在R中使用通配符匹配重命名级别

Question

是否可以在R中使用通配符匹配来替换级别？

我有一列名为“年经验值"0 YEAR, 9 MONTHS"的列，分别为"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" ，同样接近600个级别； 我希望将所有的"0 YEAR...为"<1" ，将1 YEAR为"1" ，将超过5 YEAR为">5" ，总共给出5个级别。

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

尝试mutate ，我无法精确缩小每个级别，我希望最终只能获得5或6个级别。

Answer 1

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years_of_experience根据case_when对years_of_experience进行分类

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

我们可以将df.new$bucket转换df.new$bucket具有5个级别的factor

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