根据列表应用条件并根据结果 r 创建新列答案

【问题标题】：Applying condition based on a list and create a new column based on the outcome r根据列表应用条件并根据结果 r 创建新列
【发布时间】：2020-09-05 01:27:57
【问题描述】：

我有一个清单如下：

c1 <-("apple", "tree", "husband")

还有这个数据：

df <-data.frame(
  ID = c("b","b","b","a","a","c"),
  col = c("husband", "apple", "juice", "happy", "husband", "white"),
)

我想要这个输出：

df <-data.frame(
  ID = c("b","b","b","a","a","c"),
  col = c("husband", "apple", "juice", "happy", "husband", "white"),
  c1 = c("1","1","0","0","1","0")
)

通过应用列表 (c1) 作为条件并且不使用

mutate(c1= ifelse(col==happy | col==tree | col==husband,1,0))

谢谢

【问题讨论】：

标签： r data-cleaning

【解决方案1】：

您可以使用%in% 来检查col 中的c1 值

transform(df, c1 = as.integer(col %in% c1))
#Even shorter
#transform(df, c1 = +(col %in% c1))

#  ID     col c1
#1  b husband  1
#2  b   apple  1
#3  b   juice  0
#4  a   happy  0
#5  a husband  1
#6  c   white  0

在逻辑值上使用as.integer 比使用ifelse 更快：

transform(df, c1 = ifelse(col %in% c1, 1, 0))

【讨论】：

【解决方案2】：

你可以通过factor来玩花样，例如，

within(df, out <- +!is.na(factor(col,levels = c1)))

或通过%in%

within(df, out <- +(col %in%c1))

或通过match

within(df,out <- 1-is.na(match(col,c1)))

这样

  ID     col out
1  b husband   1
2  b   apple   1
3  b   juice   0
4  a   happy   0
5  a husband   1
6  c   white   0

【讨论】：

【解决方案3】：

您还可以使用grepl() 来检查c1 中的任何值并直接分配给新变量：

#Data 1
c1 <- c("apple", "tree", "husband")
#Data 2
df <-data.frame(
    ID = c("b","b","b","a","a","c"),
    col = c("husband", "apple", "juice", "happy", "husband", "white"),stringsAsFactors = F)
#Match and create new variable
df$NewVar <- as.numeric(grepl(paste0(c1,collapse = '|'),df$col))

输出：

  ID     col NewVar
1  b husband      1
2  b   apple      1
3  b   juice      0
4  a   happy      0
5  a husband      1
6  c   white      0

【讨论】：

【解决方案4】：

case_when 的选项

library(dplyr)
df %>%
     mutate(c1 = case_when(col %in% c1, 1, 0))

或者另一种选择是

df %>%
    mutate(c1 = +(col %in% c1))

【讨论】：