我正在尝试将数据框列中的所有数值替换为有序类别。下面是一个虚拟数据框：
df <- data.frame(a = c(1:100), b = sample(c(0,20), size = 100, replace = TRUE), c = c(1:100))
请注意，有问题的实际数据框是使用haven::read_dta()导入的 dta 文件。实际数据框可以在GSS上找到 这里.我正在处理 2018 年的文件，并希望用一组类别替换 b 中的所有值，即 0 到 20，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。我尝试了几种方法一步到位，但没有成功。 一种方法是这样的：
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).
其他两种方式相似，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"
任何人都可以对此给出解决方案吗？
由于某些原因，我无法读取dta文件，因此下面我模拟数据以向您展示我的建议。您从educ_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，它已经是您想要的格式
# 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，它就会起作用：
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
您可以看到新列是具有预期类别的因素。
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 中，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 Match 在您的educ_vec中找不到相应的名称，它将返回 NA
解决此问题的另一种方法是使用命名向量并在以后进行因子排序。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
# -------------------------------------------------------------------------