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r - 有条件地更改数据框中分类调查响应列的值

问题描述

试图创建一个将某些类别合并到变量中的对象

background <- NULL

data$y11[data$y11 == "English/Welsh/Scottish/Northern Irish/British"] <-"White"

data$y11[data$y11 == "Gypsy or Irish Traveller"] <-"White"

data$y11[data$y11 == "Any other White background, please describe"] <-"White"

data$y11[data$y11 == "Irish"] <-"White"

data$y11[data$y11 == "Any other Mixed/Multiple ethnic background, please describe"] <-"Mixed"

data$y11[data$y11 == "White and Asian "] <-"Mixed"

data$y11[data$y11 == "White and Black African "] <-"Mixed"

data$y11[data$y11 == "White and Black Caribbean"] <-"Mixed"

data$y11[data$y11 == "Any other Asian background, please describe"] <-"Asian"

data$y11[data$y11 == "Bangladeshi"] <-"Asian"

data$y11[data$y11 == "Chinese"] <-"Asian"

data$y11[data$y11 == "Indian"] <-"Asian"

data$y11[data$y11 == "Pakistani"] <-"Asian"

data$y11[data$y11 == "Arab"] <-"Arab & Other"

data$y11[data$y11 == "Any other ethnic group, please describ"] <-"Arab & Other"

data$y11[data$y11 == "African"] <-"Black"

data$y11[data$y11 == "Any other Black/African/Caribbean background, please describe"] <-"Black"

data$y11[data$y11 == "Caribbean"] <-"Black"

但我保留有关“无效因子水平，NA 生成”的警告消息

请帮忙！

标签： robject

解决方案

stringsAsFactors = FALSE您的主要问题是您在读取数据时没有使用（可能使用read.csv）。因此，您应该将其添加到read.csv通话中。

还有一种更好的方法来做你正在做的事情。一种方法是创建一个从一个类别到另一个类别的“查找”或“翻译”表，然后使用merge基础 R 或left_join“tidyverse”自动为您进行替换，而无需所有这些条件分配。

我们将制作翻译表：

data.frame(
  answer = c(
    "African", "Any other Asian background, please describe",
    "Any other Black/African/Caribbean background, please describe",
    "Any other ethnic group, please describ",
    "Any other Mixed/Multiple ethnic background, please describe",
    "Any other White background, please describe", "Arab", "Bangladeshi",
    "Caribbean", "Chinese", "English/Welsh/Scottish/Northern Irish/British",
    "Gypsy or Irish Traveller", "Indian", "Irish", "Pakistani", "White and Asian ",
    "White and Black African ", "White and Black Caribbean"
  ),
  subst = c(
    "Black", "Asian", "Black", "Arab & Other", "Mixed", "White",
    "Arab & Other", "Asian", "Black", "Asian", "White", "White", "Asian",
    "White", "Asian", "Mixed", "Mixed", "Mixed"
  ),
  stringsAsFactors = FALSE
) -> trans_tbl

现在我们将模拟一些数据（我使用datvsdata作为变量名，因为使用data最终会在某天给你带来痛苦，因为它是一个 R 函数名）：

set.seed(2018-11-30)
data.frame(
  y11 = sample(trans_tbl$answer, 100, replace = TRUE),
  stringsAsFactors = FALSE
) -> dat

str(dat)
## 'data.frame':    100 obs. of  1 variable:
##  $ y11: chr  "Caribbean" "Chinese" "Indian" "Any other Black/African/Caribbean background, please describe" ...

您的数据框有不止一列，但您没有向我们展示，所以我只是用y11. 现在，我们只需调用merge：

dat <- merge(dat, trans_tbl, by.x="y11", by.y="answer", all.x=TRUE)

str(dat)
## 'data.frame':    100 obs. of  2 variables:
##  $ y11  : chr  "African" "African" "African" "African" ...
##  $ subst: chr  "Black" "Black" "Black" "Black" ...

然后，执行一些基本操作以将subst列转换为y11您的代码所做的那样：

dat$y11 <- dat$subst
dat$subst <- NULL

str(dat)
## 'data.frame':    100 obs. of  1 variable:
##  $ y11: chr  "Black" "Black" "Black" "Black" ...

我们也可以dplyr从“tidyverse”中使用：

library(tidyverse)

set.seed(2018-11-30)
data_frame( # this is the `data_frame()` function from dplyr, NOT `data.frame()` from base R
  y11 = sample(trans_tbl$answer, 100, replace = TRUE)
) -> dat

left_join(dat, trans_tbl, by = c("y11"="answer")) %>%
  select(y11 = subst)
## # A tibble: 100 x 1
##    y11         
##    <chr>       
##  1 Black       
##  2 Asian       
##  3 Asian       
##  4 Black       
##  5 Asian       
##  6 Mixed       
##  7 Arab & Other
##  8 Asian       
##  9 Arab & Other
## 10 Asian       
## # ... with 90 more rows

另一种方法是使用因子运算。

我们将使用相同的代码来制作模拟数据框：

possible_answers <- c(
  "African", "Any other Asian background, please describe",
  "Any other Black/African/Caribbean background, please describe",
  "Any other ethnic group, please describ",
  "Any other Mixed/Multiple ethnic background, please describe",
  "Any other White background, please describe", "Arab", "Bangladeshi",
  "Caribbean", "Chinese", "English/Welsh/Scottish/Northern Irish/British",
  "Gypsy or Irish Traveller", "Indian", "Irish", "Pakistani", "White and Asian ",
  "White and Black African ", "White and Black Caribbean"
)

what_they_should_be <- c(
  "Black", "Asian", "Black", "Arab & Other", "Mixed", "White",
  "Arab & Other", "Asian", "Black", "Asian", "White", "White", "Asian",
  "White", "Asian", "Mixed", "Mixed", "Mixed"
)

set.seed(2018-11-30)
data.frame(
  y11 = sample(possible_answers, 100, replace = TRUE)
) -> dat

请注意，我没有使用它stringsAsFactors = FALSE，这使它更像您在 R 会话中已经拥有的。

现在我们可以这样做：

dat$y11 <- as.character(factor(
  x = dat$y11,
  levels = possible_answers,
  labels = what_they_should_be
))

str(dat)
## 'data.frame':    100 obs. of  1 variable:
##  $ y11: chr  "Black" "Asian" "Asian" "Black" ...

我们将翻译后的值作为字符向量而不是因子。

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