假设我有以下矩阵:
matrix(c(1,1,2,1,2,3,2,1,3,2,2,1),ncol=3)
结果:
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 1 3 2
[3,] 2 2 2
[4,] 1 1 1
如何通过每行是否具有重复值来过滤/子化此矩阵?例如,在这种情况下,我只想保留第1行和第2行。
任何想法都会非常感激!
答案 0 :(得分:4)
试试这个:(我怀疑会比任何apply
方法更快)
mat[ rowSums(mat == mat[,1])!=ncol(mat) , ]
# ---with your object---
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 1 3 2
答案 1 :(得分:2)
#List the items and send to a.txt
ls $home\Devoluciones\Dto*.dat | select -exp Name > $home\a.txt
#From a.txt keep first 6 characters and send to b.txt
Get-Content $home\a.txt | foreach {$_.remove(6)} | Add-Content $home\b.txt
#From b.txt replace Dto with "" and send to c.txt
Get-Content $home\b.txt | foreach {$_ -replace "Dto",""} | Add-Content $home\c.txt
#From c.txt copy the files to destination
Get-Content $home\c.txt | foreach {copy-item $home\Devoluciones\*$_*.dat $Destination\$_\}
#Clean temp files
Remove-Item -ErrorAction Ignore $home\a.txt -Force
Remove-Item -ErrorAction Ignore $home\b.txt -Force
Remove-Item -ErrorAction Ignore $home\c.txt -Force
这第二个只是为了好玩。您可以按照逻辑来了解它的工作原理。
indx <- apply(m, 1, function(x) !any(duplicated(x)))
m[indx, ]
# [,1] [,2] [,3]
#[1,] 1 2 3
#[2,] 1 3 2
答案 2 :(得分:2)
使用anyDuplicated
函数,我的方法稍微缩短一点,这应该更快。
mat[!apply(mat, 1, anyDuplicated), ]
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 1 3 2