确定列值是否在data.table中是唯一的

时间:2016-02-22 02:07:00

标签: r data.table

我使用data.table来存储数据。我想弄清楚每行中的某些列是否唯一。我想在data.table中添加一个列,如果存在重复值,则将保留值“Duplicated Values”,如果没有重复值,则为NA。我想要检查重复的列的名称存储在字符向量中。例如,我创建了data.table:

tmpdt<-data.table(a=c(1,2,3,4,5), b=c(2,2,3,4,5), c=c(4,2,2,4,4), d=c(3,3,1,4,5))
> tmpdt
   a b c d
1: 1 2 4 3
2: 2 2 2 3
3: 3 3 2 1
4: 4 4 4 4
5: 5 5 4 5

我有另一个变量,指示我需要检查哪些列的重复项。重要的是,我能够将列名存储在字符向量中,而不需要“知道”它们(因为它们将作为参数传递给函数)。

dupcheckcols<-c("a", "c", "d")

我希望输出为:

> tmpdt
   a b c d     Dups
1: 1 2 4 3     <NA>
2: 2 2 2 3 Has Dups
3: 3 3 2 1     <NA>
4: 4 4 4 4 Has Dups
5: 5 5 4 5 Has Dups

如果我使用的是data.frame,这很容易。我可以简单地使用:

tmpdt<-data.frame(a=c(1,2,3,4,5), b=c(2,2,3,4,5), c=c(4,2,2,4,4), d=c(3,3,1,4,5))
tmpdt$Dups<-NA
tmpdt$Dups[apply(tmpdt[,dupcheckcols], 1, function(x) {return(sum(duplicated(x))>0)})]<-"Has Dups"
> tmpdt
  a b c d     Dups
1 1 2 4 3     <NA>
2 2 2 2 3 Has Dups
3 3 3 2 1     <NA>
4 4 4 4 4 Has Dups
5 5 5 4 5 Has Dups

但我无法弄清楚如何使用data.table完成相同的任务。非常感谢任何帮助。

5 个答案:

答案 0 :(得分:5)

我确信还有其他方法

tmpdt[, dups := tmpdt[, dupcheckcols, with=FALSE][, apply(.SD, 1, function(x){sum(duplicated(x))>0})] ]
#   a b c d  dups
#1: 1 2 4 3 FALSE
#2: 2 2 2 3  TRUE
#3: 3 3 2 1 FALSE
#4: 4 4 4 4  TRUE
#5: 5 5 4 5  TRUE

更复杂但更快(在计算方面)的方法是在i中构建过滤条件,然后通过引用在j中更新

expr <- paste(apply(t(combn(dupcheckcols,2)), 1, FUN=function(x){ paste0(x, collapse="==") }), collapse = "|")
# [1] "a==c|a==d|c==d"

expr <- parse(text=expr)
tmpdt[ eval(expr), dups := TRUE ]
#   a b c d dups
#1: 1 2 4 3   NA
#2: 2 2 2 3 TRUE
#3: 3 3 2 1   NA
#4: 4 4 4 4 TRUE
#5: 5 5 4 5 TRUE

我对速度优势感兴趣,所以我对这两个以及Ananda的解决方案进行了基准测试:

library(microbenchmark)

tmpdt<-data.table(a=c(1,2,3,4,5), b=c(2,2,3,4,5), c=c(4,2,2,4,4), d=c(3,3,1,4,5))
t1 <- tmpdt
t2 <- tmpdt
t3 <- tmpdt

expr <- paste(apply(t(combn(dupcheckcols,2)), 1, FUN=function(x){ paste0(x, collapse="==") }), collapse = "|")
expr <- parse(text=expr)

microbenchmark(
#Ananda's solution
t1[, dups := any(duplicated(unlist(.SD))), by = 1:nrow(tmpdt), .SDcols = dupcheckcols],

t2[, dups := t2[, dupcheckcols, with=FALSE][, apply(.SD, 1, function(x){sum(duplicated(x))>0})] ],

t3[ eval(expr), dups := TRUE ]
)
 #     min        lq      mean   median        uq      max neval cld
 # 531.416  552.5760  577.0345  565.182  573.2015 1761.863   100  b 
 #1277.569 1333.2615 1389.5857 1358.021 1387.9860 2694.951   100   c
 # 265.872  283.3525  293.9362  292.487  301.1640  520.436   100 a  

答案 1 :(得分:3)

你应该可以这样做:

tmpdt[, dups := any(duplicated(unlist(.SD, use.names = FALSE))), 
      by = 1:nrow(tmpdt), .SDcols = dupcheckcols]
tmpdt
#    a b c d  dups
# 1: 1 2 4 3 FALSE
# 2: 2 2 2 3  TRUE
# 3: 3 3 2 1 FALSE
# 4: 4 4 4 4  TRUE
# 5: 5 5 4 5  TRUE

如果您真的想要单词&#34; Has Dups&#34;,请相应调整,但请注意,使用逻辑值可能更容易,如我在此处的回答。

答案 2 :(得分:3)

我找到了一种方法来使用Rcpp,following an example by hadley (under "Sets")

// [[Rcpp::plugins(cpp11)]]
#include <Rcpp.h>
#include <unordered_set>
using namespace Rcpp;

// [[Rcpp::export]]
LogicalVector anyDupCols(IntegerMatrix x) {
    int nr = x.nrow();
    int nc = x.ncol();
    LogicalVector out(nr, false);

    std::unordered_set<int> seen;
    for (int i = 0; i < nr; i++) {
        seen.clear();
        for (int j = 0; j < nc; j++){
            int xij = x(i,j);
            if (seen.count(xij)){ out[i] = true; break; }
            else seen.insert(xij);
        }
    }

    return out;
}

要使用它,请将其放在cpp文件中并运行

library(Rcpp)
sourceCpp("anyDupCols.cpp")
anyDupCols(as.matrix(DT))

在基准测试中表现相当不错:

nc = 30
nv = nc^2
n  = 1e4

set.seed(1)
DT = setDT( replicate(nc, sample(nv, n, replace = TRUE), simplify=FALSE) )

library(microbenchmark)
microbenchmark(
    ananda = DT[, any(duplicated(unlist(.SD, use.names = FALSE))), by = 1:nrow(DT)]$V1,
    tospig = {
        expr = parse(text=paste(apply(t(combn(names(DT),2)),1,FUN = 
          function(x){ paste0(x, collapse="==") }), collapse = "|"))
        DT[, eval(expr)]
    },
    cpp = anyDupCols(as.matrix(DT)),
    alex = ff(DT),
    tscharf = apply(DT,1,function(row) any(duplicated(row))),
    unit = "relative", times = 10
)

Unit: relative
    expr      min       lq     mean   median       uq      max neval  cld
  ananda 2.462739 2.596990 2.774660 2.659898 2.869048 3.352547    10   c 
  tospig 3.118158 3.253102 3.606263 3.424598 3.885561 4.583268    10    d
     cpp 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000    10 a   
    alex 1.295415 1.927802 1.914883 1.982580 2.029868 2.538143    10  b  
 tscharf 2.112286 2.204654 2.385318 2.234963 2.322206 2.978047    10  bc 

如果我去nc = 50,@ tospig&#39; expr变得太长,R无法处理,我得到node stack overflow,这很有趣。

答案 3 :(得分:1)

带有一些优雅的单行

  
      
  1. 定义列

  2.   
  3. 循环行

  4.   
  5. 查看是否有任何欺骗

  6.   
tmpdt[,dups:=apply(.SD,1,function(row) any(duplicated(row))),.SDcols = dupcheckcols]

> tmpdt
   a b c d  dups
1: 1 2 4 3 FALSE
2: 2 2 2 3  TRUE
3: 3 3 2 1 FALSE
4: 4 4 4 4  TRUE
5: 5 5 4 5  TRUE

答案 4 :(得分:1)

另一种方法是沿着行列出“tmpdt”并找出哪些行包含多个元素:

tmpdt2 = tmpdt[, dupcheckcols, with = FALSE] # subset tmpdt
colSums(table(unlist(tmpdt2), row(tmpdt2)) > 1L) > 0L
#    1     2     3     4     5 
#FALSE  TRUE FALSE  TRUE  TRUE 

table偷看我们可以通过以下方式加快速度:

ff = function(x)
{
    lvs = Reduce(union, lapply(x, function(X) if(is.factor(X)) levels(X) else unique(X)))
    x = lapply(x, function(X) match(X, lvs))
    nr = length(lvs); nc = length(x[[1L]])
    tabs = "dim<-"(tabulate(unlist(x, use.names = FALSE) + (0:(nc - 1L)) * nr, nr * nc), 
                   c(nr, nc))
    colSums(tabs > 1L) > 0L
}
ff(tmpdt2)
#[1] FALSE  TRUE FALSE  TRUE  TRUE