Populate a new column if a value is found in any column

问题

I want to check every row in data frame. I need to check all columns in that row to see if it contains a 1, if it does I want to populate another column that summarizes if any of the columns had a 1 or not.

So far I have tried using grepl to return a logical index by matching the '1', and then with ifelse, change the logical vector to 'yes' or 'no'

dat1$imputed_data <- ifelse(grepl("1", imputed_columns), "yes", "no")

I have also tried

for(i in nrow(imputed_columns)){
   if (any(imputed_columns[i,])==1)
   {
       dat1$imputed_data[i] <- "yes"
   }else{
       dat1$imputed_data[i] <- "no"
   }
}

Both of my attemps have not worked, I think the problem with both might be the way I specify the columns to do the check in.

have:
A B C 
0 0 0
0 1 1
1 0 0
0 0 0

want:
A B C   imputed_data
0 0 0   no
0 1 1   yes
1 0 0   yes
0 0 0   no

Please help me figure out how to make this work.Thank you in advance.

回答1:

One of many solutions:

a1 = data.frame(A = c(0,0,1,0), B = c(0,1,0,0), C = c(0,1,0,0))

a1$imputed = apply(a1, 1, function(x) ifelse(any(x == 1), 'yes', 'no'))

  A B C imputed
1 0 0 0      no
2 0 1 1     yes
3 1 0 0     yes
4 0 0 0      no

回答2:

Using:

dat$imputed <- c('no','yes')[1 + (rowSums(dat == 1) > 0)]

gives:

> dat
  A B C imputed
1 0 0 0      no
2 0 1 1     yes
3 1 0 0     yes
4 0 0 0      no

What this does:

rowSums(dat == 1) > 0 creates a logical vector indicating whether a row contains a 1
Adding 1 to that gives an integer vector
Which in turn can be used to create the appropriate vector of yes and no values.

Used data:

dat <- structure(list(A = c(0L, 0L, 1L, 0L), B = c(0L, 1L, 0L, 0L), C = c(0L, 1L, 0L, 0L)),
                 .Names = c("A", "B", "C"), class = "data.frame", row.names = c(NA, -4L))

回答3:

A dplyr way would be

a1 %>% mutate(imputed_data = ifelse(rowSums(. == 1) != 0, "yes", "no"))

来源：https://stackoverflow.com/questions/46456265/populate-a-new-column-if-a-value-is-found-in-any-column

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