如何根据r中的有序向量替换列中的所有值

Question

我正在尝试用有序类别替换数据框列中的所有数值。这是一个虚拟数据框：

df <- data.frame(a = c(1:100), b = sample(c(0,20), size = 100, replace = TRUE), c = c(1:100))

请注意，实际的数据框是使用haven::read_dta() 导入的数据文件。实际数据框可以在 GSS here 上找到。我正在处理 2018 年的文件，并希望将 b 中的所有值（即 0 到 20）替换为一组类别，如下所示：

educ_vec <- c("No formal schooling", "1st grade", "2nd grade", "3rd grade", "4th grade", "5th grade", "6th grade", "7th grade", "8th grade", "9th grade", "10th grade", "11th grade", "12th grade", "1 year of college", "2 years of college", "3 years of college", "4 years of college", "5 years of college", "6 years of college", "7 years of college", "8 years of college")
educ_fac <- factor(educ_vec, ordered = TRUE, levels = educ_vec)

如果我对每个单独的类别都使用mutate和ifelse，则过程太长，并且不会保留educ_fac中的顺序。我尝试了几种方法来一步完成，但没有成功。 一种方法是：

gss_df %>% 
  mutate(educ = fct_recode(educ, 
                           "No formal schooling" = 0, 
                           "1st grade" = 1, 
                           "2nd grade" = 2, 
                           "3rd grade" = 3, 
                           "4th grade" = 4, 
                           "5th grade" = 5, 
                           "6th grade" = 6, 
                           "7th grade" = 7, 
                           "8th grade" = 8, 
                           "9th grade" = 9, 
                           "10th grade" = 10, 
                           "11th grade" = 11, 
                           "12th grade" = 12, 
                           "1 year of college" = 13, 
                           "2 years of college" = 14, 
                           "3 years of college" = 15, 
                           "4 years of college" = 16, 
                           "5 years of college" = 17, 
                           "6 years of college" = 18, 
                           "7 years of college" = 19, 
                           "8 years of college" = 20))

Error: `f` must be a factor (or character vector or numeric vector).

其他两种方法类似，但也没有成功：

gss_df %>% 
  mutate(educ = fct_recode(educ, educ_fac))

Error: `f` must be a factor (or character vector or numeric vector).

gss_df %>% 
  mutate(educ = recode_factor(educ, educ_vec, ordered = TRUE))

Error in UseMethod("recode") : no applicable method for 'recode' applied to an object of class "haven_labelled"

谁能解决这个问题？

Answer 1

由于某些原因，我无法读取 dta 文件，所以下面我模拟数据向您展示我的建议。你从你的 edu_vec 向量开始。

educ_vec <- c("No formal schooling", "1st grade", 
"2nd grade", "3rd grade", "4th grade", "5th grade", 
"6th grade", "7th grade", "8th grade", "9th grade", 
"10th grade", "11th grade", "12th grade", "1 year of college", 
"2 years of college", "3 years of college", "4 years of college", 
"5 years of college", "6 years of college", "7 years of college", 
"8 years of college")

如果你看educ_vec，它已经是你想要的格式了

# this is meant for 0
educ_vec[1]
[1] "No formal schooling"
# this is meant for 20
educ_vec[21]
[1] "8 years of college"

如果你的分数是 i，新的分类值将是 educ_vec[i+1];所以我们可以在下面使用它：

set.seed(100)
gss_df <- data.frame(educ=sample(0:20,30,replace=TRUE))
gss_df %>% 
mutate(new=factor(educ_vec[educ+1],ordered = TRUE, levels = educ_vec))

   educ                new
1     9          9th grade
2     5          5th grade
3    15 3 years of college
4    18 6 years of college
5    13  1 year of college
6    11         11th grade
7     5          5th grade
8     3          3rd grade
9     5          5th grade
10    1          1st grade
11    6          6th grade
12    6          6th grade
13   10         10th grade
14   17 5 years of college
15   11         11th grade
16    2          2nd grade
17   18 6 years of college
18    7          7th grade
19   17 5 years of college
20    1          1st grade
21   18 6 years of college
22    3          3rd grade
23    3          3rd grade
24   19 7 years of college
25   15 3 years of college
26   20 8 years of college
27    6          6th grade
28   15 3 years of college
29   10         10th grade
30   19 7 years of college

是的，如果在数据中找不到某些因素，它会起作用：

gss_df <- data.frame(educ=0:5)%>%
mutate(new=factor(educ_vec[educ+1],ordered = TRUE, levels = educ_vec))

  educ                 new
1    0 No formal schooling
2    1           1st grade
3    2           2nd grade
4    3           3rd grade
5    4           4th grade
6    5           5th grade

您可以看到新列是预期类别的一个因素。

str(gss_df)
'data.frame':   6 obs. of  2 variables:
 $ educ: int  0 1 2 3 4 5
 $ new : Ord.factor w/ 21 levels "No formal schooling"<..: 1 2 3 4 5 6

如果您的分数不在 0-20 之间，例如 -1、-2 或 21,22 等。那么我建议您执行以下操作：

names(educ_vec) = 0:20
gss_df <- data.frame(educ=c(-1,0,20,21))
# you can also use mutate
gss_df$new <- educ_vec[match(gss_df$educ,names(educ_vec))]
gss_df

  educ                 new
1   -1                <NA>
2    0 No formal schooling
3   20  8 years of college
4   21                <NA>

如果在你的 educ_vec 中找不到对应的名字，Match 会返回一个 NA

Answer 2

解决该问题的另一种方法是使用命名向量并稍后进行因子排序。将.dta 文件读取到工作区后，有多种方法可以解决此问题。

set.seed(777)
library(tidyverse)
df <- data.frame(a = c(1:100), b = sample(c(0:20), size = 100, replace = TRUE), c = c(1:100))

# -------------------------------------------------------------------------
head(df)
#   a  b c
# 1 1  0 1
# 2 2 18 2
# 3 3 11 3
# 4 4  9 4
# 5 5 11 5
# 6 6  8 6

# -------------------------------------------------------------------------

# this will be used as name istead
educ_vec <- c("No formal schooling", "1st grade", "2nd grade", "3rd grade", "4th grade", "5th grade", "6th grade", "7th grade", "8th grade", "9th grade", "10th grade", "11th grade", "12th grade", "1 year of college", "2 years of college", "3 years of college", "4 years of college", "5 years of college", "6 years of college", "7 years of college", "8 years of college")

# alues as char from 0 to 20
value_vec <- as.character(seq(21)-1)

# assign educ_vec as names 
names(value_vec) <- educ_vec

# fct_recode b
df$educ <- fct_recode(factor(df$b), !!!value_vec)

# set educ as ordered factor using educ_vec as levels
df$educ <- factor(df$educ, ordered = TRUE, levels = educ_vec)

# -------------------------------------------------------------------------
head(df)
#   a  b c                educ
# 1 1  0 1 No formal schooling
# 2 2 18 2  6 years of college
# 3 3 11 3          11th grade
# 4 4  9 4           9th grade
# 5 5 11 5          11th grade
# 6 6  8 6           8th grade

# -------------------------------------------------------------------------