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如果数据集每个单元格至少包含三个数据点,则 fiddle 的躲避会按预期工作。请参见下图。
道奇如期而至
但是,在随后的代码中,“Verbal Class B”单元格中只有 2 个数据点。只有两个数据点,ggplot2 拒绝构建 fiddle ,我同意。但作为副作用,“Verbal Class A”条件下的 fiddle 水平未对齐,导致 fiddle 也与 geom_point 生成的数据点未对齐。请参见下图。
Violin 闪避失败并导致与其他闪避元素错位
是否有解决方法可以使 fiddle 正确闪避以与数据点保持对齐?
Score = c( 9,12,6,12,11,10,4,12,11,10,9,9,14,8,12,11,10,11,4,10,11,17,6,15,8,12,14,1,16,3,18,16,15,11,10,14,8,8,12,15)
Topic = c( "Math","Math","Math","Math","Math","Math","Math","Math","Math","Math","Math","Math","Math","Math","Math","Math","Math","Math","Math","Math","Verbal","Verbal","Verbal","Verbal","Verbal","Verbal","Verbal","Verbal","Verbal","Verbal","Verbal","Verbal","Verbal","Verbal","Verbal","Verbal","Verbal","Verbal","Verbal","Verbal")
Class = c( "A","A","A","A","A","A","A","A","A","A","A","A","B","B","B","B","B","B","B","B","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","B","B","B")
#Class = c( "A","A","A","A","A","A","A","A","A","A","A","A","B","B","B","B","B","B","B","B","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","B","B")
DataSet = data.frame(Topic,Class,Score)
mywidth <- 1.0
mydodge <- 0.90
myjitteramount <- 0.35
ggplot (data = DataSet, aes(x = Topic, y = Score, color = Class))+
geom_violin (draw_quantiles = c(0.25, 0.5, 0.75), fill = NA, width = mywidth, position = position_dodge(mydodge), alpha = 1.0, size = 0.47, scale = "area", show.legend = FALSE) +
geom_point (position = position_jitterdodge(dodge.width = mydodge, jitter.height = 0, jitter.width = myjitteramount), shape = 21, size = 1.5, stroke = 0.7, fill = NA, alpha = 1.0, show.legend = TRUE) +
ggsave ("TempPlot1.png", width = 11, height = 11, units = "in", dpi = 600)
最佳答案
我觉得这可能不太容易实现 - 当然,如果有一些糟糕的黑客攻击,它就会成为可能。
如果你想保持闪避,一个不太令人满意的解决方法是用一组不同的数据创建 fiddle 图(给最后一组假数据),用一个矩形覆盖它,然后用你的点叠加.
library(ggplot2)
Score <- c(9, 12, 6, 12, 11, 10, 4, 12, 11, 10, 9, 9, 14, 8, 12, 11, 10, 11, 4, 10, 11, 17, 6, 15, 8, 12, 14, 1, 16, 3, 18, 16, 15, 11, 10, 14, 8, 8, 12, 15)
Topic <- c("Math", "Math", "Math", "Math", "Math", "Math", "Math", "Math", "Math", "Math", "Math", "Math", "Math", "Math", "Math", "Math", "Math", "Math", "Math", "Math", "Verbal", "Verbal", "Verbal", "Verbal", "Verbal", "Verbal", "Verbal", "Verbal", "Verbal", "Verbal", "Verbal", "Verbal", "Verbal", "Verbal", "Verbal", "Verbal", "Verbal", "Verbal", "Verbal", "Verbal")
Class1 <- c( "A","A","A","A","A","A","A","A","A","A","A","A","B","B","B","B","B","B","B","B","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","B","B","B")
Class2 <- c( "A","A","A","A","A","A","A","A","A","A","A","A","B","B","B","B","B","B","B","B","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","A","B","B")
DataSet1 <- data.frame(Topic, Class1, Score)
DataSet2 <- data.frame(Topic, Class2, Score)
ggplot() +
geom_violin(data = DataSet1, aes(x = Topic, y = Score, color = Class1), draw_quantiles = c(0.25, 0.5, 0.75), position = position_dodge()) +
annotate(geom = 'rect', xmin = 2, xmax = Inf, ymin = -Inf, ymax = Inf, fill = 'white') +
geom_point(data = DataSet2, aes(x = Topic, y = Score, color = Class2), position = position_jitterdodge())
更好的选择可能是使用 facet
分隔数据。你只能真正按类分面,这可能会使比较变得困难,但至少数据点与 fiddle 重叠:
ggplot(data = DataSet2, aes(x = Topic, y = Score, color = Class2)) +
geom_violin(draw_quantiles = c(0.25, 0.5, 0.75), position = position_dodge()) +
geom_point(position = position_jitterdodge()) +
facet_grid(~Class2, scales = 'free_x')
另一种选择是重新考虑您的可视化效果,例如使用 ggbeeswarm。
library(ggbeeswarm)
ggplot(DataSet2, aes(x = Topic, y = Score, color = Class2)) +
geom_beeswarm(dodge.width = 0.5)
关于r - 在 geom_violin 中躲闪失败有解决方法吗?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59960048/
我想了解 Ruby 方法 methods() 是如何工作的。 我尝试使用“ruby 方法”在 Google 上搜索,但这不是我需要的。 我也看过 ruby-doc.org,但我没有找到这种方法。
Test 方法 对指定的字符串执行一个正则表达式搜索,并返回一个 Boolean 值指示是否找到匹配的模式。 object.Test(string) 参数 object 必选项。总是一个
Replace 方法 替换在正则表达式查找中找到的文本。 object.Replace(string1, string2) 参数 object 必选项。总是一个 RegExp 对象的名称。
Raise 方法 生成运行时错误 object.Raise(number, source, description, helpfile, helpcontext) 参数 object 应为
Execute 方法 对指定的字符串执行正则表达式搜索。 object.Execute(string) 参数 object 必选项。总是一个 RegExp 对象的名称。 string
Clear 方法 清除 Err 对象的所有属性设置。 object.Clear object 应为 Err 对象的名称。 说明 在错误处理后,使用 Clear 显式地清除 Err 对象。此
CopyFile 方法 将一个或多个文件从某位置复制到另一位置。 object.CopyFile source, destination[, overwrite] 参数 object 必选
Copy 方法 将指定的文件或文件夹从某位置复制到另一位置。 object.Copy destination[, overwrite] 参数 object 必选项。应为 File 或 F
Close 方法 关闭打开的 TextStream 文件。 object.Close object 应为 TextStream 对象的名称。 说明 下面例子举例说明如何使用 Close 方
BuildPath 方法 向现有路径后添加名称。 object.BuildPath(path, name) 参数 object 必选项。应为 FileSystemObject 对象的名称
GetFolder 方法 返回与指定的路径中某文件夹相应的 Folder 对象。 object.GetFolder(folderspec) 参数 object 必选项。应为 FileSy
GetFileName 方法 返回指定路径(不是指定驱动器路径部分)的最后一个文件或文件夹。 object.GetFileName(pathspec) 参数 object 必选项。应为
GetFile 方法 返回与指定路径中某文件相应的 File 对象。 object.GetFile(filespec) 参数 object 必选项。应为 FileSystemObject
GetExtensionName 方法 返回字符串,该字符串包含路径最后一个组成部分的扩展名。 object.GetExtensionName(path) 参数 object 必选项。应
GetDriveName 方法 返回包含指定路径中驱动器名的字符串。 object.GetDriveName(path) 参数 object 必选项。应为 FileSystemObjec
GetDrive 方法 返回与指定的路径中驱动器相对应的 Drive 对象。 object.GetDrive drivespec 参数 object 必选项。应为 FileSystemO
GetBaseName 方法 返回字符串,其中包含文件的基本名 (不带扩展名), 或者提供的路径说明中的文件夹。 object.GetBaseName(path) 参数 object 必
GetAbsolutePathName 方法 从提供的指定路径中返回完整且含义明确的路径。 object.GetAbsolutePathName(pathspec) 参数 object
FolderExists 方法 如果指定的文件夹存在,则返回 True;否则返回 False。 object.FolderExists(folderspec) 参数 object 必选项
FileExists 方法 如果指定的文件存在返回 True;否则返回 False。 object.FileExists(filespec) 参数 object 必选项。应为 FileS
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