获取连续日值的总和

我有一个大数据集，如下所示：

Date       rain code
2009-04-01  0.0 0 
2009-04-02  0.0 0 
2009-04-03  0.0 0 
2009-04-04  0.7 1 
2009-04-05 54.2 1  
2009-04-06  0.0 0 
2009-04-07  0.0 0 
2009-04-08  0.0 0 
2009-04-09  0.0 0 
2009-04-10  0.0 0 
2009-04-11  0.0 0 
2009-04-12  5.3 1  
2009-04-13 10.1 1  
2009-04-14  6.0 1  
2009-04-15  8.7 1  
2009-04-16  0.0 0 
2009-04-17  0.0 0 
2009-04-18  0.0 0 
2009-04-19  0.0 0 
2009-04-20  0.0 0 
2009-04-21  0.0 0 
2009-04-22  0.0 0 
2009-04-23  0.0 0 
2009-04-24  0.0 0 
2009-04-25  4.3 1  
2009-04-26 42.2 1  
2009-04-27 45.6 1  
2009-04-28 12.6 1  
2009-04-29  6.2 1  
2009-04-30  1.0 1  

当代码为“1”时，我试图计算连续降雨值的总和，并且我需要分别计算它们的总和。例如我想获得降雨值的总和2009-04-12 to 2009-04-15。所以我试图找到方法来定义代码何时等于 1 并且有连续的降雨值，我得到它们的总和。

对上述问题的任何帮助将不胜感激。

一种简单的解决方案是使用rle。但我怀疑可能有更“优雅”的解决方案。

# assuming dd is your data.frame
dd.rle <- rle(dd$code)
# get start pos of each consecutive 1's
start  <- (cumsum(dd.rle$lengths) - dd.rle$lengths + 1)[dd.rle$values == 1]
# how long do each 1's extend?
ival   <- dd.rle$lengths[dd.rle$values == 1]
# using these two, compute the sum
apply(as.matrix(seq_along(start)), 1, function(idx) {
    sum(dd$rain[start[idx]:(start[idx]+ival[idx]-1)])
})

# [1]  54.9  30.1 111.9

Edit:一个更简单的方法rle and tapply.

dd.rle <- rle(dd$code)
# get the length of each consecutive 1's
ival <- dd.rle$lengths[dd.rle$values == 1]
# using lengths, construct a `factor` with levels = length(ival)
levl  <- factor(rep(seq_along(ival), ival))
# use these levels to extract `rain[code == 1]` and compute sum
tapply(dd$rain[dd$code == 1], levl, sum)

#    1     2     3 
# 54.9  30.1 111.9