性能

Question

我有一个名为mymat的矩阵。我想创建另一个矩阵，其中mymat中所有项目的成对组合和values加在一起，得到类似result的内容。

mymat<- structure(c("AOGC-03-0122", "AOGC-05-0009", "AOGC-08-0006", "AOGC-08-0032", 
"AOGC-08-0054", "0.000971685122254438", "0.00114138129544444", 
"0.000779586347096811", "0.00132807674454652", "0.000867219894408284"
), .Dim = c(5L, 2L), .Dimnames = list(NULL, c("samples", "value"
)))

结果

combination                      total.value
AOGC-03-0122+AOGC-03-0122         0.00194337
AOGC-03-0122+AOGC-05-0009         0.002113066
.
.
.
AOGC-08-0054+AOGC-08-0054         0.00173444

Answer 1

矩阵是同类数据对象。它基本上是一个matrix - 分类的原子向量，具有维度属性（忽略列表矩阵的情况）。您不能在单个矩阵中使用字符串和数字的组合。如果要存储具有异构列类型的数据表，则应使用data.frame。绝对看来samples和value列的适当类型分别是字符串和数字。因此，您的输入矩阵应该是data.frame，并且您的输出也应该是data.frame，因为它只是置换输入记录。
您不需要在此处拨打merge()，当然也不需要两次;矢量化索引可以完成这项工作。使用merge()将导致置换顺序依赖于samples值的词典顺序，而不是它们在输入中出现的顺序，这可能是不合需要的。

values <- as.double(mymat[,'value']);
with(expand.grid(rep(list(seq_len(nrow(mymat))),2L)),
    data.frame(
        combination=paste(mymat[Var2,'samples'],mymat[Var1,'samples'],sep='+'),
        total.value=values[Var2]+values[Var1]
    )
);
##                  combination total.value
## 1  AOGC-03-0122+AOGC-03-0122 0.001943370
## 2  AOGC-03-0122+AOGC-05-0009 0.002113066
## 3  AOGC-03-0122+AOGC-08-0006 0.001751271
## 4  AOGC-03-0122+AOGC-08-0032 0.002299762
## 5  AOGC-03-0122+AOGC-08-0054 0.001838905
## 6  AOGC-05-0009+AOGC-03-0122 0.002113066
## 7  AOGC-05-0009+AOGC-05-0009 0.002282763
## 8  AOGC-05-0009+AOGC-08-0006 0.001920968
## 9  AOGC-05-0009+AOGC-08-0032 0.002469458
## 10 AOGC-05-0009+AOGC-08-0054 0.002008601
## 11 AOGC-08-0006+AOGC-03-0122 0.001751271
## 12 AOGC-08-0006+AOGC-05-0009 0.001920968
## 13 AOGC-08-0006+AOGC-08-0006 0.001559173
## 14 AOGC-08-0006+AOGC-08-0032 0.002107663
## 15 AOGC-08-0006+AOGC-08-0054 0.001646806
## 16 AOGC-08-0032+AOGC-03-0122 0.002299762
## 17 AOGC-08-0032+AOGC-05-0009 0.002469458
## 18 AOGC-08-0032+AOGC-08-0006 0.002107663
## 19 AOGC-08-0032+AOGC-08-0032 0.002656153
## 20 AOGC-08-0032+AOGC-08-0054 0.002195297
## 21 AOGC-08-0054+AOGC-03-0122 0.001838905
## 22 AOGC-08-0054+AOGC-05-0009 0.002008601
## 23 AOGC-08-0054+AOGC-08-0006 0.001646806
## 24 AOGC-08-0054+AOGC-08-0032 0.002195297
## 25 AOGC-08-0054+AOGC-08-0054 0.001734440

性能

bgoldst <- function(mymat) { values <- as.double(mymat[,'value']); with(expand.grid(rep(list(seq_len(nrow(mymat))),2L)),data.frame(combination=paste(mymat[Var2,'samples'],mymat[Var1,'samples'],sep='+'),total.value=values[Var2]+values[Var1])); };
akrun <- function(mymat) { d1 <- expand.grid(rep(list(mymat[, "samples"]),2)); d2 <-  data.frame(samples=mymat[,1], value = as.numeric(mymat[,2]), stringsAsFactors=FALSE); d3 <- merge(merge(d1, d2, by.x="Var1", by.y="samples", all.x=TRUE), d2, by.x="Var2", by.y= "samples"); res <- data.frame(combination = do.call(paste, c(d3[1:2], sep="+")), total.value = d3[,3]+d3[,4]); };
identical(bgoldst(mymat),akrun(mymat));
## [1] TRUE

library(microbenchmark);
microbenchmark(bgoldst(mymat),akrun(mymat));
## Unit: microseconds
##            expr      min       lq      mean    median       uq      max neval
##  bgoldst(mymat)  390.875  412.685  444.4554  433.8535  457.589  662.434   100
##    akrun(mymat) 1603.697 1658.009 1789.0585 1692.0075 1824.793 3227.921   100

N <- 1e3; mymat <- matrix(c(sprintf('sample_%d',seq_len(N)),runif(N)),ncol=2L,dimnames=list(NULL,c('samples','value')));
x <- bgoldst(mymat); y <- akrun(mymat); identical(structure(transform(x[order(x$combination),],combination=as.character(combination)),row.names=seq_len(nrow(x))),structure(transform(y[order(y$combination),],combination=as.character(combination)),row.names=seq_len(nrow(y)))); ## annoyingly involved line of code to obviate row order, factor levels order, and row names differences
## [1] TRUE
microbenchmark(bgoldst(mymat),akrun(mymat),times=3L);
## Unit: seconds
##            expr       min        lq      mean    median        uq       max neval
##  bgoldst(mymat)  8.103589  8.328722  8.418285  8.553854  8.575633  8.597411     3
##    akrun(mymat) 30.777301 31.152458 31.348615 31.527615 31.634272 31.740929     3

Answer 2

我们可以将expand.grid与merge

一起使用

d1 <- expand.grid(rep(list(mymat[, "samples"]),2))
d2 <-  data.frame(samples=mymat[,1], value = as.numeric(mymat[,2]), 
                 stringsAsFactors=FALSE)
d3 <- merge(merge(d1, d2, by.x="Var1", by.y="samples", all.x=TRUE),
                     d2, by.x="Var2", by.y= "samples")
res <- data.frame(combination = do.call(paste, c(d3[1:2], sep="+")), 
                     total.value = d3[,3]+d3[,4])
head(res,3)
#                combination total.value
#1 AOGC-03-0122+AOGC-03-0122 0.001943370
#2 AOGC-03-0122+AOGC-05-0009 0.002113066
#3 AOGC-03-0122+AOGC-08-0006 0.001751271

如何在列中添加项目的组合矩阵

2 个答案:

性能