我是R（和stackoverflow）的新手，感谢您的帮助。我想计算矩阵中每个唯一列的出现次数。我已经编写了以下代码，但是它非常慢：
frequencyofequalcolumnsinmatrix = function(matrixM){
# returns a matrix columnswithfrequencyofmtxM that contains each distinct column and the frequency of each distinct columns on the last row. Hence if the last row is c(3,5,3,2), then matrixM has 3+5+3+2=13 columns; there are 4 distinct columns; and the first distinct column appears 3 times, the second distinct column appears 5 times, etc.
n = nrow(matrixM)
columnswithfrequencyofmtxM = c()
while (ncol(matrixM)>0){
indexzero = which(apply(matrixM-matrixM[,1], 2, function(x) identical(as.vector(x),rep(0,n))));
indexnotzero = setdiff(seq(1:ncol(matrixM)),indexzero);
frequencyofgivencolumn = c(matrixM[,1], length(indexzero)); #vector of length n. Coordinates 1 to nrow(matrixM) contains the coordinates of the given distinct column while coordinate nrow(matrixM)+1 contains the frequency of appearance of that column
columnswithfrequencyofmtxM = cbind(columnswithfrequencyofmtxM,frequencyofgivencolumn, deparse.level=0);
matrixM=matrixM[,indexnotzero];
matrixM = as.matrix(matrixM);
}
return(columnswithfrequencyofmtxM)
}
如果我们对矩阵“ testmtx”进行应用，则可以获得：
> testmtx = matrix(c(1,2,4,0,1,1,1,2,1,1,2,4,0,1,1,0,1,1), nrow=3, ncol=6)
> frequencyofequalcolumnsinmatrix(testmtx)
[,1] [,2] [,3]
[1,] 1 0 1
[2,] 2 1 2
[3,] 4 1 1
[4,] 2 3 1
最后一行包含以上列的出现次数。
对我的代码不满意，我浏览了stackoverflow。我发现以下问题：
Fastest way to count occurrences of each unique element
结果表明，对向量的每个唯一元素的出现进行计数的最快方法是使用data.table（）包。这是代码：
f6 <- function(x){
data.table(x)[, .N, keyby = x]
}
当我们运行它时，我们获得：
> vtr = c(1,2,3,1,1,2,4,2,4)
> f6(vtr)
x N
1: 1 3
2: 2 3
3: 3 1
4: 4 2
我试图修改此代码以便在我的情况下使用它。这要求能够将vtr创建为向量，其中每个元素都是向量。但是我还没有做到这一点（很可能是因为在R中，c（c（1,2），c（3,4））与c（1,2,3,4）相同）。
我应该尝试修改功能f6吗？如果是这样，怎么办？
还是应该采用完全不同的方法？如果是这样，哪一个？
谢谢！
最佳答案
一种简单的方法是将行一起粘贴到向量中，然后使用该函数。
mat <- matrix(c(1,2,4,0,1,1,1,2,1,1,2,4,0,1,1,0,1,1), nrow=3, ncol=6)
vec <- apply(mat, 2, paste, collapse=" ")
f6(vec)
x N
1: 011 3
2: 121 1
3: 124 2
编辑
@RohitDas的回答让我思考，在考虑性能时，最好始终进行检查。如果我采用了问题中先前显示的所有功能，OP链接了here并添加
f7 <- table还添加了@DavidArenburg的f10建议
f10 <- function(x){
table(unlist(data.table(x)[, lapply(.SD, paste, collapse = "")]))
}
结果如下：
通过@MaratTalipov添加解决方案后，它无疑是赢家。直接应用于矩阵，它比所有矢量解都快。
set.seed(1)
testmx <- matrix(sample(1:10, 3 * 1e3, rep=T), nrow=1000)
microbenchmark(
f1(apply(testmx, 2, paste, collapse=" ")),
f2(apply(testmx, 2, paste, collapse=" ")),
f3(apply(testmx, 2, paste, collapse=" ")),
f4(apply(testmx, 2, paste, collapse=" ")),
f5(apply(testmx, 2, paste, collapse=" ")),
f6(apply(testmx, 2, paste, collapse=" ")),
f7(apply(testmx, 2, paste, collapse=" ")),
f8(apply(testmx, 2, paste, collapse=" ")),
f9(apply(testmx, 2, paste, collapse=" ")),
f10(testmx),
f11(testmx),
f12(testmx)
)
Unit: microseconds
expr min lq mean median uq max neval
f1(apply(testmx, 2, paste, collapse = " ")) 3311.770 3511.5620 3901.0020 3612.035 3849.3600 9569.987 100
f2(apply(testmx, 2, paste, collapse = " ")) 3044.997 3263.6515 3667.9232 3430.914 3847.2430 6721.318 100
f3(apply(testmx, 2, paste, collapse = " ")) 2032.179 2118.0245 2371.8638 2213.301 2430.4155 6631.624 100
f4(apply(testmx, 2, paste, collapse = " ")) 2119.949 2218.3050 2497.1513 2286.442 2425.0260 6258.987 100
f5(apply(testmx, 2, paste, collapse = " ")) 2131.498 2221.5775 2459.9300 2309.925 2530.3115 4222.575 100
f6(apply(testmx, 2, paste, collapse = " ")) 3121.217 3367.7815 3738.3239 3486.155 3835.1175 7979.352 100
f7(apply(testmx, 2, paste, collapse = " ")) 1766.175 1832.9650 2040.5483 1889.169 2032.1795 3784.110 100
f8(apply(testmx, 2, paste, collapse = " ")) 2085.303 2169.2240 2435.6932 2237.168 2404.2380 5002.109 100
f9(apply(testmx, 2, paste, collapse = " ")) 2802.090 2988.0230 3449.0685 3056.930 3373.1710 17640.957 100
f10(testmx) 4027.017 4251.6385 4865.7036 4399.461 4848.7035 11811.581 100
f11(testmx) 500.058 549.1395 624.9526 576.279 636.1395 1176.809 100
f12(testmx) 1827.769 1886.4740 1957.0555 1902.834 1964.4270 3600.487 100