Question

我想用ifelse（）条件创建几列。这是我的示例代码：

df <- tibble( 
date = lubridate::today() +0:9,
return= c(1,2.5,2,3,5,6.5,1,9,3,2))

现在，我想添加条件递增的新列（从1到8）。第一列应仅包含“ return”列中的值，且该值均大于1，第二列应仅包含其值，其中均值大于2，依此类推...

我可以使用mutate（）函数计算每一列：

df <- df %>% mutate( `return>1`= ifelse(return > 1, return, NA))
df <- df %>% mutate( `return>2`= ifelse(return > 2, return, NA))
df <- df %>% mutate( `return>3`= ifelse(return > 3, return, NA))
df <- df %>% mutate( `return>4`= ifelse(return > 4, return, NA))
df <- df %>% mutate( `return>5`= ifelse(return > 5, return, NA))
df <- df %>% mutate( `return>6`= ifelse(return > 6, return, NA))
df <- df %>% mutate( `return>7`= ifelse(return > 7, return, NA))
df <- df %>% mutate( `return>8`= ifelse(return > 8, return, NA))


> head(df)
# A tibble: 6 x 10
date       return `return>1` `return>2` `return>3` `return>4` `return>5` `return>6` `return>7` `return>8`
<date>      <dbl>      <dbl>      <dbl>      <dbl>      <dbl>      <dbl>      <dbl>      <dbl>      <dbl>
1 2019-03-08    1         NA         NA         NA         NA         NA         NA           NA         NA
2 2019-03-09    2.5        2.5        2.5       NA         NA         NA         NA           NA         NA
3 2019-03-10    2          2         NA         NA         NA         NA         NA           NA         NA
4 2019-03-11    3          3          3         NA         NA         NA         NA           NA         NA
5 2019-03-12    5          5          5          5          5         NA         NA           NA         NA
6 2019-03-13    6.5        6.5        6.5        6.5        6.5        6.5        6.5         NA         NA

有没有更简单的方法来创建所有这些列并减少所有这些代码？ Maby具有map_function？有没有一种方法可以自动命名新列？

Answer 1

带有lapply

的选项

n <- seq(1, 8)
df[paste0("return > ", n)] <- lapply(n, function(x) 
                    replace(df$return, df$return <= x, NA))


#       date       return `return > 1` `return > 2` `return > 3` .....
#  <date>      <dbl>        <dbl>        <dbl>        <dbl> 
#1 2019-03-08    1           NA           NA           NA  
#2 2019-03-09    2.5          2.5          2.5         NA    
#3 2019-03-10    2            2           NA           NA    
#4 2019-03-11    3            3            3           NA    
#5 2019-03-12    5            5            5            5    
#6 2019-03-13    6.5          6.5          6.5          6.5  
#...

Answer 2

这是一个for循环解决方案：

for(i in 1:8){
  varname =paste0("return>",i)
  df[[varname]] <- with(df, ifelse(return > i, return, NA))
}

Answer 3

使用purrr :: map_df

> bind_cols(df,purrr::map_df(setNames(1:8,paste0('return>',1:8)),
+               function(x) ifelse(df$return > x, df$return, NA)))
# A tibble: 6 x 10
#   date       return `return>1` `return>2` `return>3` `return>4` `return>5` `return>6` `return>7` `return>8`
#   <date>      <dbl>      <dbl>      <dbl>      <dbl>      <dbl>      <dbl>      <dbl>      <dbl>      <dbl>
# 1 2019-03-08    1         NA         NA         NA         NA         NA         NA           NA         NA
# 2 2019-03-09    2.5        2.5        2.5       NA         NA         NA         NA           NA         NA
# 3 2019-03-10    2          2         NA         NA         NA         NA         NA           NA         NA
# 4 2019-03-11    3          3          3         NA         NA         NA         NA           NA         NA
# 5 2019-03-12    5          5          5          5          5         NA         NA           NA         NA
# 6 2019-03-13    6.5        6.5        6.5        6.5        6.5        6.5        6.5         NA         NA

R使用ifelse（）条件对多列进行突变

3 个答案: