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I have a df like this:

id <- c("defoo","ghfoo","abfoo")
abc <- c(.3,.1,.4)
ghi <- c(.4,.2,.2)
abc_dif <- c(.4,.3,.8)
def_dif <- c(.5,.7,.6)
ghi_dif <- c(.2,.1,.9)
df <- data.frame(id,abc,ghi,abc_dif,def_dif,ghi_dif)

I want to look for columns whose names contain the first two characters of the value in the id row and also include "dif," and create a new column containing the corresponding values in those columns for each row.

In this sample data, the new column would be

df$result <- c(.5,.1,.8)

My numerous attempts involved various versions of sapply and apply, like the following attempt to simply get the column index:

df$result <- apply(substr(df[,which(colnames(df)=="id")],1,2),1,function(x) grep(x,colnames(df[which(grepl("dif",colnames(df),fixed=TRUE))]),fixed = TRUE))

This gives the error:

"Error in apply(substr(df[, which(colnames(df) == "id")], 1, 2), 1, function(x) grep(x,  : 
  dim(X) must have a positive length"

What is the best way to go about doing this?

We could create a row/column index to get the values

df$result <- df[4:6][cbind(1:nrow(df), match( substr(df$id, 1, 2),
                substr(names(df)[4:6], 1, 2)))]

#[1] 0.5 0.1 0.8

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You can try a tidyverse


df %>% 
  gather(k,v, -id:-ghi) %>% 
  filter(str_sub(id,1,2) == str_sub(k,1,2)) %>% 
  select(1,result=v) %>% 
  left_join(df, .)
     id abc ghi abc_dif def_dif ghi_dif result
1 defoo 0.3 0.4     0.4     0.5     0.2 0.5
2 ghfoo 0.1 0.2     0.3     0.7     0.1 0.1
3 abfoo 0.4 0.2     0.8     0.6     0.9 0.8

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You could loop through the df$id and then for each one, select the relevant cell in df:

df$result <- sapply(df$id, function(x) df[df$id == x,
                                          grepl(paste0(substring(x,1,2),".*dif"), names(df))])

#[1] 0.5 0.1 0.8

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