我在1951年至2007年的数据框架中按国家/地区划分了一个平衡的面板。我想将其转换为我的其他变量的五年平均值的新数据框架。当我坐下来做这个时，我意识到我能想到的唯一方法是进行for循环，然后决定是时候来stackoverflow寻求帮助了。
那么，是否有一种简单的方法来转换如下所示的数据：
country country.isocode year POP ci grgdpch
Argentina ARG 1951 17517.34 18.445022145 3.4602044759
Argentina ARG 1952 17876.96 17.76066507 -7.887407586
Argentina ARG 1953 18230.82 18.365255769 2.3118720688
Argentina ARG 1954 18580.56 16.982113434 1.5693778844
Argentina ARG 1955 18927.82 17.488907008 5.3690276523
Argentina ARG 1956 19271.51 15.907756547 0.3125559183
Argentina ARG 1957 19610.54 17.028450999 2.4896639667
Argentina ARG 1958 19946.54 17.541597134 5.0025894968
Argentina ARG 1959 20281.15 16.137310492 -6.763501447
Argentina ARG 1960 20616.01 20.519539628 8.481742144
...
Venezuela VEN 1997 22361.80 21.923577413 5.603872759
Venezuela VEN 1998 22751.36 24.451736863 -0.781844721
Venezuela VEN 1999 23128.64 21.585034168 -8.728234466
Venezuela VEN 2000 23492.75 20.224310777 2.6828641218
Venezuela VEN 2001 23843.87 23.480311721 0.2476965412
Venezuela VEN 2002 24191.77 16.290691319 -8.02535946
Venezuela VEN 2003 24545.43 10.972153646 -8.341989049
Venezuela VEN 2004 24904.62 17.147693312 14.644028806
Venezuela VEN 2005 25269.18 18.805970212 7.3156977879
Venezuela VEN 2006 25641.46 22.191098769 5.2737381326
Venezuela VEN 2007 26023.53 26.518210052 4.1367897561
这样的事情：
country country.isocode period AvPOP Avci Avgrgdpch
Argentina ARG 1 18230 17.38474 1.423454
...
Venezuela VEN 12 25274 21.45343 5.454334
我是否需要使用特定的面板数据包转换此数据框？还是有另一种简单的方法可以做到这一点，我错过了吗？
答案 0 :(得分：11)
这是为aggregate制作的东西。 ：
Df <- data.frame(
year=rep(1951:1970,2),
country=rep(c("Arg","Ven"),each=20),
var1 = c(1:20,51:70),
var2 = c(20:1,70:51)
)
Level <-cut(Df$year,seq(1951,1971,by=5),right=F)
id <- c("var1","var2")
> aggregate(Df[id],list(Df$country,Level),mean)
Group.1 Group.2 var1 var2
1 Arg [1951,1956) 3 18
2 Ven [1951,1956) 53 68
3 Arg [1956,1961) 8 13
4 Ven [1956,1961) 58 63
5 Arg [1961,1966) 13 8
6 Ven [1961,1966) 63 58
7 Arg [1966,1971) 18 3
8 Ven [1966,1971) 68 53
您唯一想做的就是重命名类别和变量名称。
答案 1 :(得分：3)
对于这类问题，plyr包真的非常惊人。下面是一些代码，它们基本上只需一行代码就可以提供您想要的内容以及一个小辅助函数。
library(plyr)
library(zoo)
library(pwt)
# First recreate dataset, using package pwt
data(pwt6.3)
pwt <- pwt6.3[
pwt6.3$country %in% c("Argentina", "Venezuela"),
c("country", "isocode", "year", "pop", "ci", "rgdpch")
]
# Use rollmean() in zoo as basis for defining a rolling 5-period rolling mean
rollmean5 <- function(x){
rollmean(x, 5)
}
# Use ddply() in plyr package to create rolling average per country
pwt.ma <- ddply(pwt, .(country), numcolwise(rollmean5))
以下是此输出：
> head(pwt, 10)
country isocode year pop ci rgdpch
ARG-1950 Argentina ARG 1950 17150.34 13.29214 7736.338
ARG-1951 Argentina ARG 1951 17517.34 18.44502 8004.031
ARG-1952 Argentina ARG 1952 17876.96 17.76067 7372.721
ARG-1953 Argentina ARG 1953 18230.82 18.36526 7543.169
ARG-1954 Argentina ARG 1954 18580.56 16.98211 7661.550
ARG-1955 Argentina ARG 1955 18927.82 17.48891 8072.900
ARG-1956 Argentina ARG 1956 19271.51 15.90776 8098.133
ARG-1957 Argentina ARG 1957 19610.54 17.02845 8299.749
ARG-1958 Argentina ARG 1958 19946.54 17.54160 8714.951
ARG-1959 Argentina ARG 1959 20281.15 16.13731 8125.515
> head(pwt.ma)
country year pop ci rgdpch
1 Argentina 1952 17871.20 16.96904 7663.562
2 Argentina 1953 18226.70 17.80839 7730.874
3 Argentina 1954 18577.53 17.30094 7749.694
4 Argentina 1955 18924.25 17.15450 7935.100
5 Argentina 1956 19267.39 16.98977 8169.456
6 Argentina 1957 19607.51 16.82080 8262.250
请注意，默认情况下，rollmean（）会计算居中的移动平均值。您可以通过将此参数传递给辅助函数来修改此行为以获得左移动或右移动平均值。
修改
@Joris Meys温和地指出，你可能实际上是在平均五年后。以下是修改后的代码：
pwt$period <- cut(pwt$year, seq(1900, 2100, 5))
pwt.ma <- ddply(pwt, .(country, period), numcolwise(mean))
pwt.ma
输出：
> pwt.ma
country period year pop ci rgdpch
1 Argentina (1945,1950] 1950.0 17150.336 13.29214 7736.338
2 Argentina (1950,1955] 1953.0 18226.699 17.80839 7730.874
3 Argentina (1955,1960] 1958.0 19945.149 17.42693 8410.610
4 Argentina (1960,1965] 1963.0 21616.623 19.09067 9000.918
5 Argentina (1965,1970] 1968.0 23273.736 18.89005 10202.665
6 Argentina (1970,1975] 1973.0 25216.339 19.70203 11348.321
7 Argentina (1975,1980] 1978.0 27445.430 23.34439 11907.939
8 Argentina (1980,1985] 1983.0 29774.778 17.58909 10987.538
9 Argentina (1985,1990] 1988.0 32095.227 15.17531 10313.375
10 Argentina (1990,1995] 1993.0 34399.829 17.96758 11221.807
11 Argentina (1995,2000] 1998.0 36512.422 19.03551 12652.849
12 Argentina (2000,2005] 2003.0 38390.719 15.22084 12308.493
13 Argentina (2005,2010] 2006.5 39831.625 21.11783 14885.227
14 Venezuela (1945,1950] 1950.0 5009.006 41.07972 7067.947
15 Venezuela (1950,1955] 1953.0 5684.009 44.60849 8132.041
16 Venezuela (1955,1960] 1958.0 6988.078 37.87946 9468.001
17 Venezuela (1960,1965] 1963.0 8451.073 26.93877 9958.935
18 Venezuela (1965,1970] 1968.0 10056.910 28.66512 11083.242
19 Venezuela (1970,1975] 1973.0 11903.185 32.02671 12862.966
20 Venezuela (1975,1980] 1978.0 13927.882 36.35687 13530.556
21 Venezuela (1980,1985] 1983.0 16082.694 22.21093 10762.718
22 Venezuela (1985,1990] 1988.0 18382.964 19.48447 10376.123
23 Venezuela (1990,1995] 1993.0 20680.645 19.82371 10988.096
24 Venezuela (1995,2000] 1998.0 22739.062 20.93509 10837.580
25 Venezuela (2000,2005] 2003.0 24550.973 17.33936 10085.322
26 Venezuela (2005,2010] 2006.5 25832.495 24.35465 11790.497
答案 2 :(得分：1)
在年变量上使用cut来设置期间变量，然后使用重塑包中的melt和cast来获取平均值。还有很多其他答案可以告诉你如何;见https://stackoverflow.com/questions/tagged/r+reshape
答案 3 :(得分：0)
有基本统计信息和plyr答案，因此为了完整起见，这是一个基于dplyr的答案。使用Joris提供的玩具数据，我们有
Df <- data.frame(
year=rep(1951:1970,2),
country=rep(c("Arg","Ven"),each=20),
var1 = c(1:20,51:70),
var2 = c(20:1,70:51)
)
现在，使用cut创建句点，然后我们可以对它们进行分组并获取方法：
Df %>% mutate(period = cut(Df$year,seq(1951,1971,by=5),right=F)) %>%
group_by(country, period) %>% summarise(V1 = mean(var1), V2 = mean(var2))
Source: local data frame [8 x 4]
Groups: country
country period V1 V2
1 Arg [1951,1956) 3 18
2 Arg [1956,1961) 8 13
3 Arg [1961,1966) 13 8
4 Arg [1966,1971) 18 3
5 Ven [1951,1956) 53 68
6 Ven [1956,1961) 58 63
7 Ven [1961,1966) 63 58
8 Ven [1966,1971) 68 53