独立计算每个面的色阶范围

Question

我在

dur

中有

Utterance

的音节，并希望将它们绘制在

geom_tile

中。我在下面执行此操作的方式不是最佳的，因为色标面向数据帧中 all

Utterance

中的最大音节持续时间数。

  library(ggplot2)
  df %>%
    ggplot(aes(x = start + dur/2, y = MM, width = dur, height = 0.3, fill = rank)) +
    geom_tile(size = 0.65, color = "white") +
    facet_wrap(~ Utterance, scales = "free_x", ncol = 1)+
    labs(fill = "SyllDur")

我想要的是一个色标，无论它包含多少个音节，每个

Utterance

的基本范围都是相同的。也就是说，即使

Utterance

只有 3 个音节，具有最小和最大持续时间的音节也将与具有 10 个音节的

Utterance

中最小和最大持续时间具有相同的色调。

如何实现？

  df <- structure(list(Utterance = structure(c(1L, 4L, 4L, 1L, 4L, 1L, 
    1L, 4L, 1L, 4L, 4L, 1L, 4L, 4L, 4L, 4L, 4L, 4L), levels = c("C: [what's] a !mountain! for you:", 
    "NA: (0.554)", "C: ((v: laughs))=", "A: =I don't know the (Schauinsland) is a small mount[ain]"
    ), class = "factor"), MM = c("syll", "syll", "syll", "syll", 
    "syll", "syll", "syll", "syll", "syll", "syll", "syll", "syll", 
    "syll", "syll", "syll", "syll", "syll", "syll"), start = c(407, 
    3967, 4827, 1147, 3682, 577, 847, 4697, 0, 4317, 3807, 1340, 
    5287, 4457, 3437, 4037, 4917, 5517), dur = c(170L, 70L, 90L, 
    193L, 125L, 270L, 300L, 130L, 407L, 140L, 160L, 480L, 230L, 240L, 
    245L, 280L, 370L, 538L), rank = c(1, 1, 2, 2, 3, 3, 4, 4, 5, 
    5, 6, 6, 7, 8, 9, 10, 11, 12)), class = c("tbl_df", "tbl", "data.frame"
    ), row.names = c(NA, -18L))

Answer 1

一种方法可能是将每个排名标准化为该话语的最大排名的百分比：

df |>
  dplyr::mutate(rank_norm = rank / max(rank), .by = Utterance) |>
  ggplot(aes(x = start + dur/2, y = MM, width = dur, height = 0.3, fill = rank_norm)) +
  geom_tile(size = 0.65, color = "white") +
  facet_wrap(~ Utterance, scales = "free_x", ncol = 1)+
  labs(fill = "SyllDur")