如何使用变量列名进行非equi连接

时间:2016-09-18 06:01:08

标签: r data.table

如何在data.table 1.9.7中使用变量列名进行非equi连接?例如,

已知列名:

library(data.table)
dt <- data.table(x=round(rnorm(10)), y=rnorm(10))
binDT <- data.table(LB=c(-Inf, -1, 0, .2, .7, 1.5, 3), RB=c(-1, 0, .2, .7, 1.5, 3, Inf))
dt[binDT, on=.(x>=LB, x<RB)]
       x       y  x.1
 1: -Inf  2.2669 -1.0
 2: -1.0 -0.5453  0.0
 3: -1.0  0.5125  0.0
 4:  0.0  1.4151  0.2
 5:  0.0 -0.1440  0.2
 6:  0.0 -1.1802  0.2
 7:  0.0  0.3338  0.2
 8:  0.0 -1.8220  0.2
 9:  0.2      NA  0.7
10:  0.7  0.3155  1.5
11:  0.7 -0.6284  1.5
12:  1.5      NA  3.0
13:  3.0      NA  Inf

变量列名:

colName <- "x"
dt[binDT, on=.(get(colName)>=LB, get(colName)<RB)]  # Error
dt[binDT, on=eval(parse(text="list(x>=LB, x<RB)"))]  # Error

2 个答案:

答案 0 :(得分:7)

@Shape答案很好,但有更简单的方法来实现它。 on参数可以采用字符向量,因此可以将粘贴预期的列和运算符。

colName="x"
on=sprintf(c("%s>=LB","%s<RB"), colName)
print(on)
#[1] "x>=LB" "x<RB"
dt[binDT, on=on]
#       x           y  x.1
# 1: -Inf          NA -1.0
# 2: -1.0  0.48127355  0.0
# 3:  0.0  0.11779604  0.2
# 4:  0.0 -0.97891522  0.2
# 5:  0.0 -0.05969859  0.2
# 6:  0.0 -0.05625401  0.2
# 7:  0.2          NA  0.7
# 8:  0.7 -0.84438216  1.5
# 9:  0.7  0.80151913  1.5
#10:  1.5 -0.11013456  3.0
#11:  1.5  0.82139242  3.0
#12:  3.0 -1.24386831  Inf

答案 1 :(得分:4)

使用替换为虚拟变量名称并将其提供给命名列表:

res <- substitute(dt[binDT,on=.(A>=LB,B<RB)],
                  list(A = as.name(colName), 
                       B = as.name(colName)))

# the values get replaced in the call
> res
dt[binDT, on = .(x >= LB, x < RB)]

eval(res)

       x           y  x.1
 1: -Inf          NA -1.0
 2: -1.0  0.69668714  0.0
 3: -1.0 -0.03824623  0.0
 4:  0.0  0.91269554  0.2
 5:  0.0  0.42322463  0.2
 6:  0.0 -0.22891670  0.2
 7:  0.0  0.61413004  0.2
 8:  0.2          NA  0.7
 9:  0.7 -1.47526635  1.5
10:  0.7 -1.12899562  1.5
11:  0.7  1.05462948  1.5
12:  1.5 -0.04467894  3.0
13:  3.0          NA  Inf