我有一个问题,我有一堆长度,想从原点开始(假装我正对着y轴的正端),我做一个右并沿着x轴正向移动了length_i的距离。这时我再右拐,走length_i的距离并重复n次。我可以做到,但是我认为有一种更有效的方法,而且我缺乏数学背景:
## Fake Data
set.seed(11)
dat <- data.frame(id = LETTERS[1:6], lens=sample(2:9, 6),
x1=NA, y1=NA, x2=NA, y2=NA)
## id lens x1 y1 x2 y2
## 1 A 4 NA NA NA NA
## 2 B 2 NA NA NA NA
## 3 C 5 NA NA NA NA
## 4 D 8 NA NA NA NA
## 5 E 6 NA NA NA NA
## 6 F 9 NA NA NA NA
## Add a cycle of 4 column
dat[, "cycle"] <- rep(1:4, ceiling(nrow(dat)/4))[1:nrow(dat)]
##For loop to use the information from cycle column
for(i in 1:nrow(dat)) {
## set x1, y1
if (i == 1) {
dat[1, c("x1", "y1")] <- 0
} else {
dat[i, c("x1", "y1")] <- dat[(i - 1), c("x2", "y2")]
}
col1 <- ifelse(dat[i, "cycle"] %% 2 == 0, "x1", "y1")
col2 <- ifelse(dat[i, "cycle"] %% 2 == 0, "x2", "y2")
dat[i, col2] <- dat[i, col1]
col3 <- ifelse(dat[i, "cycle"] %% 2 != 0, "x2", "y2")
col4 <- ifelse(dat[i, "cycle"] %% 2 != 0, "x1", "y1")
mag <- ifelse(dat[i, "cycle"] %in% c(1, 4), 1, -1)
dat[i, col3] <- dat[i, col4] + (dat[i, "lens"] * mag)
}
这样可以得到预期的结果:
> dat
id lens x1 y1 x2 y2 cycle
1 A 4 0 0 4 0 1
2 B 2 4 0 4 -2 2
3 C 5 4 -2 -1 -2 3
4 D 8 -1 -2 -1 6 4
5 E 6 -1 6 5 6 1
6 F 9 5 6 5 -3 2
这是一个情节:
library(ggplot2); library(grid)
ggplot(dat, aes(x = x1, y = y1, xend = x2, yend = y2)) +
geom_segment(aes(color=id), size=3, arrow = arrow(length = unit(0.5, "cm"))) +
ylim(c(-10, 10)) + xlim(c(-10, 10))
这似乎很慢而且笨拙。我想有一种比
for
循环中的项目更好的方法。保持程序权利的一种更有效的方法是什么?
最佳答案
(如@DWin所建议)这是一个使用复数的解决方案,它对任何类型的turn
都具有灵活性,而不仅仅是90度(-pi / 2弧度)直角。一切都是矢量化的:
set.seed(11)
dat <- data.frame(id = LETTERS[1:6], lens = sample(2:9, 6),
turn = -pi/2)
dat <- within(dat, { facing <- pi/2 + cumsum(turn)
move <- lens * exp(1i * facing)
position <- cumsum(move)
x2 <- Re(position)
y2 <- Im(position)
x1 <- c(0, head(x2, -1))
y1 <- c(0, head(y2, -1))
})
dat[c("id", "lens", "x1", "y1", "x2", "y2")]
# id lens x1 y1 x2 y2
# 1 A 4 0 0 4 0
# 2 B 2 4 0 4 -2
# 3 C 5 4 -2 -1 -2
# 4 D 8 -1 -2 -1 6
# 5 E 6 -1 6 5 6
# 6 F 9 5 6 5 -3
实际上,应将
turn
变量与lens
一起视为输入。现在所有转弯都是-pi/2
弧度,但是您可以将每个转角设置为任意值。所有其他变量均为输出。现在有一点乐趣:
trace.path <- function(lens, turn) {
facing <- pi/2 + cumsum(turn)
move <- lens * exp(1i * facing)
position <- cumsum(move)
x <- c(0, Re(position))
y <- c(0, Im(position))
plot.new()
plot.window(range(x), range(y))
lines(x, y)
}
trace.path(lens = seq(0, 1, length.out = 200),
turn = rep(pi/2 * (-1 + 1/200), 200))
(我在此处尝试复制图形:http://en.wikipedia.org/wiki/Turtle_graphics)
我也让您尝试以下方法:
trace.path(lens = seq(1, 10, length.out = 1000),
turn = rep(2 * pi / 10, 1000))
trace.path(lens = seq(0, 1, length.out = 500),
turn = seq(0, pi, length.out = 500))
trace.path(lens = seq(0, 1, length.out = 600) * c(1, -1),
turn = seq(0, 8*pi, length.out = 600) * seq(-1, 1, length.out = 200))
随时添加您的!
关于r - 右转,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20361768/