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I am new to R and I would really appreciate your assistance in this.

I have a dataframe,with 2 levels being 'Y' AND 'N' indicators on 11 variables.enter image description here

I would like to have a new column, which concatenates column names when row value equals to 'Y'

i.e. enter image description here

  • 3
    Images are a really bad way of posting data (or code). Can you post sample data in `dput` format? Please edit **the question** with the output of `dput(df)`. Or, if it is too big with the output of `dput(head(df, 20))`. (`df` stands for the names of each of your datasets.) – Rui Barradas Feb 03 '20 at 12:54
  • I will take the advie into consideration. Thank you so much for assisting Rui Barradas! I appreciate it. – Siyabonga Mbonambi Feb 03 '20 at 13:07

2 Answers2

4

In base R, we can create a row/column index matrix where value is "Y" using which. Using tapply, we can paste the column names for each row.

cols <- paste0('col', 1:9)
mat <- which(df[cols] == 'Y', arr.ind = TRUE)
df$new_col <- as.character(tapply(names(df)[mat[, 2]], mat[, 1], 
               paste, collapse = "_"))
df
#  col1 col2 col3 col4 col5 col6 col7 col8 col9 col10 col11                       new_col
#1    N    Y    N    Y    Y    Y    N    Y    Y     1   624 col2_col4_col5_col6_col8_col9
#2    N    Y    N    Y    Y    Y    N    Y    N     7   548      col2_col4_col5_col6_col8

Using tidyverse we can get the data in long format, filter rows where value is "Y" and for each row paste column values.

library(dplyr)

df %>%
  mutate(row = row_number()) %>%
  tidyr::pivot_longer(cols = -c(col10, col11, row)) %>%
  filter(value == 'Y') %>%
  group_by(row, col10, col11) %>%
  summarise(newcol = toString(name)) %>%
  ungroup() %>%
  select(-row)

data

df <- structure(list(col1 = structure(c(1L, 1L), .Label = "N", class = "factor"), 
col2 = structure(c(1L, 1L), .Label = "Y", class = "factor"), 
col3 = structure(c(1L, 1L), .Label = "N", class = "factor"), 
col4 = structure(c(1L, 1L), .Label = "Y", class = "factor"), 
col5 = structure(c(1L, 1L), .Label = "Y", class = "factor"), 
col6 = structure(c(1L, 1L), .Label = "Y", class = "factor"), 
col7 = structure(c(1L, 1L), .Label = "N", class = "factor"), 
col8 = structure(c(1L, 1L), .Label = "Y", class = "factor"), 
col9 = structure(2:1, .Label = c("N", "Y"), class = "factor"), 
col10 = c(1, 7), col11 = c(623.53, 548.028)), row.names = c(NA, -2L),
class = "data.frame")
Ronak Shah
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4

A simple, base R, way of doing it is

df1$newcol <- apply(df1, 1, function(x){
  paste(names(df1)[x == "Y"], collapse = "_")
})

Test data creation code.

set.seed(1234)
df1 <- t(replicate(2, sample(c("N", "Y"), 10, TRUE)))
df1 <- as.data.frame(df1)
df1 <- cbind(df1, matrix(1:4, 2))
names(df1) <- paste0("col", 1:ncol(df1))
Rui Barradas
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