简体   繁体   中英

Add additional column based on the content of others (in R)

I have a data frame with 3 qc's. I would like to create 4th column where the result will depend on all 3 qc's. Condition: at least 2 have to be "bad" to display in qc4 bad otherwise good.

  ID qc1  qc2  qc3
1 AV bad  good bad
2 AX bad  bad  bad
3 AS good good good
4 AW good bad  bad

result

  ID qc1  qc2  qc3  q4
1 AV bad  good bad  bad
2 AX bad  bad  bad  bad
3 AS good good good good
4 AW good bad  bad  bad

We can construct a logical vector with rowSums and use ifelse to create the 'bad', 'good' based on the value from the logical vector

nm1 <- grep("^qc\\d+$", names(df), value = TRUE)
df$q4 <- ifelse(rowSums(df[nm1] == "bad") >=2, "bad", "good")
df
#  ID  qc1  qc2  qc3   q4
#1 AV  bad good  bad  bad
#2 AX  bad  bad  bad  bad
#3 AS good good good good
#4 AW good  bad  bad  bad

Or another option is with rowwise/c_across from dplyr

library(dplyr)
df %>% 
   rowwise %>%
   mutate(q4 = c("good", "bad")[1+(sum(c_across(starts_with('qc')) ==
       'bad') >= 2)]) %>%
   ungroup

data

df <- structure(list(ID = c("AV", "AX", "AS", "AW"), qc1 = c("bad", 
"bad", "good", "good"), qc2 = c("good", "bad", "good", "bad"), 
    qc3 = c("bad", "bad", "good", "bad")), class = "data.frame",
    row.names = c("1", 
"2", "3", "4"))

Another base R option using rowSums

df$qc4 <- c("good", "bad")[1 + (rowSums(df == "bad") >= 2)]

such that

> df
  ID  qc1  qc2  qc3  qc4
1 AV  bad good  bad  bad
2 AX  bad  bad  bad  bad
3 AS good good good good
4 AW good  bad  bad  bad

Data

> dput(df)
structure(list(ID = c("AV", "AX", "AS", "AW"), qc1 = c("bad", 
"bad", "good", "good"), qc2 = c("good", "bad", "good", "bad"),
    qc3 = c("bad", "bad", "good", "bad"), qc4 = c("bad", "bad",
    "good", "bad")), row.names = c("1", "2", "3", "4"), class = "data.frame")

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM