Question

关于ddply的简单任务可能是一个愚蠢的问题，但奇怪的是我无法找到解决方案。所以，假设我有一个数据框，包含国内的受访者，以及受访者在其职业生涯中所担任的一些工作：

mydata <- structure(list(country = structure(c(11L, 6L, 7L, 12L, 12L, 3L, 
7L, 10L, 6L, 4L, 5L, 12L, 3L, 1L, 4L, 13L, 2L, 4L, 7L, 3L), contrasts = structure(c(1, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 1, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, -1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 
0, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 
0, 0, 0, -1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 
0, 0, 0, 1, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 
0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, -1, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 
0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1), .Dim = c(13L, 
12L), .Dimnames = list(c("Austria", "Germany", "Sweden", "Netherlands", 
"Spain", "Italy", "France", "Denmark", "Greece", "Switzerland", 
"Belgium", "Czechia", "Poland"), c("AT", "DE", "SE", "NL", "ES", 
"IT", "FR", "DK", "GR", "CH", "BE", "CZ"))), .Label = c("Austria", 
"Germany", "Sweden", "Netherlands", "Spain", "Italy", "France", 
"Denmark", "Greece", "Switzerland", "Belgium", "Czechia", "Poland"
), class = "factor"), njobs = c(2, 2, 3, 2, 1, 2, 4, 2, 1, 3, 
2, 3, 3, 2, 8, 3, 1, 2, 9, 3)), .Names = c("country", "njobs"
), class = "data.frame", row.names = c(NA, -20L))

我想添加第三个列变量，其中包含该特定国家/地区中职业的平均工作数。这很容易分为两行：

ctry.means <- ddply(mydata,.(country),summarize,avejobs=mean(njobs)) result <- merge(mydata,ctry.means,by="country")

然而，这是一个如此简单且经常使用的操作，我觉得必须有一个更简单的方法来一步完成它，一些技巧ddply。在更一般的情况下，这涉及在单个summarize或mutate语句中组合组级和案例级变量。

Answer 1

如果您对简单的基本解决方案感到满意，

mydata$new = ave(mydata$njobs, mydata$country)

也会这样做。

用相应的组填充所有行的平均值（ddply？）

1 个答案: