回复第5楼的 zjgslxh：

outer 函数:

<br />
function (X, Y, FUN = "*", ...)<br />
{<br />
    if (is.array(X)) {<br />
        dX <- dim(X)<br />
        nx <- dimnames(X)<br />
        no.nx <- is.null(nx)<br />
    }<br />
    else {<br />
        dX <- length(X)<br />
        no.nx <- is.null(names(X))<br />
        if (!no.nx)<br />
            nx <- list(names(X))<br />
    }<br />
    if (is.array(Y)) {<br />
        dY <- dim(Y)<br />
        ny <- dimnames(Y)<br />
        no.ny <- is.null(ny)<br />
    }<br />
    else {<br />
        dY <- length(Y)<br />
        no.ny <- is.null(names(Y))<br />
        if (!no.ny)<br />
            ny <- list(names(Y))<br />
    }<br />
    if (is.character(FUN) && FUN == "*") {<br />
        robj <- as.vector(X) %*% t(as.vector(Y))<br />
        dim(robj) <- c(dX, dY)<br />
    }<br />
    else {<br />
        FUN <- match.fun(FUN)<br />
        Y <- rep(Y, rep.int(length(X), length(Y)))<br />
        if (length(X))<br />
            X <- rep(X, times = ceiling(length(Y)/length(X)))<br />
        robj <- FUN(X, Y, ...)<br />
        dim(robj) <- c(dX, dY)<br />
    }<br />
    if (no.nx)<br />
        nx <- vector("list", length(dX))<br />
    else if (no.ny)<br />
        ny <- vector("list", length(dY))<br />
    if (!(no.nx && no.ny))<br />
        dimnames(robj) <- c(nx, ny)<br />
    robj<br />
}<br />
<bytecode: 0x07f2d728><br />
<environment: namespace:base><br />

不好意思,知识浅薄我也看不粗来它是不是向量化运算.
</p>