R 从大型相关矩阵中提取唯一对

如果你有如你所说的那样多的内存，那么对于比6大得多的n来说，这真的应该是没有问题的.

set.seed(0)
n <- 6L
x <- provideDimnames(cor(matrix(rnorm(as.double(n) * n), n, n)))
x

            A           B            C           D           E            F
A  1.00000000  0.42679900  0.113100027 -0.03952030 -0.02406114 -0.693427730
B  0.42679900  1.00000000  0.519377903  0.06136646 -0.51713799 -0.331961466
C  0.11310003  0.51937790  1.000000000 -0.43996491 -0.32225557 -0.006199606
D -0.03952030  0.06136646 -0.439964909  1.00000000 -0.42053358  0.537301520
E -0.02406114 -0.51713799 -0.322255571 -0.42053358  1.00000000 -0.367595700
F -0.69342773 -0.33196147 -0.006199606  0.53730152 -0.36759570  1.000000000

s <- seq_len(n) - 1L
nms <- dimnames(x)
dat <- data.frame(val = x[sequence(s, seq.int(1L, length(x), n))],
                  row = gl(n, 1L, labels = nms[[1L]])[sequence(s, 1L)], 
                  col = rep.int(gl(n, 1L, labels = nms[[2L]]), s))
dat

            val row col
1   0.426798998   A   B
2   0.113100027   A   C
3   0.519377903   B   C
4  -0.039520302   A   D
5   0.061366463   B   D
6  -0.439964909   C   D
7  -0.024061141   A   E
8  -0.517137993   B   E
9  -0.322255571   C   E
10 -0.420533577   D   E
11 -0.693427730   A   F
12 -0.331961466   B   F
13 -0.006199606   C   F
14  0.537301520   D   F
15 -0.367595700   E   F

如果您使用的R版本早于4.0.0，其中sequence的定义不同，那么您将需要类似于:

sequence <- function(nvec, from = 1L, by = 1L)
    unlist(.mapply(seq.int, list(from = from, by = by, length.out = nvec), NULL),
           recursive = FALSE, use.names = FALSE)

请注意，seq.int同时接受integer和double参数，如果确定结果将溢出integer，则可以方便地返回double而不是integer，这在本例中为n * n > .Machine$integer.max.

M <- .Machine$integer.max
typeof(seq.int(from = 1L, to = 1 + M, by = M))

[1] "double"