Question

我正在绘制离散CDF。我有一些关于geom_step的问题，使用Google找不到。

是否可以使代表跳跃的线段变为虚线而不是为了更好地显示正在发生的事情？
是否可以比我更有效地添加geom_point？（减 c / p）。

以下是我当前的解决方案：

library(tidyverse)
library(ggthemes)
theme_set(theme_few())

x0 <- seq(-0.5, -0.01, by = 0.01)
x1 <- seq(0, 0.99, by = 0.02)
x2 <- seq(1, 1.99, by = 0.02)
x3 <- seq(2, 2.99, by = 0.02)
x35 <- seq(3, 3.49, by = 0.01)
x4 <- seq(3.5, 3.99, by = 0.01)

tibble_ex <- tibble(
  x0 = x0,
  x1 = x1,
  x2 = x2,
  x3 = x3,
  x35 = x35,
  x4 = x4
)

tibble_ex %>%
  gather(x, xax, x0:x4) %>%
  mutate(cdf = case_when(x == 'x0' ~ 0,
                         x == 'x1' ~ 1/2,
                         x == 'x2' ~ 3/5,
                         x == 'x3' ~ 4/5,
                         x == 'x35' ~ 9/10,
                         x == 'x4' ~ 1)) %>%
  ggplot(aes(x = xax, y = cdf)) +
  geom_step() +
  geom_point(aes(x = 0, y = 0), size = 3, shape = 21, fill = 'white') +
  geom_point(aes(x = 1, y = 0.5), size = 3, shape = 21, fill = 'white') +
  geom_point(aes(x = 2, y = 3/5), size = 3, shape = 21, fill = 'white') +
  geom_point(aes(x = 3, y = 4/5), size = 3, shape = 21, fill = 'white') +
  geom_point(aes(x = 3.5, y = 9/10), size = 3, shape = 21, fill = 'white') +
  geom_point(aes(x = 0, y = 0.5), size = 3, shape = 21, fill = 'black') +
  geom_point(aes(x = 1, y = 3/5), size = 3, shape = 21, fill = 'black') +
  geom_point(aes(x = 2, y = 4/5), size = 3, shape = 21, fill = 'black') +
  geom_point(aes(x = 3, y = 9/10), size = 3, shape = 21, fill = 'black') +
  geom_point(aes(x = 3.5, y = 1), size = 3, shape = 21, fill = 'black') +
  labs(x = 'x', y = 'F(x)')

Answer 1

如果您可以将数据放入数据帧并对其进行结构化，以便可以直接映射数据的特征，那么

ggplot将更强大。

通过将每个x与先前的cdf值进行匹配，这是一种获取数据并用代表连接点的其他行进行扩充的方法。我添加了一个列type，以跟踪哪个是哪个。我还安排了df，以便geom_segment以正确的顺序绘制点。

new_steps <- 
  tibble(x = c(0:3, 3.5, 4),
         cdf = c(0, .5, .6, .8, .9, 1))


df <- new_steps %>%
  mutate(type = "cdf") %>%
  bind_rows(new_steps %>%
              mutate(type = "prior",
                     cdf = lag(cdf))) %>%
  drop_na() %>%
  arrange(x, desc(type))

然后，我们可以将点的fill和geom_segments的线型映射到type。

ggplot(df) + 
  geom_point(aes(x, cdf, fill = type),
             shape = 21) +
  scale_fill_manual(values = c("black", "white")) +
  geom_segment(aes(x = lag(x), y = lag(cdf),
                   xend = x, yend = cdf,
                   lty = type)) +
  scale_linetype_manual(values = c("dashed", "solid"))

plot with dashed vertical lines

Answer 2

（1）不，没有内置的方法可以使geom_step半点划线。但是，如果您将其作为一个单独的问题发布，也许有人会帮助您为此创建一个新的几何。

（2）答案是将要绘制的点放在数据框中，就像您可能要绘制的其他任何东西一样：

point_data = data.frame(x = rep(c(0, 1, 2, 3, 3.5), 2),
                        y = c(0, rep(c(.5, .6, .8, .9), 2), 1),
                        z = rep(c("a", "b"), each = 5))

# calling your gathered/mutated version of tibble_ex df
ggplot(df, aes(x = xax, y = cdf)) +
  geom_step() +
  geom_point(data = point_data, aes(x = x, y = y, fill = z), shape = 21) +
  scale_fill_manual(values = c("white", "black"), guide = FALSE) +
  labs(x = 'x', y = 'F(x)')

Answer 3

对于问题的第二部分，您可以将所有坐标放在单独的数据框中并仅调用geom_point：

ddf <- data.frame(xax = rep(c(0:3, 3.5), 2),
                  cdf = c(0, .5, .6, .8, .9, .5, .6, .8, .9, 1),
                  col = rep(c("white", "black"), each = 5))
dev.new()
tibble_ex %>%
  gather(x, xax, x0:x4) %>%
  mutate(cdf = case_when(x == 'x0' ~ 0,
                         x == 'x1' ~ 1/2,
                         x == 'x2' ~ 3/5,
                         x == 'x3' ~ 4/5,
                         x == 'x35' ~ 9/10,
                         x == 'x4' ~ 1)) %>%
  ggplot(aes(x = xax, y = cdf)) +
  geom_step() +
  geom_point(data = ddf, aes(fill = I(col)), size = 3, shape = 21) +
  labs(x = 'x', y = 'F(x)')

R-（ggplot）使geom_step跳跃为虚线

3 个答案: