Replace value in column with predefined value from another column in R

Question

I have a dataframe with data organised like this:

data <- structure(list(group = c("a", "a", "b", "b"), yr1 = c(1927L, 1953L, 1957L, 1961L), yr2 = c(1934L, 1955L, 1958L, 1965L), val = c("1", "1",  "1", "NA"), col2 = c("x", " -x", " ", " x ")), .Names = c("group", "yr1", "yr2", "val", "col2"), class = "data.frame", row.names = c(NA, -4L))

 group  yr1  yr2 val col2
1     a 1927 1934   1   x   
2     a 1953 1955   1  -x
3     b 1957 1958   1     
4     b 1961 1965  NA   x 

What I'd like to do is to check column val and if there is NA replace it with the value 1 if a certain string in column col2 is true (in this case x).

There might be other strings for the condition which I could maybe sumarise.

values <- c("x", "test", "nts", "kfc") 

The result for the sample data I am aiming for looks like this:

      group  yr1  yr2 val col2
    1     a 1927 1934   1   x   
    2     a 1953 1955   1  -x
    3     b 1957 1958   1     
    4     b 1961 1965   1   x

I have tried to do it with subsetting like this but failed.

data[data$col2 == "x", ][, "val"] <- "1"

Does anyone have an idea how to solve this?

Answer 1

In your case NA is a character 'NA' Also, there seems to be leading/lagging spaces for the col2 .

 data$col2
 #[1] "x"   " -x" " "   " x "

So,

 data$col2=='x'
 #[1]  TRUE FALSE FALSE FALSE

and because NA is character string

  is.na(data$val)
 #[1] FALSE FALSE FALSE FALSE

Because of the above mentioned problems, I would use

 data[with(data, grepl('x', col2) & val=='NA'),'val'] <- 1

EDIT

In cases with leading/lagging spaces, you could remove those for the columns of interest (suppose, group and col2 )

 library(stringr)
 data[c('group', 'col2')] <- lapply(data[c('group', 'col2')], str_trim)

Or

 data[c('group', 'col2')] <- lapply(data[c('group', 'col2')],
                              function(x) gsub('^ +| +$', '', x))

and if you change the 'NA' to NA

  data$val[data$val=='NA'] <- NA

then you could use @Neal Fultz's solution as well

Answer 2

您可以使用&组合条件。

data[data$col2 == "x" & is.na(data$val), "val"] <- 1