Question

是否有现有的便利功能可以过滤data.table中的行，给定搜索模式，查看所有列？

names(DT)

[1] "Name" "LongName" "SomeOtherCol" "NumericCol" "bar" "foo"

像这样的东西，适用于任意数量的列：

DT[Name %like% pattern | LongName %like% pattern | SomeOtherCol %like% pattern | bar %like% pattern | foo %like% pattern]

Answer 1

一种方法是循环遍历列，应用正则表达式，然后返回逻辑数据表。您可以使用rowSums来获取行。

dt <- data.table(a=c("Aa1","bb","1c"),b=c("A1","a1","1C"), c=letters[1:3])
# "a1" is the pattern to search for
ldt <- dt[, lapply(.SD, function(x) grepl("a1", x, perl=TRUE))] 
dt[rowSums(ldt)>0]
#      a  b c
# 1: Aa1 A1 a
# 2:  bb a1 b

Answer 2

我不认为这是最好的方法。但它的目的是：

> dt <- data.table(a=c("a1","bb","1c"),b=c("A1","BB","1C"))
> dt
    a  b
1: a1 A1
2: bb BB
3: 1c 1C

> combined <- apply(dt,1,function(r) paste(r,collapse="/%/"))
> combined
[1] "a1/%/A1" "bb/%/BB" "1c/%/1C"

> grepped <- grepl("[a-z][0-9]",apply(dt,1,function(r) paste(r,collapse="/")))
> grepped
[1]  TRUE FALSE FALSE

> dt[grepped,]
    a  b
1: a1 A1

＆＃34; /％/＆＃34;必须是与模式无关的东西，并可靠地分隔列。

当然，这些步骤可以组合成一个表达式。

Answer 3

解决方案3：

首先构建附加所有列的逻辑grep expression。然后eval一次性表达整个表达式：

dt <- data.table(a=c("a1","bb","1c"),b=c("A1","BB","1C"))

search.data.table <- function(x, pattern) {
  nms <- names(x)
  string <- eval(expression(paste0("grepl('",
                                   pattern, 
                                   "', ",
                                   nms,",
                                   ignore.case=TRUE, perl=FALSE)",
                                   collapse = " | ")))
  x[eval(as.call(parse(text=string))[[1]])]
}

search.data.table(dt, "a1")
#      a  b c
# 1: Aa1 A1 a
# 2:  bb a1 b

<强>基准

# functions

Raffael <- function(x, pattern) {
# unfortunately this implementation throws an error so I can't run the benchmark test. 
# Any help?
  combined <- apply(x,1,function(r) paste(r,collapse="/%/"))
  grepped <- grepl(pattern,apply(x,1,function(r) paste(r,collapse="/")))
  x[grepped,]
}

Arun <- function(x, pattern) {
  ldt <- x[, lapply(.SD, function(x) grepl(pattern, x, perl=TRUE, ignore.case=TRUE))] 
  x[rowSums(ldt)>0]
}

DanielKrizian <- function(x, pattern) {
  nms <- names(x)
  string <- eval(expression(paste0("grepl('", pattern, "', ",nms,", ignore.case=TRUE,      perl=FALSE)",collapse = " | ")))
  x[eval(as.call(parse(text=string))[[1]])]
}

# generate 1000 x 1000 benchmark data.table

require(data.table)
expr <- quote(paste0(sample(c(LETTERS,tolower(LETTERS),0:9),12, replace=T)
                 ,collapse=""))
set.seed(1)
BIGISH <- data.table(matrix(replicate(1000*1000,eval(expr)),nrow = 1000))
object.size(BIGISH) # 68520912 bytes

# test

benchmark(
  DK <- DanielKrizian(BIGISH,"qx"),
  A <- Arun(BIGISH,"qx"),
  replications=100)

<强>结果

                               test replications elapsed relative user.self sys.self user.child sys.child
2           A <- Arun(BIGISH, "qx")          100   57.72    1.000     51.95     0.44         NA        NA
1 DK <- DanielKrizian(BIGISH, "qx")          100   59.28    1.027     53.72     0.50         NA        NA

identical(DK,A)
[1] TRUE

`data.table`全局搜索 - 在`any`列中过滤给定模式匹配的行

3 个答案: