我正在从data.table的“ j”参数中调用一个函数,但是得到的结果与直接调用它的结果不同。这似乎与my function called in data.table j not returning expected results
有关get.lower.bound <- function (x) {
rex <-
regexpr (pattern = "((?<lower>[\\-+\\d*\\.,]*)%\\s*<\\s*)?X(\\s<\\s(?<uppper>[\\-+\\d\\.,]*)%)?",
text = x,
perl = TRUE)
lower_bound <-
substring(
text = x,
first = attr(rex, "capture.start")[2],
last = attr(rex, "capture.start")[2] + attr(rex, "capture.length")[2] -1
)
lower_bound
}
dat <- data.table(
A = c('1% < X < 2%', '4% < X', 'X < 8%' ),
B = c('2% < X < 3%', '5% < X < 6%', '8% < X < 9%' ),
C = c('3% < X < 4%', '6% < X < 7%', 'X < 10%' )
)
get.lower.bound(dat[1,'A']) # this returns 1; as I expect
get.lower.bound(dat[2,'A']) # this returns 4; as I expect
get.lower.bound(dat[3,'A']) # this returns a blank string; as I expect
dat[i = 1, j = .(lb1 = get.lower.bound(A))] # this returns a data table of just one cell with 1 in it; as I expect
dat[i = 1:3, j = A] # this returns a character vector with the original strings in it; as I expect
dat[i = 1:3, j = .(lb1 = c(A))] # this returns a data table with the original strings in it; as I expect
dat[i = 1:3, j = .(lb1 = get.lower.bound(A))] # this returns "1% <", "4% <", "X < 8"; but I expect a "1", "4" & ""
为什么最后一行没有给我我所期望的?我需要做些什么才能得到想要的东西?
答案 0 :(得分:2)
我认为这是因为您的函数get.lower.bound
没有被向量化。如果按每一行将其分组然后传递函数,它将给出预期的输出。
library(data.table)
dat[, j = .(lb1 = get.lower.bound(A)), by = 1:nrow(dat)]
# nrow lb1
#1: 1 1
#2: 2 4
#3: 3