如何使用应用于多列数据的条件创建新变量?

问题描述 投票:1回答:2

我是R新手,目前我的代码遇到了一些困难。基本上,我在数据集中有几个变量,其中包含个人经常参与的活动类型的信息(例如,1 =阅读,2 =艺术和手工艺,3 =园艺,等等)。

一些模拟数据:

df = data.frame(ID = c(1001, 1002, 1003, 1004, 1005,1006,1007,1008,1009,1010,1011),
                    orig_1 = c('-7', '2','1','1','NA','2', '3','NA','NA','2', '2'),
                    orig_2 = c('1','1','2','1','3','2', '2', '3','NA','2', '2'),
                    orig_3 = c('-7','3','NA','1','NA','2','NA','1','NA','2', '2'))

基于这些变量,我想创建新的变量,例如,反映一个人是否参与特定的变量(例如0 =否,1 =是)。我做的第一件事就是代码值对应于“不知道”为NA:

#Recode variables
df$orig_1[df$orig_1==-7] <- NA

df$orig_2[df$orig_2==-7] <- NA

df$orig_3[df$orig_3==-7] <- NA

然后我创建了我的新“活动”变量:

# create new activity variable
df$activity_1 <- NA
df$activity_2 <- NA
df$activity_3 <- NA

接下来,我调整了一个函数(由@Sonny友好建议)搜索以下列并返回“1”(对于那些报告参与活动的人)或者“0”:

df$activity_1 <- na.omit(apply(df[, 2:4], 1, function(x) {
  if(any(x %in% c(1))) {
    return(1)
  } else {
    return(0)
  }
}))

df$activity_2 <- na.omit(apply(df[, 2:4], 1, function(x) {
  if(any(x %in% c(2))) {
    return(1)
  } else {
    return(0)
  }
}))

df$activity_3 <- na.omit(apply(df[, 2:4], 1, function(x) {
  if(any(x %in% c(3))) {
    return(1)
  } else {
    return(0)
  }
}))

这部分不起作用,但这里的想法是如果所有原始变量都等于“NA”,则将Na引入新变量:

df$activity_1[df$orig_1==NA & df$orig_2==NA & df$orig_3==NA] <- NA

理想情况下,生成的数据框应如下所示:

     ID orig_1 orig_2 orig_3 activity_1 activity_2 activity_3
1  1001      NA      1    NA          1          0          0
2  1002      2      1     NA          1          1          0
3  1003      1      2     NA          1          1          0
4  1004      1      1      1          1          0          0
5  1005     NA      3     NA          0          0          1
6  1006      2      2      2          0          1          0
7  1007      3      2     NA          0          1          1
8  1008     NA     NA      1          1          0          0
9  1009     NA     NA     NA          NA         NA         NA
10 1010      2      2      2          0          1          0
11 1011      2      2      2          0          1          0

我非常感谢您对改进此代码的任何建议!

r function na recode
2个回答
2
投票

首先,你需要制作真正的NAs。你正在做'NA'这是一个字符串,不同于NA。我们可以解决这个问题:

df[df == "NA"] <- NA

然后我们可以查看apply,其中所有列2:4都是NA并相应地设置activity_*列。

df[apply(df[2:4], 1, function(x) all(is.na(x))), 5:7] <- NA

或矢量化,如@akrun建议:

df[!rowSums(!is.na(df[2:4])), 5:7] <- NA

df
#      ID orig_1 orig_2 orig_3 activity_1 activity_2 activity_3
# 1  1001     -7      1     -7          1          0          0
# 2  1002      2      1      3          1          1          1
# 3  1003      1      2   <NA>          1          1          0
# 4  1004      1      1      1          1          0          0
# 5  1005   <NA>      3   <NA>          0          0          1
# 6  1006      2      2      2          0          1          0
# 7  1007      3      2   <NA>          0          1          1
# 8  1008   <NA>      3      1          1          0          1
# 9  1009   <NA>   <NA>   <NA>         NA         NA         NA
# 10 1010      2      2      2          0          1          0
# 11 1011      2      2      2          0          1          0

数据

df <- structure(list(ID = c(1001, 1002, 1003, 1004, 1005, 1006, 1007, 
1008, 1009, 1010, 1011), orig_1 = structure(c(NA, 3L, 2L, 2L, 
5L, 3L, 4L, 5L, 5L, 3L, 3L), .Label = c("-7", "1", "2", "3", 
"NA"), class = "factor"), orig_2 = structure(c(1L, 1L, 2L, 1L, 
3L, 2L, 2L, 3L, 4L, 2L, 2L), .Label = c("1", "2", "3", "NA"), class = "factor"), 
    orig_3 = structure(c(NA, 4L, 5L, 2L, 5L, 3L, 5L, 2L, 5L, 
    3L, 3L), .Label = c("-7", "1", "2", "3", "NA"), class = "factor"), 
    activity_1 = c(1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0), activity_2 = c(0, 
    1, 1, 0, 0, 1, 1, 0, 0, 1, 1), activity_3 = c(0, 1, 0, 0, 
    1, 0, 1, 1, 0, 0, 0)), .Names = c("ID", "orig_1", "orig_2", 
"orig_3", "activity_1", "activity_2", "activity_3"), row.names = c(NA, 
-11L), class = "data.frame")

1
投票

使用dplyr:

library(dplyr)
#df[df == "NA"] <- NA
df %>%   mutate(activity_1 = case_when( orig_1 == 1 | orig_2 == 1 | orig_3 == 1 ~ 1,
                                 TRUE ~ 0),
         activity_2 = case_when( orig_1 == 2 | orig_2 == 2 | orig_3 == 2 ~ 1,
                                 TRUE ~ 0),
         activity_3 = case_when( orig_1 == 3 | orig_2 == 3 | orig_3 == 3 ~ 1,
                                 TRUE ~ 0)) %>%

  #mutate_at(.vars = c(5:7), list(~ifelse(is.na(orig_1) & is.na(orig_2) &is.na(orig_3), NA, .)))

  mutate_at(.vars = c(5:7), list(~ifelse(orig_1 =="NA" & orig_2  =="NA" & orig_3  =="NA", NA, .)))

要么

df %>% na_if(.,"NA") %>% #na_if replaces specified value (this case "NA") to NA
                mutate(activity_1 = case_when( orig_1 == 1 | orig_2 == 1 | orig_3 == 1 ~ 1,
                                        TRUE ~ 0),
                activity_2 = case_when( orig_1 == 2 | orig_2 == 2 | orig_3 == 2 ~ 1,
                                        TRUE ~ 0),
                activity_3 = case_when( orig_1 == 3 | orig_2 == 3 | orig_3 == 3 ~ 1,
                                        TRUE ~ 0)) %>%
  mutate_at(.vars = c(5:7), list(~ifelse(is.na(orig_1) & is.na(orig_2) &is.na(orig_3), NA, .)))    

          ID orig_1 orig_2 orig_3 activity_1 activity_2 activity_3
1  1001     -7      1     -7          1          0          0
2  1002      2      1      3          1          1          1
3  1003      1      2     NA          1          1          0
4  1004      1      1      1          1          0          0
5  1005     NA      3     NA          0          0          1
6  1006      2      2      2          0          1          0
7  1007      3      2     NA          0          1          1
8  1008     NA      3      1          1          0          1
9  1009     NA     NA     NA         NA         NA         NA
10 1010      2      2      2          0          1          0
11 1011      2      2      2          0          1          0
> 
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