10

我在获取斜体标签中使用的某些单词时遇到问题。我使用以下代码为标签创建新行：

    levels(length_subject$CONSTRUCTION) <- 
c("THAT \n Extraposed", "THAT \n Post-predicate", "TO \n Extraposed \n for-subject", "TO \n Post-predicate \n for-subject", "THAT \n Extraposed \n that-omission", "THAT \n Post-predicate \n that-omission")

但是，我希望“that”和“for”这两个词以斜体显示。我试过类似的东西

"TO \n Extraposed \n (italics(for))-subject"

位它不起作用。

这是情节的样子：

在此处输入图像描述

使用以下代码生成：

ggplot( length_subject, aes( x = SUBJECT ) ) +
  geom_histogram(binwidth=.6, colour="black", fill="grey") +
  ylab("Frequency") +  
  xlab("Subject length") +  
  scale_x_discrete(breaks=c(2,4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30)) + #
  facet_grid( SUBJECT_TYPE~CONSTRUCTION, scales="free_x", space="free") +
  theme(strip.text.x = element_text(size = 8))

这是数据的简化变体：

structure(list(ID = structure(1:86, .Label = c("A05_122_01", 
"A05_253_01", "A05_277_07", "A05_400_01", "A05_99_01", "A06_1076_01", 
"A06_1261_01", "A06_1283_01", "A06_1283_02", "A06_1317_01", "A06_1326_01", 
"A06_1389_01", "A06_1390_01", "A06_1437_01", "A06_1441_02", "A06_1441_03", 
"A06_1442_03", "A06_1456_01", "A06_1461_01", "A06_830_01", "A06_868_01", 
"A06_884_01", "A06_884_03", "A0K_1057_02", "A0K_1144_07", "A0K_1177_01", 
"A0K_1190_03", "A0K_1214_03", "A0K_1216_01", "A0K_950_02", "A0K_986_01", 
"A1A_102_02", "A1A_163_01", "A1A_199_01", "A1A_45_01", "A1A_97_01", 
"A1B_1008_02", "A1B_1013_01", "A1B_1028_02", "A1B_1042_01", "A1B_1064_01", 
"A1B_1126_03", "A1B_1152_01", "A1B_1174_01", "A1B_1271_01", "A1B_997_01", 
"A1J_487_01", "A1J_544_02", "A1J_555_03", "A1J_569_01", "A1J_601_01", 
"A1N_422_04", "A1N_70_02", "A1S_191_01", "A1S_329_01", "A1S_330_01", 
"A1S_465_04", "A1Y_248_01", "A1Y_278_02", "A1Y_292_01", "A1Y_466_01", 
"A1Y_521_01", "A1Y_612_01", "A1Y_634_01", "A26_139_03", "A26_142_01", 
"A26_148_01", "A26_289_01", "A26_345_02", "A26_439_01", "A26_441_02", 
"A26_463_01", "A28_171_01", "A28_244_01", "A28_245_01", "A28_30_01", 
"A28_341_01", "A28_42_01", "A28_494_03", "A2A_301_01", "A2A_396_01", 
"A2A_599_01", "A2A_637_01", "A2A_676_01", "A2E_22_01", "A2E_25_03"
), class = "factor"), SUBJECT = c(3L, 2L, 6L, 2L, 2L, 1L, 1L, 
1L, 1L, 2L, 4L, 1L, 4L, 2L, 3L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 7L, 1L, 3L, 2L, 2L, 1L, 6L, 7L, 4L, 1L, 5L, 4L, 2L, 9L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 5L, 3L, 4L, 1L, 1L, 1L, 1L, 5L, 
2L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 5L, 2L, 1L, 2L, 2L, 1L, 7L, 1L, 
4L, 1L, 2L, 1L, 1L, 3L, 1L, 13L, 2L, 1L, 1L, 1L, 3L, 1L, 1L), 
    CONSTRUCTION = structure(c(1L, 3L, 1L, 1L, 1L, 4L, 4L, 1L, 
    1L, 5L, 5L, 1L, 1L, 5L, 1L, 3L, 5L, 1L, 5L, 4L, 3L, 3L, 1L, 
    5L, 3L, 5L, 1L, 1L, 2L, 3L, 1L, 1L, 3L, 1L, 1L, 1L, 3L, 1L, 
    4L, 3L, 1L, 3L, 1L, 1L, 1L, 1L, 4L, 2L, 4L, 1L, 1L, 3L, 2L, 
    5L, 1L, 1L, 1L, 3L, 1L, 1L, 4L, 4L, 3L, 1L, 2L, 3L, 3L, 1L, 
    3L, 1L, 1L, 1L, 6L, 1L, 1L, 2L, 4L, 4L, 3L, 5L, 3L, 3L, 3L, 
    3L, 5L, 1L), .Label = c("THAT_EXT", "THAT_EXT_NT", "THAT_POST", 
    "THAT_POST_NT", "TO_EXT_FOR", "TO_POST_FOR"), class = "factor"), 
    SUBJECT_TYPE = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 
    2L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 2L, 2L, 2L, 2L, 
    1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 3L, 1L, 1L, 
    2L, 3L, 1L, 2L, 2L, 3L, 1L, 3L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 
    1L, 1L, 1L, 2L, 2L, 3L, 2L, 2L, 2L, 3L, 1L, 1L, 2L, 1L, 1L, 
    2L, 1L, 2L, 1L, 2L, 1L, 2L, 2L, 1L, 2L, 1L, 1L, 2L, 2L, 2L, 
    1L, 3L, 3L), .Label = c("NP", "PRO", "PROPER"), class = "factor")), .Names = c("ID", 
"SUBJECT", "CONSTRUCTION", "SUBJECT_TYPE"), class = "data.frame", row.names = c(NA, 
-86L))
4

3 回答 3

9

要获得斜体，您需要中描述的格式plotmath（然后将其解析为表达式）。但是，该plotmath语法没有换行操作。不过，你可以得到类似的东西atop。通过您给定的示例，您可以将标签设置为

levels(length_subject$CONSTRUCTION) <- 
  c("atop(textstyle('THAT'),textstyle('Extraposed'))", 
    "atop(textstyle('THAT'),textstyle('Post-predicate'))",
    "atop(atop(textstyle('TO'),textstyle('Extraposed')),italic('for')*textstyle('-subject'))",
    "atop(atop(textstyle('TO'),textstyle('Post-predicate')),italic('for')*textstyle('-subject'))",
    "atop(atop(textstyle('THAT'),textstyle('Extraposed')),italic('that')*textstyle('-omission'))",
    "atop(atop(textstyle('THAT'),textstyle('Post-predicate')),italic('that')*textstyle('-omission'))")

然后labeller=label_parsed加入facet_grid通话

ggplot( length_subject, aes( x = SUBJECT ) ) +
  geom_histogram(binwidth=.6, colour="black", fill="grey") +
  ylab("Frequency") +  
  xlab("Subject length") +  
  scale_x_discrete(breaks=c(2,4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30)) + #
  facet_grid( SUBJECT_TYPE~CONSTRUCTION, scales="free_x", space="free", 
              labeller=label_parsed) +
  theme(strip.text.x = element_text(size = 8))

给

在此处输入图像描述

它并不完美（行间距不一样，而且行数越多，差异只会变得更糟），但这是我发现将两者结合起来的唯一方法（绘图表达式中的换行符）。

于 2013-05-10T21:10:49.947 回答
4

编辑 (2016)

使用新的刻面标签系统，该解决方案不再适用。继承自element_blank以制作自定义 grob的技巧现在已明确禁用。我想教训是要接受某些事情不能在设计上在 ggplot2 中完成，并且不要浪费太多精力来解决将来可能随时被破坏的变通办法。

原始答案

您可以尝试创建一个合适的自定义element放置在主题设置中。不幸的是，主题设计并不容易，

require(ggplot2)
require(gridExtra) # tableGrob

element_grob.element_custom <- function(element, label="", ...)  {

  mytheme <- ttheme_minimal(core = list(fg_params = list(parse=TRUE)))
  disect <- strsplit(label, "\\n")[[1]]

  g1 <- tableGrob(as.matrix(disect), theme=mytheme)
  # wrapping into a gTree only because grobHeight.gtable would be too tight
  # cf. absolute.units() squashing textGrobs
  gTree(children=gList(g1), height=sum(g1$heights), 
        cl = "custom_strip")
}

# gTrees don't know their size and ggplot would squash it, so give it room
grobHeight.custom_strip = heightDetails.custom_axis = function(x, ...)
  x$height
# silly wrapper to fool ggplot2's inheritance check...
facet_custom <- function(...){
  structure(
    list(...), # this ... information is not used, btw
    class = c("element_custom","element_blank", "element") # inheritance test workaround
  ) 

}


title <- c("First~line \n italic('wait, a second')", 
           "this~is~boring",
           "integral(f(x)*dx, a, b)")

iris2 <- iris
iris2$Species <- factor(iris$Species, labels=title)
ggplot(iris2, aes(Sepal.Length, Sepal.Width)) +
  geom_line() + facet_grid(.~Species) +
  theme(strip.text.x = facet_custom())

在此处输入图像描述

于 2013-05-10T21:59:19.127 回答
0

当你们中的几个人在寻找如何固定间距时，我找到了一个解决方案。在 3 行（使这 4 行）或任何后续行的最后一行之前添加一行，atop(scriptscriptstyle("")并且不要忘记在)之后添加

于 2018-02-20T19:37:30.767 回答