how to “translate” values of a vector into another vector in R

Question

How can I translate the column CLASS, so that I get a new column CLASS2 with "1" = "positive", "-1" = "negative", "0" = "neutral". I know this is a very basic question, and I think one can use ifelse() for this. But I just don't know how to use that function correctly.

DATE <- c("01.01.2000","02.01.2000","03.01.2000","06.01.2000","07.01.2000","09.01.2000","10.01.2000","01.01.2000","02.01.2000","04.01.2000","06.01.2000","07.01.2000","09.01.2000","10.01.2000")
RET <- c(-2.0,1.1,3,1.4,-0.2, 0.6, 0.1, -0.21, -1.2, 0.9, 0.3, -0.1,0.3,-0.12)
CLASS <- c("1","-1","0","1","1","-1","0","1","-1","-1","1","0","0","0")
df <- data.frame(DATE, RET, CLASS)

df

The output should look like this:

DATE <- c("01.01.2000","02.01.2000","03.01.2000","06.01.2000","07.01.2000","09.01.2000","10.01.2000","01.01.2000","02.01.2000","04.01.2000","06.01.2000","07.01.2000","09.01.2000","10.01.2000")
RET <- c(-2.0,1.1,3,1.4,-0.2, 0.6, 0.1, -0.21, -1.2, 0.9, 0.3, -0.1,0.3,-0.12)
CLASS <- c("1","-1","0","1","1","-1","0","1","-1","-1","1","0","0","0")
CLASS2 <- c("positive", "negative", "neutral", "positive", "positive", "negative", "neutral", "positive", "negative", "negative", "positive", "neutral", "neutral", "neutral")
df <- data.frame(DATE, RET, CLASS, CLASS2)

df

#          DATE   RET CLASS   CLASS2
# 1  01.01.2000 -2.00     1 positive
# 2  02.01.2000  1.10    -1 negative
# 3  03.01.2000  3.00     0  neutral
# 4  06.01.2000  1.40     1 positive
# 5  07.01.2000 -0.20     1 positive
# 6  09.01.2000  0.60    -1 negative
# 7  10.01.2000  0.10     0  neutral
# 8  01.01.2000 -0.21     1 positive
# 9  02.01.2000 -1.20    -1 negative
# 10 04.01.2000  0.90    -1 negative
# 11 06.01.2000  0.30     1 positive
# 12 07.01.2000 -0.10     0  neutral
# 13 09.01.2000  0.30     0  neutral
# 14 10.01.2000 -0.12     0  neutral

Thank You!

Answer 1

Here's a simple way to do it using a helper function and sapply :

translate <- function(x) {
  if (x == '1') {
    'positive'
  } else if (x == '-1') {
    'negative'
  } else {
    'neutral'
  }
}
df <- data.frame(DATE, RET, CLASS, CLASS2=sapply(CLASS, translate))

Or you can rewrite translate using ifelse to make it more compact:

translate <- function(x) {
  ifelse(x == '1', 'positive', ifelse(x == '-1', 'negative', 'neutral'))
}

Both of these will produce the output you asked for. But there might be a better way.

...like the one @joran suggested, if CLASS is of factor type (which probably it is):

df$CLASS2 <- c('negative','neutral','positive')[df$CLASS]

As @beginneR pointed out, you don't need a function in my first two proposals. But I like to use functions for better readability.

Answer 2

这是一种一般的方法，使用match可以将其应用于更多级别：

CLASS2 <- c('positive','negative','neutral')[ match(CLASS, c('1','-1','0') ) ]

Answer 3

您甚至不需要定义一个函数并使用sapply ，只需创建一个新列并直接在其上使用ifelse即可：

df$Class2 <- with(df, ifelse(CLASS == '1', 'positive', ifelse(CLASS == '-1', 'negative', 'neutral')))

Answer 4

dplyr::case_when is an option:

df %>%
  mutate(
    CLASS2 = case_when(
      CLASS ==  1 ~ 'positive',
      CLASS ==  0 ~ 'neutral',
      CLASS == -1 ~ 'negative',
      TRUE ~ '?'
    )
  )

Super readable, isn't it?

Although if you had more levels in CLASS , typing all those CLASS == conditions would've been cumbersome. In that case, IMHO, sapply is indeed the best option. Or purrr::map for that matter!

> x <- c(-1, -1, 0, 1, -1) %>% as.character()
> x %>% map(~ list(`-1` = 'negative', `0` = 'neutral', `1` = 'positive')[[.x]]) %>% unlist()
[1] "negative" "negative" "neutral"  "positive" "negative"