跨R中zoo对象的行应用函数
我想计算这个zoo对象每一行的平均值，所以在最后我需要一个新的zoo对象，每分钟的平均值。 实际上，我正在尝试应用更高级的统计数据，但从计算角度来看，它应该与平均值相同 主管（zs） 试试这个：跨R中zoo对象的行应用函数,r,function,zoo,R,Function,Zoo,我想计算这个zoo对象每一行的平均值，所以在最后我需要一个新的zoo对象，每分钟的平均值。 实际上，我正在尝试应用更高级的统计数据，但从计算角度来看，它应该与平均值相同 主管（zs） 试试这个： zoo(rowMeans(zs), time(zs)) 或 或 或 zmean我遇到了这个问题，想要一个更健壮的解决方案。具体来说，我想要一个与apply工作相同的函数，除非zoo对象（或继承自zoo的对象，例如xts）在参数X中传递。在这种情况下，函数应尽可能返回zoo（或xts等）对象。结果还应保
zoo(rowMeans(zs), time(zs))
zmean我遇到了这个问题，想要一个更健壮的解决方案。具体来说，我想要一个与apply
工作相同的函数，除非zoo
对象（或继承自zoo
的对象，例如xts
）在参数X
中传递。在这种情况下，函数应尽可能返回zoo
（或xts
等）对象。结果还应保留用户可能附加到原始zoo
对象的任何附加属性
有几个挑战。首先，根据通过参数MARGIN
和FUN
传递给apply
的内容，apply
可能返回维度大于2的列表、向量或数组。有关更多详细信息，请参阅apply
文档
因此，如果您仔细考虑一下，apply
的唯一时间结果可以作为zoo
对象返回，如果结果是一个nrow（result）==nrow（originalZooObject）
的矩阵
，或者是一个length（result）==nrow（originalZooObject）
的向量。否则，您无法确定如何将原始zoo
对象的索引值添加到apply
的结果中
apply.zoo = function(X, MARGIN, FUN, ...) {
result = apply(as.matrix(X), MARGIN, FUN, ...)
# if result is not a list and X is a zoo object, we can try to convert result
# to zoo.
if(!is.list(result) | is.zoo(X)) {
# if FUN was applied across rows and has the right length but simplfied to a
# vector, convert result back to a matrix
if(is.null(dim(result)) & MARGIN[1] == 1 & length(result) == nrow(X)) {
result = matrix(result, ncol = 1, dimnames = list(NULL, "result"))
}
# if we don't have a matrix at this point, we can't convert back to a zoo object.
if(is.matrix(result)) {
# the matrix returned by apply sometimes has dates as columns and series as
# rows. Check for this and transpose.
if(identical(dimnames(result)[[1]], dimnames(X)[[2]])) result = t(result)
# if result has maintained the same number of rows, it can be restored to a
# zoo object.
if(identical(nrow(X), nrow(result))){
# first ensure the dimname of rows is NULL
dimnames(result)[1] = list(NULL)
# then restore all attributes, other than dim and dimnames
result =
do.call(
structure,
c(
list(result),
attributes(X)[-which(names(attributes(X)) %in% c("dim", "dimnames"))]
)
)
}
}
}
result
}
我提出了以下代码，它在结果中保留了原始zoo
对象的所有适用属性。因此，它还可以用于构建在zoo
之上的大多数S3类，例如xts
我所知道的唯一一件事是，用户可能附加了一个应用于原始zoo
对象的列的属性。如果FUN
合并列，则用户属性可能不适合结果的列
虽然我使用命名约定从泛型函数进行分派，apply
不是泛型函数，因此apply.zoo
被设计为直接调用。如果要用通用版本的apply
覆盖apply
函数，则apply.zoo
将在zoo
对象传递给apply
时自动运行
apply.zoo = function(X, MARGIN, FUN, ...) {
result = apply(as.matrix(X), MARGIN, FUN, ...)
# if result is not a list and X is a zoo object, we can try to convert result
# to zoo.
if(!is.list(result) | is.zoo(X)) {
# if FUN was applied across rows and has the right length but simplfied to a
# vector, convert result back to a matrix
if(is.null(dim(result)) & MARGIN[1] == 1 & length(result) == nrow(X)) {
result = matrix(result, ncol = 1, dimnames = list(NULL, "result"))
}
# if we don't have a matrix at this point, we can't convert back to a zoo object.
if(is.matrix(result)) {
# the matrix returned by apply sometimes has dates as columns and series as
# rows. Check for this and transpose.
if(identical(dimnames(result)[[1]], dimnames(X)[[2]])) result = t(result)
# if result has maintained the same number of rows, it can be restored to a
# zoo object.
if(identical(nrow(X), nrow(result))){
# first ensure the dimname of rows is NULL
dimnames(result)[1] = list(NULL)
# then restore all attributes, other than dim and dimnames
result =
do.call(
structure,
c(
list(result),
attributes(X)[-which(names(attributes(X)) %in% c("dim", "dimnames"))]
)
)
}
}
}
result
}
非常感谢，将zoo zs对象的索引从index（zs）更改为dot-use POSIXlt之后，它工作得很好。POSIXct确实有效。如果看起来不是这样，那么你就犯了其他一些错误。
Reduce(`+`, as.list(zs)) / ncol(zs)
zmean <- zs[, 1]
for(i in 2:ncol(zs)) zmean <- zmean + zs[, i]
zmean <- zmean / ncol(zs)
apply.zoo = function(X, MARGIN, FUN, ...) {
result = apply(as.matrix(X), MARGIN, FUN, ...)
# if result is not a list and X is a zoo object, we can try to convert result
# to zoo.
if(!is.list(result) | is.zoo(X)) {
# if FUN was applied across rows and has the right length but simplfied to a
# vector, convert result back to a matrix
if(is.null(dim(result)) & MARGIN[1] == 1 & length(result) == nrow(X)) {
result = matrix(result, ncol = 1, dimnames = list(NULL, "result"))
}
# if we don't have a matrix at this point, we can't convert back to a zoo object.
if(is.matrix(result)) {
# the matrix returned by apply sometimes has dates as columns and series as
# rows. Check for this and transpose.
if(identical(dimnames(result)[[1]], dimnames(X)[[2]])) result = t(result)
# if result has maintained the same number of rows, it can be restored to a
# zoo object.
if(identical(nrow(X), nrow(result))){
# first ensure the dimname of rows is NULL
dimnames(result)[1] = list(NULL)
# then restore all attributes, other than dim and dimnames
result =
do.call(
structure,
c(
list(result),
attributes(X)[-which(names(attributes(X)) %in% c("dim", "dimnames"))]
)
)
}
}
}
result
}