冲积地块的边际
Margin for alluvial plot
我用这个例子
library(alluvial)
tit <- as.data.frame(Titanic)
# only two variables: class and survival status
tit2d <- aggregate( Freq ~ Class + Survived, data=tit, sum)
alluvial( tit2d[,1:2], freq=tit2d$Freq, xw=0.0, alpha=0.8,
gap.width=0.1, col= "steelblue", border="white",
layer = tit2d$Survived != "Yes" , cex.axis =8)
注意我使用 cex.axis =8
然后我得到
轴标签超越
我尝试使用 par(mar=c(10, 10, 10, 10))
但没有结果
谢谢你的想法
alluvial
函数的源代码中存在错误
函数集 par(mar=c(2,1,1,1))
已硬编码,因此在外部使用 par()
没有任何效果。
您可以在本地将函数的源代码更改为 2 个选项之一:
- 添加参数mar_并传递边距,并设置在正确的位置
par(mar=mar_)
.
- 只需在本地将行覆盖到所需的边距
我发现第一个选项更有吸引力,因为您可以从函数外部设置值并更轻松地进行优化。
源代码:
function (..., freq, col = "gray", border = 0, layer, hide = FALSE,
alpha = 0.5, gap.width = 0.05, xw = 0.1, cw = 0.1, blocks = TRUE,
ordering = NULL, axis_labels = NULL, cex = par("cex"), cex.axis = par("cex.axis"))
{
p <- data.frame(..., freq = freq, col, alpha, border, hide,
stringsAsFactors = FALSE)
np <- ncol(p) - 5
if (!is.null(ordering)) {
stopifnot(is.list(ordering))
if (length(ordering) != np)
stop("'ordering' argument should have ", np, " components, has ",
length(ordering))
}
n <- nrow(p)
if (missing(layer)) {
layer <- 1:n
}
p$layer <- layer
d <- p[, 1:np, drop = FALSE]
p <- p[, -c(1:np), drop = FALSE]
p$freq <- with(p, freq/sum(freq))
col <- col2rgb(p$col, alpha = TRUE)
if (!identical(alpha, FALSE)) {
col["alpha", ] <- p$alpha * 256
}
p$col <- apply(col, 2, function(x) do.call(rgb, c(as.list(x),
maxColorValue = 256)))
isch <- sapply(d, is.character)
d[isch] <- lapply(d[isch], as.factor)
if (length(blocks) == 1) {
blocks <- if (!is.na(as.logical(blocks))) {
rep(blocks, np)
}
else if (blocks == "bookends") {
c(TRUE, rep(FALSE, np - 2), TRUE)
}
}
if (is.null(axis_labels)) {
axis_labels <- names(d)
}
else {
if (length(axis_labels) != ncol(d))
stop("`axis_labels` should have length ", names(d),
", has ", length(axis_labels))
}
getp <- function(i, d, f, w = gap.width) {
a <- c(i, (1:ncol(d))[-i])
if (is.null(ordering[[i]])) {
o <- do.call(order, d[a])
}
else {
d2 <- d
d2[1] <- ordering[[i]]
o <- do.call(order, d2[a])
}
x <- c(0, cumsum(f[o])) * (1 - w)
x <- cbind(x[-length(x)], x[-1])
gap <- cumsum(c(0L, diff(as.numeric(d[o, i])) != 0))
mx <- max(gap)
if (mx == 0)
mx <- 1
gap <- gap/mx * w
(x + gap)[order(o), ]
}
dd <- lapply(seq_along(d), getp, d = d, f = p$freq)
rval <- list(endpoints = dd)
===============================================
===============Need to edit====================
op <- par(mar = c(2, 1, 1, 1))
===============================================
plot(NULL, type = "n", xlim = c(1 - cw, np + cw), ylim = c(0,
1), xaxt = "n", yaxt = "n", xaxs = "i", yaxs = "i", xlab = "",
ylab = "", frame = FALSE)
ind <- which(!p$hide)[rev(order(p[!p$hide, ]$layer))]
for (i in ind) {
for (j in 1:(np - 1)) {
xspline(c(j, j, j + xw, j + 1 - xw, j + 1, j + 1,
j + 1 - xw, j + xw, j) + rep(c(cw, -cw, cw),
c(3, 4, 2)), c(dd[[j]][i, c(1, 2, 2)], rev(dd[[j +
1]][i, c(1, 1, 2, 2)]), dd[[j]][i, c(1, 1)]),
shape = c(0, 0, 1, 1, 0, 0, 1, 1, 0, 0), open = FALSE,
col = p$col[i], border = p$border[i])
}
}
for (j in seq_along(dd)) {
ax <- lapply(split(dd[[j]], d[, j]), range)
if (blocks[j]) {
for (k in seq_along(ax)) {
rect(j - cw, ax[[k]][1], j + cw, ax[[k]][2])
}
}
else {
for (i in ind) {
x <- j + c(-1, 1) * cw
y <- t(dd[[j]][c(i, i), ])
w <- xw * (x[2] - x[1])
xspline(x = c(x[1], x[1], x[1] + w, x[2] - w,
x[2], x[2], x[2] - w, x[1] + w, x[1]), y = c(y[c(1,
2, 2), 1], y[c(2, 2, 1, 1), 2], y[c(1, 1),
1]), shape = c(0, 0, 1, 1, 0, 0, 1, 1, 0, 0),
open = FALSE, col = p$col[i], border = p$border[i])
}
}
for (k in seq_along(ax)) {
text(j, mean(ax[[k]]), labels = names(ax)[k], cex = cex)
}
}
axis(1, at = rep(c(-cw, cw), ncol(d)) + rep(seq_along(d),
each = 2), line = 0.5, col = "white", col.ticks = "black",
labels = FALSE)
axis(1, at = seq_along(d), tick = FALSE, labels = axis_labels,
cex.axis = cex.axis)
par(op)
invisible(rval)
}
我把问题发生的地方标记为:
====================================== ========
==============需要编辑======================
op <- par(mar = c(2, 1, 1, 1))
====================================== ========
将行更改为 par(mar=c(5, 5, 3, 10))
后,我得到:
我用这个例子
library(alluvial)
tit <- as.data.frame(Titanic)
# only two variables: class and survival status
tit2d <- aggregate( Freq ~ Class + Survived, data=tit, sum)
alluvial( tit2d[,1:2], freq=tit2d$Freq, xw=0.0, alpha=0.8,
gap.width=0.1, col= "steelblue", border="white",
layer = tit2d$Survived != "Yes" , cex.axis =8)
注意我使用 cex.axis =8
然后我得到
轴标签超越
我尝试使用 par(mar=c(10, 10, 10, 10))
但没有结果
谢谢你的想法
alluvial
函数的源代码中存在错误
函数集 par(mar=c(2,1,1,1))
已硬编码,因此在外部使用 par()
没有任何效果。
您可以在本地将函数的源代码更改为 2 个选项之一:
- 添加参数mar_并传递边距,并设置在正确的位置
par(mar=mar_)
. - 只需在本地将行覆盖到所需的边距
我发现第一个选项更有吸引力,因为您可以从函数外部设置值并更轻松地进行优化。
源代码:
function (..., freq, col = "gray", border = 0, layer, hide = FALSE,
alpha = 0.5, gap.width = 0.05, xw = 0.1, cw = 0.1, blocks = TRUE,
ordering = NULL, axis_labels = NULL, cex = par("cex"), cex.axis = par("cex.axis"))
{
p <- data.frame(..., freq = freq, col, alpha, border, hide,
stringsAsFactors = FALSE)
np <- ncol(p) - 5
if (!is.null(ordering)) {
stopifnot(is.list(ordering))
if (length(ordering) != np)
stop("'ordering' argument should have ", np, " components, has ",
length(ordering))
}
n <- nrow(p)
if (missing(layer)) {
layer <- 1:n
}
p$layer <- layer
d <- p[, 1:np, drop = FALSE]
p <- p[, -c(1:np), drop = FALSE]
p$freq <- with(p, freq/sum(freq))
col <- col2rgb(p$col, alpha = TRUE)
if (!identical(alpha, FALSE)) {
col["alpha", ] <- p$alpha * 256
}
p$col <- apply(col, 2, function(x) do.call(rgb, c(as.list(x),
maxColorValue = 256)))
isch <- sapply(d, is.character)
d[isch] <- lapply(d[isch], as.factor)
if (length(blocks) == 1) {
blocks <- if (!is.na(as.logical(blocks))) {
rep(blocks, np)
}
else if (blocks == "bookends") {
c(TRUE, rep(FALSE, np - 2), TRUE)
}
}
if (is.null(axis_labels)) {
axis_labels <- names(d)
}
else {
if (length(axis_labels) != ncol(d))
stop("`axis_labels` should have length ", names(d),
", has ", length(axis_labels))
}
getp <- function(i, d, f, w = gap.width) {
a <- c(i, (1:ncol(d))[-i])
if (is.null(ordering[[i]])) {
o <- do.call(order, d[a])
}
else {
d2 <- d
d2[1] <- ordering[[i]]
o <- do.call(order, d2[a])
}
x <- c(0, cumsum(f[o])) * (1 - w)
x <- cbind(x[-length(x)], x[-1])
gap <- cumsum(c(0L, diff(as.numeric(d[o, i])) != 0))
mx <- max(gap)
if (mx == 0)
mx <- 1
gap <- gap/mx * w
(x + gap)[order(o), ]
}
dd <- lapply(seq_along(d), getp, d = d, f = p$freq)
rval <- list(endpoints = dd)
===============================================
===============Need to edit====================
op <- par(mar = c(2, 1, 1, 1))
===============================================
plot(NULL, type = "n", xlim = c(1 - cw, np + cw), ylim = c(0,
1), xaxt = "n", yaxt = "n", xaxs = "i", yaxs = "i", xlab = "",
ylab = "", frame = FALSE)
ind <- which(!p$hide)[rev(order(p[!p$hide, ]$layer))]
for (i in ind) {
for (j in 1:(np - 1)) {
xspline(c(j, j, j + xw, j + 1 - xw, j + 1, j + 1,
j + 1 - xw, j + xw, j) + rep(c(cw, -cw, cw),
c(3, 4, 2)), c(dd[[j]][i, c(1, 2, 2)], rev(dd[[j +
1]][i, c(1, 1, 2, 2)]), dd[[j]][i, c(1, 1)]),
shape = c(0, 0, 1, 1, 0, 0, 1, 1, 0, 0), open = FALSE,
col = p$col[i], border = p$border[i])
}
}
for (j in seq_along(dd)) {
ax <- lapply(split(dd[[j]], d[, j]), range)
if (blocks[j]) {
for (k in seq_along(ax)) {
rect(j - cw, ax[[k]][1], j + cw, ax[[k]][2])
}
}
else {
for (i in ind) {
x <- j + c(-1, 1) * cw
y <- t(dd[[j]][c(i, i), ])
w <- xw * (x[2] - x[1])
xspline(x = c(x[1], x[1], x[1] + w, x[2] - w,
x[2], x[2], x[2] - w, x[1] + w, x[1]), y = c(y[c(1,
2, 2), 1], y[c(2, 2, 1, 1), 2], y[c(1, 1),
1]), shape = c(0, 0, 1, 1, 0, 0, 1, 1, 0, 0),
open = FALSE, col = p$col[i], border = p$border[i])
}
}
for (k in seq_along(ax)) {
text(j, mean(ax[[k]]), labels = names(ax)[k], cex = cex)
}
}
axis(1, at = rep(c(-cw, cw), ncol(d)) + rep(seq_along(d),
each = 2), line = 0.5, col = "white", col.ticks = "black",
labels = FALSE)
axis(1, at = seq_along(d), tick = FALSE, labels = axis_labels,
cex.axis = cex.axis)
par(op)
invisible(rval)
}
我把问题发生的地方标记为:
====================================== ========
==============需要编辑======================
op <- par(mar = c(2, 1, 1, 1))
====================================== ========
将行更改为 par(mar=c(5, 5, 3, 10))
后,我得到: