R - 如何计算动物园对象的“全球”每月平均值答案

【问题标题】：R - how to calculate "global" monthly means of a zoo objectR - 如何计算动物园对象的“全球”每月平均值
【发布时间】：2015-05-13 06:24:42
【问题描述】：

假设我有这个动物园对象：

library(zoo)

df <- structure(list(date = structure(c(0, 31, 59, 90, 120, 151, 181, 
212, 243, 273, 304, 334, 365, 396, 424, 455, 485, 516, 546, 577, 
608, 638, 669, 699, 730, 761, 790, 821, 851, 882, 912, 943, 974, 
1004, 1035, 1065, 1096, 1127, 1155, 1186, 1216, 1247, 1277, 1308, 
1339, 1369, 1400, 1430, 1461, 1492, 1520, 1551, 1581, 1612, 1642, 
1673, 1704, 1734, 1765, 1795), class = "Date"), cru = c(67.6552579213225, 
64.8199295882443, 67.6882136843717, 64.4932343486913, 62.0941221490892, 
59.9333537425379, 54.9891320985765, 55.7801010255625, 59.3864840042859, 
63.0966937088627, 64.384004285933, 67.4058778509108, 70.5033062911373, 
68.2652380223481, 67.5416194703811, 64.7323894076228, 62.7184907393234, 
59.8548599418338, 56.0586407469769, 56.4163171590387, 60.6882596050819, 
64.2160875554875, 64.0664013470075, 64.9156283483851, 67.7770855655901, 
65.9492882289913, 68.2622837899893, 64.3217664166539, 65.2166079902036, 
63.9083881830706, 56.3643961426603, 59.9723098117251, 60.4400887800398, 
62.8953007806521, 66.7075616102862, 69.6426603398133, 69.3881677636614, 
68.445981937854, 69.0054186438084, 67.9314556865146, 64.5561457217205, 
61.9860707178938, 59.2609980101026, 56.8957446808511, 61.1108985152304, 
64.3730904637992, 66.0301545997245, 70.4032909842339, 70.7749272922088, 
71.6810653604776, 69.300183682841, 65.3710546456452, 64.8279044849227, 
60.3732435328333, 56.3447726924843, 57.3151997550896, 60.6496709015766, 
64.0822746058472, 63.9682075616103, 69.4681922547069), model = c(70.8259646403441, 
69.1040767391383, 64.5515271993269, 64.6499744101575, 67.712215046297, 
67.1942041779122, 67.1107310531572, 64.0365506335587, 64.3861402081049, 
66.9474645232039, 70.0900007709146, 70.7322071044208, 71.3681528337099, 
70.6042579609609, 69.1904034259863, 66.3244341830203, 70.45013445949, 
68.4745405096757, 64.6434376970369, 63.6578600688845, 62.6690122245487, 
66.7726540913498, 66.9993526229524, 69.2370082486024, 72.731807414607, 
70.066272418262, 69.8665286312326, 66.7517294017474, 63.2675717559056, 
63.9713073835875, 66.8740144556168, 66.2393105120408, 67.9422440282643, 
65.7423707102056, 67.5695235158686, 69.9511480162994, 70.60938201684, 
69.6737827804075, 65.8727627084409, 69.8091812621223, 66.3834903642866, 
67.7146527454309, 66.1289215196643, 65.2953892851434, 62.4094347321203, 
66.2600390234607, 67.183259636812, 69.644620244991, 67.9219957744587, 
66.9441969493467, 63.5216819663215, 66.4959928679606, 66.1665028019955, 
66.4600239736992, 64.8322046083317, 62.6689871470412, 66.3495593398356, 
66.3105935904296, 69.9372793159708, 73.3729173325656)), .Names = c("date", 
"cru", "model"), row.names = 1441:1500, class = "data.frame")

idx <- as.yearmon(df$date, format='%Y-%m-%d')
df2 <- zoo(df[,-c(1)], idx)

我需要计算全球每月平均值 - 即系列的每个 1 月、每个 2 月、每个 3 月等的平均值。

最终结果将是一个包含 12 行的简单数据框，对应于月份（最好是上面有名称）及其平均值。

我该怎么做？

【问题讨论】：

从hydroTSM 包中尝试monthlyfunction：monthlyfunction(df, mean)。
"df" 是一个 data.frame，而不是 ts 对象。
is.data.frame(df2) == FALSE 和 is.zoo(df2) == TRUE，动物园本质上是时间序列对象的扩展
看我的基础R加法，你可以使用month、cbind和aggregate
@Pascal - 好吧，我们正在比较苹果和橙子。 OP 的df2 有一个索引，它是一个带有frequency=12 的动物园yearmon 对象。将您的 out2 索引转换为相同的形式，但仍保留在 zoo 中，将允许它工作：tapply(out2, cycle(as.yearmon(index(out2),frequency=12)), mean)

标签： r time-series mean zoo

【解决方案1】：

1 您可以使用data.table聚合

library(data.table)

# This turns all Jans to 1 and Decs to 12 for example
mth <- month(as.Date(df$date))


dt2 <- as.data.table(df) # turn df into data table dt 
dt2[, mth := mth]        # pop month into your data frame
setkey(dt2, "mth")       # data tables like a key!

# syntax of data table is dt[i, j, by] 
# “Take DT, subset rows using i, then calculate j grouped by by”

result <- dt2[, .(meancru = mean(cru), meanmodel = mean(model)), by = mth]

这让你：

  mth   meancru meanmodel
1   1   69.21975    70.69146
2   2   67.83230    69.27852
...

如果您对data.table 不满意，那么可以使用good cheat sheet here。

2 使用基础 R

使用上面的第m个代码，那么：

df <- cbind(df,mth)
result <- aggregate(df, list(mth), mean)

但格式需要一些清理。

如果您需要对数据做更多的事情，那么您将欣赏data.tables 的速度、索引和添加新变量的能力。

【讨论】：

【解决方案2】：

你可以使用hydroTSM包中的monthlyfunction函数：

# January mean
apply(df[c(1,13,25,37,49),2:3], 2, mean)
#      cru    model 
# 69.21975 70.69146 

out <- monthlyfunction(df, mean)
out <- as.data.frame(t(out))
out
#          cru    model
# Jan 69.21975 70.69146
# Feb 67.83230 69.27852
# Mar 68.35954 66.60058
# Apr 65.36998 66.80626
# May 63.88265 66.79598
# Jun 61.21118 66.76295
# Jul 56.60359 65.91786
# Aug 57.27593 64.37962
# Sep 60.45508 64.75128
# Oct 63.73269 66.40662
# Nov 65.03127 68.35588
# Dec 68.36713 70.58758

class(out)
# [1] "data.frame"

【讨论】：

这非常快并且提供了我需要的输出，但是......我更喜欢使用 R 的基本函数（已经加载了太多的包）。有什么想法吗？
@thiagoveloso - 这可能是相关的：calculate-monthly-average-of-ts-object - stackoverflow.com/questions/6529981/… - 建议 - sapply(df2, tapply, cycle(df2), mean) - 低并且看得出与以上。
@thelatemail 我想从这个问题中得到了很好的答案。我不会将其标记为重复。此外，我的数据框是 zoo 对象，而不是 ts 对象。
@thiagoveloso - 我不介意，它可能会被删除，因为我看到已经有一票赞成。不过，复制并不能杀死它。