如何在paste0（）中使用\ n或类似符号作为轴标签

Question

问题：如何利用n\或类似方法在用x-axis label编写的paste0中创建两条文本行

我有

如您所见，我当前在x-axis上有两条独立的行，如当前用labels = paste0("LN: \nLND: ", paste0(seq(0,80,5))))编写的。

我希望从seq(0,80,5)产生的数值出现在两行中，但%写在文本之一的后面。

预期输出：

奖金问题：是否可以手动右/左/居中对齐我要在x-axis label上创建的两条文本行？

撰写者

ggplot(p, aes(x=value, y=os.neck, color=name_new, fill=name_new)) + 

  geom_point(size=2, shape=21, stroke=1, fill="white") + 
  geom_quantile(quantiles=.5, size=1.3) +
  scale_x_continuous(breaks = seq(0,80,5),
                     name="Lymph nodal yield",
                     labels = paste0("LN: \nLND: ", paste0(seq(0,80,5)))) +
  scale_y_continuous(breaks = seq(0,180,20),
                     name="Months to death")

数据样本p

p <- structure(list(value = c(22.2, 18, 9.1, 11, 14, 15, 9, 6, 3, 
25, 16, 4, 6, 10, 13, 12.1, 33, 16, 6, 9, 4.3, 23, 9, 8, 13, 
5, 30, 25, 33.3, 3, 9, 33.3, 9, 12, 7, 38, 60, 5, 71.4, 7, 15, 
25, 4, 16.7, 6, 15, 11.1, 9, 8, 42.9, 7, 4, 16.7, 6, 10, 10, 
62.5, 8, 25, 4, 11.1, 9, 10, 14, 14, 3, 4, 50, 6, 6, 20, 33.3, 
3, 7.7, 26, 13, 7.7, 13, 13, 13, 3, 7, 16.7, 6, 5, 20, 10, 15, 
29, 7, 6, 11, 11.8, 17, 71.4, 14, 33.3, 18, 22, 9), os.neck = c(11.5, 
11.5, 74.38, 74.38, 17.02, 7.89, 96.03, 40.48, 17.74, 14.65, 
14.65, 62.46, 12.55, 9.92, 26.05, 45.47, 45.47, 17.38, 39.72, 
51.45, 8.61, 8.61, 76.98, 67.09, 94.79, 72.15, 93.93, 17.05, 
12.48, 12.48, 91.6, 15.87, 15.87, 11.04, 67.22, 67.02, 8.94, 
8.94, 6.6, 6.6, 5.09, 10.68, 10.68, 17.15, 17.15, 0.07, 5.19, 
5.19, 40.77, 0.2, 0.2, 170.88, 5.55, 5.55, 1.61, 1.61, 38.28, 
38.28, 10.58, 10.58, 32.99, 32.99, 110.98, 103.69, 122.32, 14.78, 
42.74, 4.04, 4.04, 8.28, 84.96, 11.7, 11.7, 49.97, 49.97, 120.48, 
52.6, 52.6, 71.26, 16.3, 100.14, 55.03, 6.51, 6.51, 89.89, 51.71, 
51.71, 24.97, 55.66, 21.91, 81.48, 30.92, 1.58, 1.58, 7.52, 7.52, 
30.75, 30.75, 3.45, 19.22), name_new = structure(c(3L, 2L, 3L, 
2L, 1L, 1L, 1L, 1L, 1L, 3L, 2L, 1L, 1L, 1L, 1L, 3L, 2L, 1L, 1L, 
1L, 3L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 2L, 1L, 3L, 2L, 1L, 1L, 
1L, 3L, 2L, 3L, 2L, 1L, 3L, 2L, 3L, 2L, 1L, 3L, 2L, 1L, 3L, 2L, 
1L, 3L, 2L, 3L, 2L, 3L, 2L, 3L, 2L, 3L, 2L, 1L, 1L, 1L, 1L, 1L, 
3L, 2L, 1L, 1L, 3L, 2L, 3L, 2L, 1L, 3L, 2L, 1L, 1L, 1L, 1L, 3L, 
2L, 1L, 3L, 2L, 1L, 1L, 1L, 1L, 1L, 3L, 2L, 3L, 2L, 3L, 2L, 1L, 
1L), .Label = c("n_fjernet0", "n_fjernet1", "lnd1"), class = "factor")), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -100L))

Answer 1

我对您的paste0语句摆弄了一​​点，对我来说，最简单的解决方案是重复paste0(seq(0,80,5))以得到所需的结果：

Answer 2

您的人可以用paste0()这样来做：

Answer 3

这是一个技巧，但我似乎可以做你想做的事（？）。不幸的是，我认为所有的文字都挤在旅馆里看起来并不好。