试图创建一个将某些类别合并到一个变量中的对象
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产生”的警告消息
请帮助！
答案 0 :(得分：1)
这意味着您的变量是因素。您可以通过以下两种方法之一解决此问题：
使用以下命令将所有因子更改为字符：
data$y11 <- as.character(data$y11)
使用以下命令将所需的新级别添加到现有因子级别：
levels(data$y11) <- c(levels(data$y11), "White", "Black", ...)
让我知道这是否没有道理
此外，以防万一您不熟悉R，也不需要像这样分散所有行。您可以将各个种族分组在一起，就像这样：
data$y11[data$y11 %in% c("English/Welsh/Scottish/Northern Irish/British",
"Gypsy or Irish Traveller",
"Any other White background, please describe",
"Irish")] <-"White"
data$y11[data$y11 %in% c("Any other Mixed/Multiple ethnic background, please describe",
"White and Asian ",
"White and Black African ",
"White and Black Caribbean")] <-"Mixed"
data$y11[data$y11 %in% c("Any other Asian background, please describe",
"Bangladeshi",
"Chinese",
"Indian",
"Pakistani")] <-"Asian"
data$y11[data$y11 %in% c("Arab",
"Any other ethnic group, please describ")] <-"Arab & Other"
data$y11[data$y11 %in% c("African",
"Any other Black/African/Caribbean background, please describe",
"Caribbean"] <-"Black"
或者还有很多其他方法，例如使用|（或）运算符，因此您不必单独写出每一行。
答案 1 :(得分：1)
您的主要问题是，在读入数据时您未使用stringsAsFactors = FALSE（可能使用read.csv）。因此，您应该将其添加到read.csv调用中。
还有一种更好的方式来做您正在做的事情。一种方法是从一个类别创建到另一个类别的“查找”或“翻译”表，然后使用基数R中的merge或“ tidyverse”中的left_join自动为您替换完成所有这些条件分配。
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dat和data作为变量名，因为使用data最终会导致某天让您感到痛苦，因为它是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，但是您没有向我们展示它，所以我只用y11制作了一个单列数据框。现在，我们只调用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列变成y11，就像您的代码一样：
dat$y11 <- dat$subst
dat$subst <- NULL
str(dat)
## 'data.frame': 100 obs. of 1 variable:
## $ y11: chr "Black" "Black" "Black" "Black" ...
我们还可以使用“ tidyverse”中的dplyr：
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$answer, 100, replace = TRUE)
) -> dat
left_join(dat, trans_tbl, by = c("y11"="answer")) %>%
select(y11 = subst)
## # A tibble: 100 x 1
## y11
## <chr>
## 1 Black
## 2 Asian
## 3 Asian
## 4 Black
## 5 Asian
## 6 Mixed
## 7 Arab & Other
## 8 Asian
## 9 Arab & Other
## 10 Asian
## # ... with 90 more rows
另一种方法是使用因子运算。
我们将使用相同的代码来制作模拟数据框：
possible_answers <- c(
"African", "Any other Asian background, please describe",
"Any other Black/African/Caribbean background, please describe",
"Any other ethnic group, please describ",
"Any other Mixed/Multiple ethnic background, please describe",
"Any other White background, please describe", "Arab", "Bangladeshi",
"Caribbean", "Chinese", "English/Welsh/Scottish/Northern Irish/British",
"Gypsy or Irish Traveller", "Indian", "Irish", "Pakistani", "White and Asian ",
"White and Black African ", "White and Black Caribbean"
)
what_they_should_be <- c(
"Black", "Asian", "Black", "Arab & Other", "Mixed", "White",
"Arab & Other", "Asian", "Black", "Asian", "White", "White", "Asian",
"White", "Asian", "Mixed", "Mixed", "Mixed"
)
set.seed(2018-11-30)
data.frame(
y11 = sample(possible_answers, 100, replace = TRUE)
) -> dat
请注意，我没有为此使用stringsAsFactors = FALSE，这使其更像您在R会话中已经拥有的东西。
现在我们可以做到：
dat$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" ...
我们将转换后的值作为字符向量而不是因子来获取。