将 xts 或 Zoo 时间序列对象的每一行除以固定行
我正在尝试划分一个包含多个时间序列（列；具有公共日期列（索引））的 xts 对象。我想将每列除以其在指定日期的值（例如“2010-09-30”）这是为了在该日期将整个对象的每列值重新调整为 1（如果它是普通矩阵 A 和行）。我想重新设定为就是说 A[6,]，我可以这样做
t(t(A)/A[6,])
并且可行，但是，尝试操作 xts 对象及其行子集 xts['2010-09-30'] 并不那么容易工作。有人可以指出我正确的方向吗？我意识到这是非常基本的，事实上，如果有更好的方法来调整时间。系列以这种方式使用一个包，我我很高兴采用这种方法。
I am trying to divide an xts object which holds a number of time series (columns; with a common date column (index). I want to divide each column by its value at a specified date (say '2010-09-30'). This is so as to re-scale the entire object with values of 1 in each column at that date (a common re-basing task). Had it been an ordinary matrix, A, and the row I wanted to rebase to was say A[6,], I could just do
t(t(A)/A[6,])
and that works. But, trying to manipulate the xts object and its row subset xts['2010-09-30'] doesn't work as easily. Could someone please point me in the right direction. I realise this is very basic and I should have found the answer on my own. In fact, if there is a better method in general for rebasing time series in this manner using a package, I am happy to adopt that approach.
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xts 和 Zoo 对象在操作之前按索引对齐。如果您想将整个对象除以单行的值，则必须使用 coredata（也可能是 drop）将值获取到原子向量（使用只有一个元素）。
例如：
xts and zoo objects are aligned by index before operations. If you want to divide an entire object by a value at a single row, you have to use
coredata(and maybedrop) to get the value to an atomic vector (with only one element).For example: