得到连续日值的总和 [英] get sum of consecutive day values

问题描述

我有大数据集如下：

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  

我正在尝试当代码为1时，计算雨的连续值的总和，我需要分别进行总和。例如，我想从 2009-04-12 到 2009-04-15 中获得雨值的总和。所以我试图找到方法来定义代码何时等于1，并且有连续的下降值我得到它们的总和。

I am trying to calculate sum of consecutive values of rain when the code is "1" and I need to have sum of them separately. For example I want to get sum of rain values from 2009-04-12 to 2009-04-15. So I am trying to find way to define when the code is equal 1 and there are consecutive rain values I get sum of them.

上述问题的任何帮助将是非常感谢。

Any help on the above problem would be greatly appreciated.

推荐答案

一个简单的解决方案是使用 rle 。但是我怀疑可能会有更多的优雅的解决方案。

One straightforward solution is to use rle. But I suspect there might be more "elegant" solutions out there.

# 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

编辑：更简单方法与 rle 和直拨。

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 

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