Question

我正在尝试使用加班结构重新编码现有数据。我的数据集如下所示：

dput(z)

structure(list(democracy = c(0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 
1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L), year.x = 1967:2008, time = c(1, 2, 3, 4, 5, 6, 7, 8, 
9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 
25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 
41, 42)), .Names = c("democracy", "year.x", "time"), row.names = 176:217, class = "data.frame")

所以我想创建一个新的变量，比如time.democ，如果democracy==0取值为零，但是再次开始计算时间段，从1开始计算，如果是democracy ==1，再次democracy==0。我将为一系列国家做这件事，但我假设如果我正确使用这个功能，使用ddply就可以很容易地推广。有什么建议？

我想得到这个：

dput(z)

structure(list(democracy = c(0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 
1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L), year.x = 1967:2008, time = c(1, 2, 3, 4, 5, 6, 7, 8, 
9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 
25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 
41, 42), new.time = c(0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 
0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 
16, 17, 18, 19, 20, 21, 22, 23, 24, 25)), .Names = c("democracy", 
"year.x", "time", "new.time"), row.names = 176:217, class = "data.frame")

谢谢！

Answer 1

您可以使用rle与sequence相结合来执行此操作。 rle执行运行长度编码，而sequence生成序列。

z$new.time <- sequence(rle(z$democracy)$lengths)
z$new.time[z$democracy==0] <- 0

head(z, 20)

    democracy year.x time new.time
176         0   1967    1        0
177         0   1968    2        0
178         0   1969    3        0
179         0   1970    4        0
180         0   1971    5        0
181         0   1972    6        0
182         1   1973    7        1
183         1   1974    8        2
184         1   1975    9        3
185         0   1976   10        0
186         0   1977   11        0
187         0   1978   12        0
188         0   1979   13        0
189         0   1980   14        0
190         0   1981   15        0
191         0   1982   16        0
192         1   1983   17        1
193         1   1984   18        2
194         1   1985   19        3
195         1   1986   20        4

Answer 2

感谢您的回复。我按照你的建议，最后编写了一个函数，以便我可以通过ddply将它应用于我的（纵向）数据集中的所有单元。我发布它可能对其他人有帮助，但我确信有更优雅的解决方案：

# is a long format data frame
new.time <- function(a){
    a <- a[order(a$year.x),]
    a$new.time <- sequence(rle(a$democracy)$lengths)-1
    a$new.time[a$democracy==0] <- 0
    return(a)
}

merged1 <- ddply(merged, .(country.x), new.time)

在r中重新编码时间序列数据

2 个答案: