我的目的是用ggplot2(作者:Hadley Wickham)重现此图 [ref]

r - ggplot dotplot : What is the proper use of geom_dotplot?-LMLPHP

这是我基于geom_point和一些丑陋的数据准备而做出的努力(请参阅下面的代码):

r - ggplot dotplot : What is the proper use of geom_dotplot?-LMLPHP

如何使用geom_dotplot()做到这一点?

在尝试中,我遇到了几个问题:(1)将geom_dotplot产生的默认密度映射到一个计数;(2)切断轴;(3)没有意外的孔。我放弃了,改用了geom_point()

我期望(并且仍然希望)它会像

ggplot(data, aes(x,y)) + geom_dotplot(stat = "identity")

但不是。因此,这是我尝试过的内容以及输出:
# Data
df <- structure(list(x = c(79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105), y = c(1, 0, 0, 2, 1, 2, 7, 3, 7, 9, 11, 12, 15, 8, 10, 13, 11, 8, 9, 2, 3, 2, 1, 3, 0, 1, 1)), class = "data.frame", row.names = c(NA, -27L))

# dotplot based on geom_dotplot
geom_dots <- function(x, count, round = 10, breaks = NULL, ...) {
    require(ggplot2)
    n = sum(count) # total number of dots to be drawn
    b = round*round(n/round) # prettify breaks
    x = rep(x, count) # make x coordinates for dots
    if (is.null(breaks))  breaks = seq(0, 1, b/4/n)
    ggplot(data.frame(x = x), aes(x = x)) +
        geom_dotplot(method = "histodot", ...) +
        scale_y_continuous(breaks = breaks,
                        #limits = c(0, max(count)+1), # doesn't work
                        labels = breaks * n)
}

geom_dots(x = df$x, count = df$y)

# dotplot based on geom_point
ggplot_dot <- function(x, count, ...) {
    require(ggplot2)
    message("The count variable must be an integer")
    count = as.integer(count) # make sure these are counts
    n = sum(count) # total number of dots to be drawn
    x = rep(x, count) # make x coordinates for dots
    count = count[count > 0]  # drop zero cases
    y = integer(0)  # initialize y coordinates for dots
    for (i in seq_along(count))
        y <- c(y, 1:(count[i]))  # compute y coordinates
    ggplot(data.frame(x = x, y = y), aes(x = x, y = y)) +
        geom_point(...)  # draw one dot per positive count
}

ggplot_dot(x = df$x, count = df$y,
    size = 11, shape = 21, fill = "orange", color = "black") + theme_gray(base_size = 18)
# ggsave("dotplot.png")
ggsave("dotplot.png", width = 12, height = 5.9)

简短的随机注释:使用geom_point()解决方案,保存图形涉及正确调整尺寸以确保点相互接触(点尺寸和图形高度/宽度)。使用geom_dotplot()解决方案,我对标签进行了四舍五入以使它们更漂亮。不幸的是,我无法在大约100处切断轴:使用limits()coord_cartesian()会导致整个图的缩放,而不是削减。还要注意,要使用geom_dotplot(),我基于计数创建了一个数据向量,因为我无法直接使用count变量(我希望stat="identity"能够做到这一点,但我无法使其工作)。

r - ggplot dotplot : What is the proper use of geom_dotplot?-LMLPHP

最佳答案

巧合的是,我也花了整整一天时间与geom_dotplot()进行斗争,并努力使它显示出来。我还没有找到使y轴显示实际数字的方法,但是我找到了一种截断y轴的方法。如您所述,coord_cartesian()limits不起作用,但是coord_fixed()起作用,因为它强制采用x:y单位的比率:

library(tidyverse)
df <- structure(list(x = c(79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105), y = c(1, 0, 0, 2, 1, 2, 7, 3, 7, 9, 11, 12, 15, 8, 10, 13, 11, 8, 9, 2, 3, 2, 1, 3, 0, 1, 1)), class = "data.frame", row.names = c(NA, -27L))
df <- tidyr::uncount(df, y)

ggplot(df, aes(x)) +
  geom_dotplot(method = 'histodot', binwidth = 1) +
  scale_y_continuous(NULL, breaks = NULL) +
  # Make this as high as the tallest column
  coord_fixed(ratio = 15)

在这里使用15作为比率是可行的,因为x轴也使用相同的单位(即单个整数)。如果x轴是百分比或对数美元或日期或其他参数,则必须修改比率,直到y轴被截断为止。

用组合情节的方法进行编辑

正如我在下面的评论中提到的那样,使用拼凑而成的结合图与coord_fixed()的方法效果不佳。但是,如果您手动将组合图的高度(或宽度)设置为与coord_fixed()中的比率相同的值,并确保每个图具有相同的x轴,则可以获得伪面化图

# Make a subset of df
df2 <- df %>% slice(1:25)

plot1 <- ggplot(df, aes(x)) +
  geom_dotplot(method = 'histodot', binwidth = 1) +
  scale_y_continuous(NULL, breaks = NULL) +
  # Make this as high as the tallest column
  # Make xlim the same on both plots
  coord_fixed(ratio = 15, xlim = c(75, 110))

plot2 <- ggplot(df2, aes(x)) +
  geom_dotplot(method = 'histodot', binwidth = 1) +
  scale_y_continuous(NULL, breaks = NULL) +
  coord_fixed(ratio = 7, xlim = c(75, 110))

# Combine both plots in a single column, with each sized incorrectly
library(patchwork)
plot1 + plot2 +
  plot_layout(ncol = 1)

# Combine both plots in a single column, with each sized appropriately
library(patchwork)
plot1 + plot2 +
  plot_layout(ncol = 1, heights = c(15, 7) / (15 + 7))

10-05 17:42