当在xts上使用rowSums时保留xts索引

Question

有没有办法在传递rowSums a xts对象时保留xts对象的索引？

目前我将结果重写为xts对象，但这看起来没有像rowSums能够简单地返回它传递的结果那样快。

xts(rowSums(abs(data)),index(data)) 

Answer 1

有趣的问题。让我们忽略abs的计算，因为与价格相关的时间并不相关。如果您关注的是性能，这里是一组定时考虑当前建议：

library(microbenchmark) 
sample.xts <- xts(order.by = as.POSIXct("2004-01-01 00:00:00") + 1:1e6, matrix(rnorm(1e6 *4), ncol = 4), dimnames = list(NULL, c("A", "B", "C", "D"))) 

# See how quickly rowSum works on just the underlying matrix of data in the timings below: 
xx <- coredata(sample.xts) 

microbenchmark(
    coredata(sample.xts), 
    rowSums(xx), 
    rowSums(sample.xts), 
    rowSums(coredata(sample.xts)), 
.xts(x = rowSums(sample.xts), .index(sample.xts)), 
xts(rowSums(coredata(sample.xts)), index(sample.xts)), 
xts(rowSums(sample.xts),index(sample.xts)), 
Reduce("+", as.list(sample.xts)), times = 100) 

# Unit: milliseconds 
#             expr  min  lq  mean median  uq  max neval 
#         coredata(sample.xts) 2.558479 2.661242 6.884048 2.817607 6.356423 104.57993 100 
#           rowSums(xx) 10.314719 10.824184 11.872108 11.289788 12.382614 18.39334 100 
#         rowSums(sample.xts) 10.358009 10.887609 11.814267 11.335977 12.387085 17.16193 100 
#       rowSums(coredata(sample.xts)) 12.916714 13.839761 18.968731 15.950048 17.836838 113.78552 100 
#  .xts(x = rowSums(sample.xts), .index(sample.xts)) 14.402382 15.764736 20.307027 17.808984 19.072600 114.24039 100 
# xts(rowSums(coredata(sample.xts)), index(sample.xts)) 20.490542 24.183286 34.251031 25.566188 27.900599 125.93967 100 
#   xts(rowSums(sample.xts), index(sample.xts)) 17.436137 19.087269 25.259143 21.923877 22.805013 119.60638 100 
#      Reduce("+", as.list(sample.xts)) 21.745574 26.075326 41.696152 27.669601 30.442397 136.38650 100 

y = .xts(x = rowSums(sample.xts), .index(sample.xts)) 
y2 = xts(rowSums(sample.xts),index(sample.xts)) 
all.equal(y, y2) 
#[1] TRUE 

coredata(sample.xts)返回底层数字矩阵。我认为您可以预期的最快性能是rowSums(xx)用于计算，这可以被认为是“基准”。问题是，在xts对象中，最快的方法是什么？看起来 .xts(x = rowSums(sample.xts), .index(sample.xts))给人体面的表现。

Answer 2

如果你的异议是有挑开，并放在一起输入的成分那么如果x是您的XTS对象，然后尝试。它直接返回一个xts对象：

Reduce("+", as.list(x))