在分组数据中用用户定义的函数替换值答案

【问题标题】：Replacing a value with user defined function in grouped data在分组数据中用用户定义的函数替换值
【发布时间】：2017-11-06 22:56:05
【问题描述】：

我遇到了一个问题，如果满足条件，则将一个值替换为另一个值。我使用自己的函数data_manip，我可以在需要时分配或添加任何其他条件。

但是，当我尝试使用此 data_manip 函数时，它会使用分配的值更改该组中的所有值。但是该特定组中的其他值不符合此条件。

这是我尝试过的，

df <- data.frame(percent = c(0.6, 0.7,1, 0.5,0.5,1,0.4,0.6,1), 
                 type = rep(c("good", "bad","ugly"),each=3), smoke=rep(c('Visky','Wine','Wine'),3),
                 sex=rep(c('male','male','female'),3))

> df
  percent type smoke    sex
1     0.6 good Visky   male
2     0.7 good  Wine   male
3     1.0 good  Wine female
4     0.5  bad Visky   male
5     0.5  bad  Wine   male
6     1.0  bad  Wine female
7     0.4 ugly Visky   male
8     0.6 ugly  Wine   male
9     1.0 ugly  Wine female


data_manip <- function(x,gr){
  if(grepl('goo|ug',gr)&&x<1){
    x[x==0.6] <- 1
  }
    else
  x
}

df%>%
  group_by(type)%>%
  mutate(percent_new=data_manip(percent,type))

给予

# A tibble: 9 x 5
# Groups:   type [3]
  percent   type  smoke    sex percent_new
    <dbl> <fctr> <fctr> <fctr>       <dbl>
1     0.6   good  Visky   male         1.0
2     0.7   good   Wine   male         1.0
3     1.0   good   Wine female         1.0
4     0.5    bad  Visky   male         0.5
5     0.5    bad   Wine   male         0.5
6     1.0    bad   Wine female         1.0
7     0.4   ugly  Visky   male         1.0
8     0.6   ugly   Wine   male         1.0
9     1.0   ugly   Wine female         1.0

如果条件不适合它们，我想保留原始 percent 值。

预期输出

 # A tibble: 9 x 5
    # Groups:   type [3]
      percent   type  smoke    sex percent_new
        <dbl> <fctr> <fctr> <fctr>       <dbl>
    1     0.6   good  Visky   male         1.0
    2     0.7   good   Wine   male         0.7
    3     1.0   good   Wine female         1.0
    4     0.5    bad  Visky   male         0.5
    5     0.5    bad   Wine   male         0.5
    6     1.0    bad   Wine female         1.0
    7     0.4   ugly  Visky   male         0.4
    8     0.6   ugly   Wine   male         1.0
    9     1.0   ugly   Wine female         1.0

【问题讨论】：

在mutate()中尝试ifelse()，ifelse()是一个向量化函数。
或在 mutate() 中使用 sapply() 任何返回矢量化输出的内容都可以在 mutate 中用于替换或创建新值。
@fhlgood 得到与sapply 相同的结果；(

标签： r if-statement dplyr

【解决方案1】：

您当前的data_manip 函数似乎没有向量化，因为它使用if (cond) { ... } else { ... }，它通常只检查一个值并且可能默认为向量的第一个元素。您的函数的矢量化版本如下所示：

data_manip <- function(x,gr){
    ifelse(grepl('goo|ug', gr) & x == 0.6, 1, x)
}

并给出了预期的结果：

> df%>%
+     group_by(type)%>%
+     mutate(percent_new=data_manip(percent,type))
# A tibble: 9 x 5
# Groups:   type [3]
  percent   type  smoke    sex percent_new
    <dbl> <fctr> <fctr> <fctr>       <dbl>
1     0.6   good  Visky   male         1.0
2     0.7   good   Wine   male         0.7
3     1.0   good   Wine female         1.0
4     0.5    bad  Visky   male         0.5
5     0.5    bad   Wine   male         0.5
6     1.0    bad   Wine female         1.0
7     0.4   ugly  Visky   male         0.4
8     0.6   ugly   Wine   male         1.0
9     1.0   ugly   Wine female         1.0

使用ifelse 进行矢量化条件检查。

【讨论】：

非常感谢您的解决方案和解释！谢谢！

【解决方案2】：

这看起来像是 case_when 可以解决的问题。

试试这个：

library(tidyverse)

df %>% 
  mutate(new_percentage = case_when(type == "good" & percent == 0.6 ~ 1,
                                    type == "ugly" & percent == 0.6 ~ 1,
                                    TRUE ~ as.double(.$percent)))

这给出了：

# A tibble: 9 x 5
  percent   type  smoke    sex new_percentage
    <dbl> <fctr> <fctr> <fctr>          <dbl>
1     0.6   good  Visky   male            1.0
2     0.7   good   Wine   male            0.7
3     1.0   good   Wine female            1.0
4     0.5    bad  Visky   male            0.5
5     0.5    bad   Wine   male            0.5
6     1.0    bad   Wine female            1.0
7     0.4   ugly  Visky   male            0.4
8     0.6   ugly   Wine   male            1.0
9     1.0   ugly   Wine female            1.0

【讨论】：