r - 将表达式与 sprintf 结合起来，在 R 中生成用于 ggplot/plot-labeling 的结果

Question

我想使用 expression结合 sprintf -ggplot2 中标记的结果或 plot .

所以我尝试了(无意义的例子，只是为了说明问题):

require(ggplot2)
test <- 2
ggplot(data = iris, aes(x = Sepal.Length)) +
  geom_bar() +
  scale_x_continuous(name = expression(""<=sprintf(paste0("%.", test, "f"), pi)))

显然，这不起作用(仅评估 <= 和 pi )，即使两个单独的元素都起作用。是否有可能重写表达式，使其与组合一起使用？我已经尝试包括 ~或 eval(parse(text = ...))但这些也不起作用。最后:必须集成 test -对象进入sprintf对我来说。

提前致谢!

编辑所以这是一个更好的例子。我以为我把另一个做得尽可能小，但显然它不够具体。对不起。希望有可能了解我的问题。谢谢!

require(ggplot2)
require(scales)

# data
ids <- factor(c("1.1", "2.1", "1.2", "2.2", "1.3", "2.3"))
values <- data.frame(id = ids,
                 value = c(3, 3.1, 3.1, 3.2, 3.15, 3.5))
positions <- data.frame(id = rep(ids, each = 4),
                    x = c(2, 1, 1.1, 2.2, 1, 0, 0.3, 1.1, 2.2, 1.1, 1.2, 2.5, 1.1, 0.3,
                          0.5, 1.2, 2.5, 1.2, 1.3, 2.7, 1.2, 0.5, 0.6, 1.3),
                    y = c(-0.5, 0, 1, 0.5, 0, 0.5, 1.5, 1, 0.5, 1, 2.1, 1.7, 1, 1.5,
                          2.2, 2.1, 1.7, 2.1, 3.2, 2.8, 2.1, 2.2, 3.3, 3.2))
datapoly <- merge(values, positions, by = c("id"))

# plot preperation (for many ggplots is especially decimals output of a function)
decimals <- 2
lim_min <- 3.1
lim_max <- 3.4
breaks <- seq(lim_min, lim_max, (lim_max - lim_min)/4)
labels <- c(paste0("<", sprintf(paste0("%.", decimals, "f"), breaks[1])),
        sprintf(paste0("%.", decimals, "f"), breaks[2]),
        sprintf(paste0("%.", decimals, "f"), breaks[3]),
        sprintf(paste0("%.", decimals, "f"), breaks[4]),
        paste0(">", sprintf(paste0("%.", decimals, "f"), breaks[5])))

# plot
ggplot(datapoly, aes(x = x, y = y)) +
  geom_polygon(aes(fill = value, group = id)) +
  scale_fill_gradient(limits = c(3.1, 3.4),
                  low = "green", high = "orange",
                  breaks = breaks,
                  labels = labels,
                  oob = squish)

现在轴刻度标签是 "<3.10" "3.17" "3.25" "3.33" ">3.40"我想要 "<=3.10" "3.17" "3.25" "3.33" ">=3.40" , 而不是 >=和 <=使用像 here 这样的数学符号和 here

Answer 1

所以在baptiste的提示下我自己找到了一个substitute的解决方案。改变

labels <- c(substitute(""<=a, list(a =sprintf(paste0("%.", decimals, "f"), breaks[1]))),
            sprintf(paste0("%.", decimals, "f"), breaks[2]),
            sprintf(paste0("%.", decimals, "f"), breaks[3]),
            sprintf(paste0("%.", decimals, "f"), breaks[4]),
            substitute("">=a, list(a =sprintf(paste0("%.", decimals, "f"), breaks[5]))))

在上面的后一个例子中，结果是我想要的!谢谢!