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r - 在 r 中使用通配符匹配重命名级别

问题描述

是否可以在 R 中使用通配符匹配替换级别？

我有一个名为年经验值的列"0 YEAR, 9 MONTHS"， "1 YEAR, 0 MONTHS", "1 YEAR, 1 MONTHS", "1 YEAR, 10 MONTHS", , "1 YEAR, 9 MONTHS", "10 YEAR, 0 MONTHS", "10 YEAR, 1 MONTHS", "10 YEAR, 10 MONTHS", 同样近 600 个级别；我希望将 all "0 YEAR...as "<1", 1 YEARas "1", more than 5 YEARas存储">5"起来，总共给出 5 个级别。

grep("9 YEAR", data$Service, ignore.case = TRUE, value = TRUE)

试过了mutate，我不能完全缩小每个级别，我希望最终只能获得 5 或 6 个级别。

标签： r

解决方案

首先让我们生成一些随机样本数据

set.seed(2018)
x <- factor(paste(sample(0:10, 10, replace = T), "YEAR,", sample(0:11, 10, replace = T), "MONTHS"))
df <- data.frame(years_of_experience = x)
#   years_of_experience
#1     3 YEAR, 4 MONTHS
#2     5 YEAR, 7 MONTHS
#3    0 YEAR, 11 MONTHS
#4     2 YEAR, 8 MONTHS
#5     5 YEAR, 9 MONTHS
#6     3 YEAR, 7 MONTHS
#7     6 YEAR, 3 MONTHS
#8     1 YEAR, 6 MONTHS
#9    10 YEAR, 8 MONTHS
#10    6 YEAR, 9 MONTHS

然后我们可以使用case_whento binyears_of_experience基于年份

df.new <- df %>%
    mutate(
        yr = as.numeric(gsub(" YEAR.*$", "", x)),
        bucket = case_when(
            yr < 1 ~ "<1",
            yr >= 5 ~ ">=5",
            TRUE ~ as.character(yr)))
df.new
#   years_of_experience yr bucket
#1     3 YEAR, 4 MONTHS  3      3
#2     5 YEAR, 7 MONTHS  5    >=5
#3    0 YEAR, 11 MONTHS  0     <1
#4     2 YEAR, 8 MONTHS  2      2
#5     5 YEAR, 9 MONTHS  5    >=5
#6     3 YEAR, 7 MONTHS  3      3
#7     6 YEAR, 3 MONTHS  6    >=5
#8     1 YEAR, 6 MONTHS  1      1
#9    10 YEAR, 8 MONTHS 10    >=5
#10    6 YEAR, 9 MONTHS  6    >=5

我们可以转换df.new$bucket为factor具有 5 个级别的

df.new %>% mutate(bucket = as.factor(bucket)) %>% pull(bucket)
# [1] 3   >=5 <1  2   >=5 3   >=5 1   >=5 >=5
#Levels: <1 >=5 1 2 3

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