Question

如何翻译CLASS列，以便我得到一个新列CLASS2，其中“1”=“正”，“ - 1”=“负”，“0”=“中性”。我知道这是一个非常基本的问题，我认为可以使用ifelse()。但我只是不知道如何正确使用该功能。

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

输出应如下所示：

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

谢谢！

Answer 1

这是使用帮助函数sapply执行此操作的简单方法：

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

或者您可以使用translate重写ifelse以使其更紧凑：

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

这两个都会产生您要求的输出。但可能有更好的方法。

...就像@joran建议的那样，如果CLASS是因子类型（可能是它）：

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

正如@beginneR所指出的那样，您在前两个提案中并不需要一个功能。但我喜欢使用函数来提高可读性。

Answer 2

以下是一种适用于match的更多关卡的一般方法：

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使用ifelse：

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

Answer 4

dplyr::case_when是一个选项：

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

超级可读，不是吗？

尽管如果您在CLASS中有更多的关卡，那么键入所有这些CLASS ==条件会很麻烦。在这种情况下，恕我直言，sapply确实是最佳选择。或者purrr::map！

> 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"

如何＆＃34;翻译＆＃34;矢量值在R中的另一个矢量中

4 个答案: