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我想将我的大 xts 对象拆分为包含原始对象的所有观察结果的常规一秒周期列表。目标是将每个列表元素发送到集群上的节点进行处理。

我想出了这个解决方案，非常精细。我想知道是否可以简化此代码：

library(xts)
set.seed(123)
myts = xts(1:10000, as.POSIXlt(1366039619, ts="EST", origin="1970-01-01") + rnorm(10000, 1, 100))

# insure we have at least one observation per second
secs = seq(trunc(index(head(myts, 1))), trunc(index(tail(myts, 1))), by="s")

# generate second periods endpoints
myts = merge(myts, secs, fill=na.locf)
myts.aligned = align.time(myts, 1)
myts.ep = endpoints(myts.aligned, "seconds", 1)

# split large xts object in list of second periods
myts.list = lapply(1:(length(myts.ep)-1), function(x, myts, ep) { myts[ep[x]:ep[x+1],] }, myts, myts.ep)

# call to parLapply here...
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1 回答 1

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我认为这可以满足您的要求：

split(myts, "secs")

它将创建一个列表，其中每个组件是 1 秒的非重叠数据。

看?split.xts

于 2013-05-07T15:43:26.767 回答