Question

我正在尝试将函数应用于xts对象。我正在使用ave函数将函数分别应用于每一天。该函数抛出以下错误：

Error in rbind.zoo(...) : indexes overlap In addition: Warning messages:
1: In split.default(seq_len(nrow(xc)), f, drop = drop, ...) :
   data length is not a multiple of split variable
2: In zoo(value2, i2) :
  some methods for “zoo” objects do not work if the index entries in ‘order.by’ are not unique

我调试了该函数，当我尝试使用以下行将-Inf转换为NA时抛出错误：x[x == -Inf] <- NA。

这是一个最小的可重复示例，仅通过ave函数应用函数的有问题的行：

x <- as.xts(c(NA,-Inf,1,2,3,-Inf,NA,NA,NA),as.POSIXct(c(
  "2010-01-05 00:00:00", "2010-01-05 00:04:00", "2010-01-05 00:08:00", 
  "2010-01-05 00:12:00", "2010-01-05 00:16:00", "2010-01-05 00:20:00",
  "2010-01-06 00:00:00", "2010-01-06 00:04:00", "2010-01-06 00:08:00")))

out <- ave(x, as.Date(index(x)), FUN= function(x) x[x == -Inf] <- NA)

Answer 1

没有理由按天分组，因为计算是逐个元素完成的：

replace(x, x == -Inf, NA)

或者如果覆盖是好的话：

x[x == -Inf] <- NA

如果您的实际函数确实使用了分组，这只是一个示例，那么将ave应用于coredata(x)，以便我们处理普通向量并确保应用的函数实际返回结果（在问题中没有）：

fun <- function(x) replace(x, x == -Inf, NA)
x[] <- ave(coredata(x), as.Date(index(x)), FUN = fun)

我们还可以考虑使用!is.finite(x)进行测试。

Answer 2

现有答案很好，但您可以使用xts工具轻松简洁地在时间间隔（小时，星期等）中实现平均（或任何合理的功能）（将zoo转换为xts）。您是否了解?period.apply及其包装函数apply.daily？

# Solve your overall problem in one line with no error generated when Inf values included:
x.daily <- apply.daily(x, mean, na.rm = T)

# > x.daily
# [,1]
# 2010-01-05 00:20:00 -Inf
# 2010-01-06 00:08:00  NaN

# Solve your first problem compactly (Replace +-Inf values in column 1 (generalise to any column number) of `x` with NA):
x[!is.finite(x[, 1]) &!is.na(x[, 1]), 1] <- NA

# Solve your second problem compactly (average by day):
x.daily <- apply.daily(x, FUN = mean, na.rm = T)

#Optional: tidy up timestamps in x.daily for equal spaced alignment in YYMMDD HHMMSS (NB HHMMSS will vary depending on your timezone.  Should align for 86400 to 00:00:00 equivalent in UTC):
x.daily <- align.time(x.daily, 86400)
# > x.daily
# [,1]
# 2010-01-05 19:00:00    2
# 2010-01-06 19:00:00  NaN

rbind.zoo（...）：索引重叠错误

2 个答案: