将汇总值连接回原始数据框 [英] Joining aggregated values back to the original data frame

问题描述

我反复使用的一种设计模式是在数据帧上执行分组"或拆分，应用，合并(SAC)"，然后将聚合的数据重新连接回原始数据.例如，当在具有许多州和县的数据框中计算每个县与州平均值的偏差时，此功能很有用.我的汇总计算很少只是一个简单的平均值，但这是一个很好的例子.我经常通过以下方式解决此问题:

One of the design patterns I use over and over is performing a "group by" or "split, apply, combine (SAC)" on a data frame and then joining the aggregated data back to the original data. This is useful, for example, when calculating each county's deviation from the state mean in a data frame with many states and counties. Rarely is my aggregate calculation only a simple mean, but it makes a good example. I often solve this problem the following way:

require(plyr)
set.seed(1)

## set up some data
group1 <- rep(1:3, 4)
group2 <- sample(c("A","B","C"), 12, rep=TRUE) 
values <- rnorm(12)
df <- data.frame(group1, group2, values)

## got some data, so let's aggregate

group1Mean <- ddply( df, "group1", function(x) 
                     data.frame( meanValue = mean(x$values) ) )
df <- merge( df, group1Mean )
df

哪个会产生如下的汇总数据:

Which produces nice aggregate data like the following:

> df
   group1 group2   values meanValue
1       1      A  0.48743 -0.121033
2       1      A -0.04493 -0.121033
3       1      C -0.62124 -0.121033
4       1      C -0.30539 -0.121033
5       2      A  1.51178  0.004804
6       2      B  0.73832  0.004804
7       2      A -0.01619  0.004804
8       2      B -2.21470  0.004804
9       3      B  1.12493  0.758598
10      3      C  0.38984  0.758598
11      3      B  0.57578  0.758598
12      3      A  0.94384  0.758598

这行得通，但是有其他替代方法可以提高可读性，性能等吗?

This works, but are there alternative ways of doing this which improve on readability, performance, etc?

推荐答案

只需一行代码即可解决问题:

One line of code does the trick:

new <- ddply( df, "group1", transform, numcolwise(mean))
new

group1 group2      values    meanValue
1       1      A  0.48742905 -0.121033381
2       1      A -0.04493361 -0.121033381
3       1      C -0.62124058 -0.121033381
4       1      C -0.30538839 -0.121033381
5       2      A  1.51178117  0.004803931
6       2      B  0.73832471  0.004803931
7       2      A -0.01619026  0.004803931
8       2      B -2.21469989  0.004803931
9       3      B  1.12493092  0.758597929
10      3      C  0.38984324  0.758597929
11      3      B  0.57578135  0.758597929
12      3      A  0.94383621  0.758597929

identical(df, new)
[1] TRUE

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