list 从 R 中的列表绘制数据
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Plot data from lists in R
提问by Bobesh
here is the problem. I have two lists of vectors, sorted (i hope, but i thing i can sort them if they arent) where ith vector in first list has as many numbers as ith vector in the second list. I just want to plot them. But I fear R cant plot elements of lists. Any ideas how to fix that? Thx a lot. Here is the code i tried.
这是问题所在。我有两个向量列表，已排序（我希望，但如果它们不存在，我可以对它们进行排序），其中第一个列表中的第 i 个向量与第二个列表中的第 i 个向量的数字一样多。我只是想绘制它们。但我担心 R 不能绘制列表的元素。任何想法如何解决这个问题？多谢。这是我试过的代码。
a<-c(2,1,5)
b<-c(2,2,2)
f<-c(4)
g<-c(1)
k<-list(a,f)
l<-list(b,g)
for(i in 1:2){
plot(l[i],k[i])}
and the issue is
问题是
Error in xy.coords(x, y, xlabel, ylabel, log) :
(list) object cannot be coerced to type 'double'
回答by LyzandeR
The best way to do it is to avoid the for-loopand unlist the lists in order to plot them.
最好的方法是避免for-loop和取消列出列表以绘制它们。
This is a way using unlist:
这是一种使用方法unlist：
plot(unlist(l),unlist(k))
The way to do it with a for-loopwould be the following:
使用 afor-loop的方法如下：
for (i in 1:2) {
par(new=T)
plot(l[[i]], k[[i]], xlim=c(0,2), ylim=c(0,5))
}
But it is totally unnecessary as you can get the same result simply by unlisting. You would also have to use par(new=T)so that the second (or any other) plot won't overwrite the previous ones and you would have to specify x and y limits so that the two plots would have the same scales. Also, you would have to use double square brackets [[]]as @HubertL mentions in his answer to access the lists. The result would be the same as above (with the labels in a bolder format since labels would be plotted twice on top of each other).
但这完全没有必要，因为您只需取消列出即可获得相同的结果。您还必须使用，par(new=T)以便第二个（或任何其他）图不会覆盖以前的图，并且您必须指定 x 和 y 限制，以便两个图具有相同的比例。此外，您必须使用方括号[[]]作为@HubertL 在他的回答中提到的访问列表。结果将与上述相同（标签采用更粗的格式，因为标签将在彼此之上绘制两次）。
回答by HubertL
You can try to use double brackets[[]]:
您可以尝试使用双括号[[]]：
plot(l[[i]],k[[i]])
回答by Sergio
Almost there, as @HubertL, just two brackets
几乎就在那里，作为@HubertL，只有两个括号
a<-c(2,1,5)
b<-c(2,2,2)
f<-c(4)
g<-c(1)
k<-list(a,f)
l<-list(b,g)
for(i in 1:2){
plot(l[[i]],k[[i]])
}
