  }
  oc
}

res <- c()
nm <- c()

for( i in 1:(length(names(df[-1])))){

  com <- combn(1:ncol(df[-1]) , i)
  
  for(c in 1:ncol(com)){
    dfsub <- df[,com[,c]   1L , drop = F]
    coln <- names(dfsub)
    oc <- occurence(dfsub)
    nm <- append(nm , do.call(paste0 , as.list(coln)))
    res <- append(res , oc)
    names(res) <- nm
  }
}

res
#>               FirstCon               SeconCon                 Upcond 
#>                     11                      4                      3 
#>       FirstConSeconCon         FirstConUpcond         SeconConUpcond 
#>                      4                      2                      2 
#> FirstConSeconConUpcond 
#>                      2

^{Created on 2022-06-03 by the reprex package (v2.0.1)}

CodePudding user response：

Here's one way with base R but without loops:

df[df == ""] <- NA

cols <- names(df)[-1]

combins <- do.call("c", lapply(2:length(cols), function(x) combn(cols, x, FUN = list)))
combin_vals <- sapply(1:length(combins), function(x) sum(rowSums(is.na(df[,combins[[x]]])) == 0))
names(combin_vals) <-  sapply(combins, function(x) paste(x, collapse = ""))

c(colSums(!is.na(df[,cols])), combin_vals)

          FirstCon               SeconCon                 Upcond       FirstConSeconCon 
                11                      4                      3                      4 
    FirstConUpcond         SeconConUpcond FirstConSeconConUpcond 
                 2                      2                      2