根据data.table中的逻辑字符串匹配将值分配给新列

Question

我有一个庞大的学生数据集,其中有荣誉学生的非标准命名惯例.我需要创建/填充一个新列,它将根据单词"Honors"返回Y或N以进行字符串匹配

目前我的数据看起来像这样,有超过200,000名学生

library(data.table)
students<-data.table(Student_ID = c(10001:10005), 
                    Degree= c("Bachelor of Laws", "Honours Degree in Commerce", "Bachelor of Laws (with Honours)", "Bachelor of Nursing with Honours", "Bachelor of Nursing"))

我需要添加第三列,以便在创建新列'Honors'数据表方式后,它将填充如下:

students<-data.table(Student_ID = c(10001:10005), 
                      Degree= c("Bachelor of Laws", "Honours Degree in Commerce","Bachelor of Laws (with Honours)", "Bachelor of Nursing with Honours", "Bachelor of Nursing"), 
                      Honours = c("N","Y", "Y", "Y","N"))

任何帮助将不胜感激.

另外,按数据表的方式我的意思是:

students[,Honours:="N"]

Answer 1

实际上这很简单

students[, Honours := c("N", "Y")[grepl("Honours", Degree, fixed = TRUE) + 1L]]

您需要做的就是使用一些正则表达式实现函数搜索"Honors" grepl,例如(这不是一个真正的表达式,因此您可以使用增强性能fixed = TREU),然后c("N", "Y")根据您的发现进行矢量子集化( a TRUE/ FALSE逻辑向量+ 1L,它将把它转换为一个向量,1,2用于减去来自的值c("N", "Y"))

或者,如果这太难阅读,您可以ifelse改用

students[, Honours := ifelse(grepl("Honours", Degree, fixed = TRUE), "Y", "N")]

当然,如果"荣誉"可以出现在不同的案例变体中,您可以将grepl通话切换到grepl("Honours", Degree, ignore.case = TRUE)

PS

我建议坚持使用逻辑向量,因为之后你可以轻松地操作它

例如

students[, Honours := grepl("Honours", Degree, fixed = TRUE)]

现在,如果你只想选择有"荣誉"的人,你就可以做到

students[(Honours)]
#    Student_ID                           Degree Honours
# 1:      10002       Honours Degree in Commerce    TRUE
# 2:      10003  Bachelor of Laws (with Honours)    TRUE
# 3:      10004 Bachelor of Nursing with Honours    TRUE

或者没有"荣誉"的人

students[!(Honours)]
#    Student_ID              Degree Honours
# 1:      10001    Bachelor of Laws   FALSE
# 2:      10005 Bachelor of Nursing   FALSE