问题描述
我有一个人口统计数据集,其中包括一个家庭的年龄。这是通过调查收集的,允许参与者拒绝提供年龄。
结果是一个数据集,该数据集每行有一个家庭(每个家庭都有一个家庭ID码),并且各个家庭特征(例如列中的年龄)也是如此。拒绝的响应编码为“ R”,您可以使用以下代码重新创建示例:
df <- list(Household_ID = c("1A","1B","1C","1D","1E"),AGE1 = c("25","47","39","50","R"),AGE2 = c("66","23","71","R","16"),AGE3 = c("28","17","80"),AGE4 = c("81","22","48","59","R"))
df <- as_tibble(df)
> df
# A tibble: 5 x 5
Household_ID AGE1 AGE2 AGE3 AGE4
<chr> <chr> <chr> <chr> <chr>
1 1A 25 66 28 81
2 1B 47 23 17 22
3 1C 39 71 R 48
4 1D 50 R R 59
5 1E R 16 80 R
出于我们的意图和目的,我们将“ R”重新编码为“ -9”,以便随后可以将AGE列的格式转换为整数,并进行分析。我们通常在其他软件中执行此操作,而我的目标是在R中复制此过程。
我设法使用以下代码做到了这一点:
df <- df %>% mutate(AGE1 = case_when(AGE1 == "R" ~ "-9",TRUE ~ as.character(AGE1)))
df <- df %>% mutate(AGE2 = case_when(AGE2 == "R" ~ "-9",TRUE ~ as.character(AGE2)))
df <- df %>% mutate(AGE3 = case_when(AGE3 == "R" ~ "-9",TRUE ~ as.character(AGE3)))
df <- df %>% mutate(AGE4 = case_when(AGE4 == "R" ~ "-9",TRUE ~ as.character(AGE4)))
鉴于这感觉很笨拙,我尝试使用mutate_if等找到解决方案,但读到它们已被cross()取代。因此,我尝试使用cross()复制此操作:
df <- df %>%
mutate(across(AGE1:AEG4),~ (case_when(. == "R" ~ "-9")))
但是出现以下错误:
Error: Problem with `mutate()` input `..2`.
x Input `..2` must be a vector,not a `formula` object.
i Input `..2` is `~(case_when(. == "R" ~ "-9"))`.
曾经为此苦苦挣扎,现在搜索了一段时间,但无法弄清我的缺失。非常感谢您提供一些有关如何使它正常工作的意见,谢谢。
编辑:已解决!
df <- df %>%
mutate(across(AGE1:AGE4,~ (case_when(.x == "R" ~ "-9",TRUE ~ as.character(.x)))))
解决方法
为什么不简单?
df[,2:5][df[,2:5] == 'R'] <- '-9'
# A tibble: 5 x 5
Household_ID AGE1 AGE2 AGE3 AGE4
<chr> <chr> <chr> <chr> <chr>
1 1A 25 66 28 81
2 1B 47 23 17 22
3 1C 39 71 -9 48
4 1D 50 -9 -9 59
5 1E -9 16 80 -9
,
或者也许这个与亲爱的@TarJae的解释没有太大区别:
library(dplyr)
library(stringr)
df %>%
mutate(across(AGE1:AGE4,~ str_replace(.,"R","-9")),across(AGE1:AGE4,as.integer))
# A tibble: 5 x 5
Household_ID AGE1 AGE2 AGE3 AGE4
<chr> <int> <int> <int> <int>
1 1A 25 66 28 81
2 1B 47 23 17 22
3 1C 39 71 -9 48
4 1D 50 -9 -9 59
5 1E -9 16 80 -9
数据:
df <- list(Household_ID = c("1A","1B","1C","1D","1E"),AGE1 = c("25","47","39","50","R"),AGE2 = c("66","23","71","16"),AGE3 = c("28","17","80"),AGE4 = c("81","22","48","59","R"))
df <- as_tibble(df)
,
您可以将 across
与 replace
一起使用。
- 用
as_tibble()
列出来 - 用 -9 替换 R
- AGE 的整数类
df %>%
as_tibble() %>%
mutate(across(everything(),~replace(.,. == "R","-9"))) %>%
type.convert(as.is=TRUE)
输出:
Household_ID AGE1 AGE2 AGE3 AGE4
<chr> <int> <int> <int> <int>
1 1A 25 66 28 81
2 1B 47 23 17 22
3 1C 39 71 -9 48
4 1D 50 -9 -9 59
5 1E -9 16 80 -9