如何使用 R 按列计算每周数据的平均值? [英] How does one compute the mean of weekly data by column using R?

问题描述

我有一个每日数据的数据框:df，有四列:Date, A, B, C并试图按列获取每周数据的平均值.我试过了:

I have a data frame of daily data: df with four columns: Date, A, B, C and am trying to get the mean of weekly data by column. I tried:

library(xts)
xdf <- xts(df[ ,-1],as.Date(df$Date))

apply.weekly(xdf,mean)

但这给了我所有列的平均值.我需要每列的每周平均值.

but this gives me the mean of all the columns. I need the weekly means of each column.

感谢您的帮助

推荐答案

首先获取一些数据(我将使用 data.frame 因为这是您正在使用的，但我真的建议您对时间序列对象使用时间序列类)

First get some data (I'll use a data.frame since that's what you're using, but I really suggest you use a time series class for time series objects)

getSymbols("SPY", src='yahoo', from='2012-05-01', to='2012-06-15',
           return.class='data.frame')
#"SPY"

apply.weekly 将按周拆分数据并对每周应用一个函数.apply.weekly(SPY, mean) 将计算该周所有数据的平均值，但您希望它计算每周 每列的平均值.所以，你必须apply mean 到每一列

apply.weekly will split the data up by weeks and apply a function to each week. apply.weekly(SPY, mean) would calculate the mean of all the data for that week, but you want it to calculate the mean of each column for each week. So, you have to apply mean to each column

apply.weekly(SPY, function(x) apply(x, 2, mean))
#           SPY.Open SPY.High SPY.Low SPY.Close SPY.Volume SPY.Adjusted
#2012-05-04 139.6425 140.4675 138.750  139.3275  149400050     138.6075
#2012-05-11 135.9480 136.9320 135.338  136.2040  173105700     135.5020
#2012-05-18 133.3000 133.9180 132.036  132.1760  229282720     131.4960
#2012-05-25 131.7660 132.6800 130.896  132.2140  176634780     131.5340
#2012-06-01 131.7100 132.8925 130.290  131.2725  191170200     130.5950
#2012-06-08 130.2780 131.3380 129.584  130.8580  175917220     130.1820
#2012-06-15 132.8420 133.7760 131.828  132.8020  184751180     132.2540

作为参考，apply.weekly 是 period.apply 的包装器，因此您也可以使用

For reference, apply.weekly is a wrapper for period.apply so you could also get the above result with

period.apply(SPY, endpoints(SPY, "weeks"), FUN=function(x) apply(x, 2, mean))

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