如何在泊松曲线上画一条线？

Question

如何在R中的泊松曲线上画一条线？ 这是我用于绘图的代码；

plot(dogbites$daily.dogbites, dpois(dogbites$daily.dogbites, dogbites_lambda),ylab="prob(x)", main="Poisson dog bites")

这是我得到的情节：

我希望得到这样的东西：

我可以为此使用什么代码吗？

编辑：我尝试了行函数并输入=“ o”，但是我得到了它

> dput(dogbites)
structure(list(daily.dogbites = c(1L, 0L, 0L, 0L, 1L, 0L, 0L, 
0L, 0L, 1L, 0L, 0L, 0L, 1L, 3L, 0L, 6L, 9L, 15L, 3L, 4L, 3L, 
7L, 6L, 1L, 2L, 3L, 4L, 2L, 5L, 3L, 1L, 6L, 2L, 0L, 0L, 3L, 3L, 
6L, 1L, 3L, 2L, 2L, 5L, 6L, 7L, 4L, 10L, 4L, 18L, 4L, 3L, 2L, 
5L, 4L, 3L, 2L, 6L, 4L, 6L, 6L, 1L, 2L, 5L, 10L, 4L, 4L, 3L, 
0L, 3L, 4L, 2L, 3L, 3L, 5L, 5L, 5L, 8L, 13L, 10L, 12L, 4L, 5L, 
3L, 3L, 5L, 4L, 2L, 6L, 4L, 2L, 1L, 3L, 3L, 7L, 5L, 3L, 2L, 5L, 
6L, 5L, 3L, 6L, 5L, 3L, 6L, 5L, 9L, 7L, 8L, 12L, 5L, 2L, 6L, 
8L, 4L, 2L, 3L, 6L, 6L, 7L, 6L, 5L, 3L, 3L, 6L, 4L, 3L, 6L, 2L, 
2L, 6L, 2L, 4L, 5L, 3L, 4L, 5L, 9L, 12L, 9L, 16L, 7L, 3L, 2L, 
3L, 0L, 1L, 1L, 2L, 2L, 2L, 3L, 1L, 2L, 3L, 6L, 4L, 6L, 2L, 6L, 
5L, 8L, 3L, 3L, 6L, 7L, 5L, 9L, 18L, 22L, 0L, 7L, 5L, 7L, 1L, 
5L, 2L, 4L, 1L, 4L, 5L, 3L, 9L, 5L, 4L, 2L, 4L, 4L, 0L, 4L, 4L, 
5L, 4L, 9L, 8L, 9L, 7L, 4L, 13L, 12L, 24L, 7L, 4L, 5L, 10L, 2L, 
2L, 3L, 8L, 8L, 4L, 6L, 6L, 3L, 7L, 6L, 2L, 6L, 5L, 2L, 1L, 7L, 
0L, 8L, 11L, 2L, 10L, 3L, 7L, 9L, 10L, 7L, 2L, 2L, 5L, 2L, 1L, 
8L, 4L, 4L, 5L, 3L, 3L, 2L, 4L, 7L, 3L, 2L, 1L, 3L, 7L, 9L, 8L, 
2L, 4L, 8L, 7L, 4L, 9L, 21L, 3L, 2L, 1L, 5L, 3L, 4L, 3L, 3L, 
4L, 4L, 2L, 5L, 5L, 2L, 3L, 1L, 4L, 4L, 0L, 1L, 7L, 4L, 2L, 2L, 
1L, 5L, 6L, 3L, 7L, 7L, 14L, 4L, 1L, 4L, 6L, 6L, 1L, 2L, 3L, 
2L, 0L, 8L, 3L, 1L, 5L, 1L, 4L, 3L, 5L, 7L, 0L, 3L, 3L, 5L, 2L, 
4L, 7L, 6L, 7L, 9L, 19L, 5L, 0L, 3L, 0L, 1L, 3L, 4L, 1L, 5L, 
2L, 4L, 3L, 6L, 3L, 4L, 7L, 5L, 9L, 3L, 7L, 6L, 5L, 3L, 6L, 5L, 
3L, 5L, 8L, 12L, 5L, 17L, 3L, 3L, 2L, 4L, 5L, 4L, 2L, 2L, 1L, 
3L, 5L, 4L, 3L, 2L, 1L, 2L, 4L, 2L, 1L, 1L, 1L, 2L, 2L, 1L, 1L
)), class = "data.frame", row.names = c(NA, -378L))
> dput(dogbites_lambda)
4.50529100529101

Answer 1

您需要按x轴值对数据进行排序

set.seed(42)
x = sample(1:25)
y = dpois(x, 5)
graphics.off()
plot(sort(x), y[order(x)], type = "o")