我正在使用一个功能,在几个方面使用par(mfrow = c(1, 2))
在两个面板中绘制结果。
我想运行具有三个不同数据集的函数并将输出绘制在一起,以便每个输出都在一行中,就像我使用par(mfrow = c(3, 2))
一样。
如果有办法不修改函数本身吗?
可能是基本问题,但非常感谢。
该功能有点长,但相关部分绘制了一个PCoA:
# perform classical multidimensional sclaing (PCoA) of the dist matrix
acop <- cmdscale(dat.d, k = nrow(as.matrix(dat.d)) - 1, eig = TRUE) # keep n-1 eigenvalues
axes.tot <- acop$eig # eig are the n eigenvalues computed by cmdscale. Axes are ranked by their
eigenvalues, so the first axis has the highest eigenvalue, the second axis has the second highest
eigenvalue, etc.
inertia <- sum(acop$eig[acop$eig > 0])
percents <- round(as.numeric(100 * axes.tot/inertia), digits = 0) # The eigenvalues represent
the variance extracted by each axis, here they are expressed as a percentage of the sum of all
eigenvalues (i.e. total variance).
par()
par(mfrow = c(1, 2), pty = "s")
coord1 <- acop$points[, c(1, 2)] # points is a matrix whose rows give the coordinates of the
points chosen to represent the dissimilarities
col.grps <- data.frame(vect.grps, coord1)
# plot so that the maximum variance is projected along the first axis, then on the second and so
on
plot(coord1, asp = 1, cex = 0.1, xlab = paste("Axis 1 ",
"(", percents[1], " % variance explained)", sep = ""),
ylab = paste("Axis 2 ", "(", percents[2], " % variance explained)",
sep = ""), main = "", type = "n", bty = "n")
abline(h = 0, lty = 2, col = "grey")
abline(v = 0, lty = 2, col = "grey")
if (length(vect.grps) == nrow(as.matrix(dat.d))) {
for (g in 1:length(names.grps)) {
text(x = coord1[col.grps[, 1] == names.grps[g], 1],
y = coord1[col.grps[, 1] == names.grps[g], 2],
labels = names.grps[g], col = topo.col[g], cex = 0.7)
}
}
else {
points(coord1, pch = 19, col = "blue", cex = 0.5)
}
coord1 <- acop$points[, c(3, 4)]
col.grps <- data.frame(vect.grps, coord1)
plot(coord1, asp = 1, cex = 0.1, xlab = paste("Axis 3 ",
"(", percents[3], " % variance explained)", sep = ""),
ylab = paste("Axis 4 ", "(", percents[4], " % variance explained)",
sep = ""), main = "", type = "n", bty = "n")
abline(h = 0, lty = 2, col = "grey")
abline(v = 0, lty = 2, col = "grey")
if (length(vect.grps) == nrow(as.matrix(dat.d))) {
for (g in 1:length(names.grps)) {
text(x = coord1[col.grps[, 1] == names.grps[g], 1],
y = coord1[col.grps[, 1] == names.grps[g], 2],
labels = names.grps[g], col = topo.col[g], cex = 0.7)
}
}
else {
points(coord1, pch = 19, col = "blue", cex = 0.5)
答案 0 :(得分:2)
我猜你可以覆盖par()
,
foo <- function(){
par(mfrow=c(2,1), mar=c(0,0,0,0))
plot(1:10)
plot(rnorm(10))
}
par <- function(mfrow, ...) {graphics::par(mfrow=c(3, 2), ...)}
foo()
rm(par)