计算两个数据帧之间的相关性是否需要循环？ [英] Does calculating correlation between two dataframes require a loop?

问题描述

我有一组看起来像A和B的大型数据框：

I have a set of large dataframes that look like A and B:

A <- data.frame(A1=c(1,2,3,4,5),B1=c(6,7,8,9,10),C1=c(11,12,13,14,15 ))

  A1 B1 C1
1  1  6 11
2  2  7 12
3  3  8 13
4  4  9 14
5  5 10 15

B <- data.frame(A2=c(6,7,7,10,11),B2=c(2,1,3,8,11),C2=c(1,5,16,7,8))

  A2 B2 C2
1  6  2  1
2  7  1  5
3  7  3 16
4 10  8  7
5 11 11  8

我想创建一个表示皮尔逊的向量（C） A1与A2，B1和B2和C1& C2。例如，在这种情况下，这些相关性为：

I would like to create a vector (C) that denotes the Pearson correlation between A1 & A2, B1 & B2, and C1 & C2. In this case, for example, those correlations are:

[1] 0.95 0.92 0.46

推荐答案

cor 接受两个data.frames：

cor accepts two data.frames:

A<-data.frame(A1=c(1,2,3,4,5),B1=c(6,7,8,9,10),C1=c(11,12,13,14,15 ))

B<-data.frame(A2=c(6,7,7,10,11),B2=c(2,1,3,8,11),C2=c(1,5,16,7,8))

cor(A,B)

#           A2        B2       C2
# A1 0.9481224 0.9190183 0.459588
# B1 0.9481224 0.9190183 0.459588
# C1 0.9481224 0.9190183 0.459588

diag(cor(A,B))
#[1] 0.9481224 0.9190183 0.4595880

编辑：

以下是一些基准：

Unit: microseconds
                   expr     min       lq   median       uq      max neval
        diag(cor(A, B)) 230.292 238.4225 243.0115 255.0295  352.955   100
      mapply(cor, A, B) 267.076 281.5120 286.8030 299.5260  375.087   100
 unlist(Map(cor, A, B)) 250.053 259.1045 264.5635 275.9035 1146.140   100

Edit2

以及一些更好的基准测试

And some better benchmarks using

set.seed(42)
A <- as.data.frame(matrix(rnorm(10*n),ncol=n))
B <- as.data.frame(matrix(rnorm(10*n),ncol=n))

但是，我可能应该提到这些基准

However, I should probably mention that these benchmarks strongly depend on the number of rows.

Edit3：因为我被要求提供基准测试代码，所以就在这里。

Since I was asked for the benchmarking code, here it is.

b <- sapply(2^(1:12), function(n) {
    set.seed(42)
    A <- as.data.frame(matrix(rnorm(10*n),ncol=n))
    B <- as.data.frame(matrix(rnorm(10*n),ncol=n))

    require(microbenchmark)
    res <- print(microbenchmark(
                   diag(cor(A,B)),
                   mapply(cor, A, B),
                   unlist(Map(cor,A,B)),
                   times=10
                 ),unit="us")
    res$median
})

b <- t(b)

matplot(x=1:12,log10(b),type="l",
        ylab="log10(median [µs])", 
        xlab="log2(n)",col=1:3,lty=1)
legend("topleft", legend=c("diag(cor(A, B))", 
                           "mapply(cor, A, B)",
                           "unlist(Map(cor,A,B))"),lty=1, col=1:3)

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