假设我有一个矩阵foo,如下所示:
Suppose I have a matrix foo as follows:
foo <- cbind(c(1,2,3), c(15,16,17)) > foo [,1] [,2] [1,] 1 15 [2,] 2 16 [3,] 3 17我想将其变成一个看起来像
I'd like to turn it into a list that looks like
[[1]] [1] 1 15 [[2]] [1] 2 16 [[3]] [1] 3 17您可以按照以下步骤进行操作:
You can do it as follows:
lapply(apply(foo, 1, function(x) list(c(x[1], x[2]))), function(y) unlist(y))
我对没有那么复杂的替代方法感兴趣.请注意,如果您只是执行apply(foo, 1, function(x) list(c(x[1], x[2]))),它将返回列表中的一个列表,我希望避免这种情况.
I'm interested in an alternative method that isn't as complicated. Note, if you just do apply(foo, 1, function(x) list(c(x[1], x[2]))), it returns a list within a list, which I'm hoping to avoid.
推荐答案这是一个更干净的解决方案:
Here's a cleaner solution:
as.list(data.frame(t(foo)))这利用了一个事实,即数据帧实际上只是等长向量的列表(而矩阵实际上是一个显示有列和行的向量...您可以通过调用foo [5]来看到这一点) (例如).
That takes advantage of the fact that a data frame is really just a list of equal length vectors (while a matrix is really a vector that is displayed with columns and rows...you can see this by calling foo[5], for instance).
您也可以这样做,尽管并没有太大改善:
You could also do this, although it isn't much of an improvement:
lapply(1:nrow(foo), function(i) foo[i,])更多推荐
将矩阵转换为列表
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