Question

我有一个467,000乘8000的data.table。

我想用下划线替换所有冒号和空格。对于data.table中的每一行和每列。

而不是

Assignment 5: Constitutional Law

Assignment_5__Constitutional_Law

我的数据包括日期，数字和字符变量。

library(data.table)
sample<-data.table(STUDENT_ID = c("A1","A2","A3","A4","A5"), Duedate=c("2015-07-29 08:00", "2015-08-05 08:00","2015-08-12 08:00", "2015-08-19 08:00", "2015-08-26 08:00"),Assignment=c(rep("Assignment 1: Physics",5)), GRADE = c(70:74))
sample$Duedate <- as.Date(sample$Duedate,"%Y-%m-%d %H:%M")

Answer 1

找到字符变量，然后通过引用替换它们：

charvars <- sapply(sample,is.character)
sample[,
       (names(sample)[charvars]) := lapply(.SD, gsub, pat="[: ]", rep="_"),
       .SDcols=charvars
      ]
sample

#   STUDENT_ID    Duedate            Assignment GRADE
#1:         A1 2015-07-29 Assignment_1__Physics    70
#2:         A2 2015-08-05 Assignment_1__Physics    71
#3:         A3 2015-08-12 Assignment_1__Physics    72
#4:         A4 2015-08-19 Assignment_1__Physics    73
#5:         A5 2015-08-26 Assignment_1__Physics    74

用下划线r替换data.table中的所有空格和冒号

1 个答案: