我有如下大型数据集：
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
我试图在代码为“1”时计算连续雨的总和，并且我需要分别得到它们的总和。例如，我想获得从2009-04-12到2009-04-15的降雨量总和。所以我试图找到方法来定义代码何时等于1并且有连续的降雨值我得到它们的总和。
对于上述问题的任何帮助将不胜感激。
答案 0 :(得分：4)
一个简单的解决方案是使用rle。但我怀疑那里可能会有更多“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$rain[start[idx]:(start[idx]+ival[idx]-1)])
})
# [1] 54.9 30.1 111.9
修改：使用rle和tapply的更简单的方法。
dd.rle <- rle(dd$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
答案 1 :(得分：2)
以下是获得所需结果的矢量化方式。
df <- read.table(textConnection("Date rain code\n2009-04-01 0.0 0\n2009-04-02 0.0 0\n2009-04-03 0.0 0\n2009-04-04 0.7 1\n2009-04-05 54.2 1\n2009-04-06 0.0 0\n2009-04-07 0.0 0\n2009-04-08 0.0 0\n2009-04-09 0.0 0\n2009-04-10 0.0 0\n2009-04-11 0.0 0\n2009-04-12 5.3 1\n2009-04-13 10.1 1\n2009-04-14 6.0 1\n2009-04-15 8.7 1\n2009-04-16 0.0 0\n2009-04-17 0.0 0\n2009-04-18 0.0 0\n2009-04-19 0.0 0\n2009-04-20 0.0 0\n2009-04-21 0.0 0\n2009-04-22 0.0 0\n2009-04-23 0.0 0\n2009-04-24 0.0 0\n2009-04-25 4.3 1\n2009-04-26 42.2 1\n2009-04-27 45.6 1\n2009-04-28 12.6 1\n2009-04-29 6.2 1\n2009-04-30 1.0 1"),
header = TRUE)
df$cumsum <- cumsum(df$rain)
df$diff <- c(diff(df$code), 0)
df$result <- rep(NA, nrow(df))
if (nrow(df[df$diff == -1, ]) == nrow(df[df$diff == 1, ])) {
result <- df[df$diff == -1, "cumsum"] - df[df$diff == 1, "cumsum"]
df[df$diff == -1, "result"] <- result
} else {
result <- c(df[df$diff == -1, "cumsum"], df[nrow(df), "cumsum"]) - df[df$diff == 1, "cumsum"]
df[df$diff == -1, "result"] <- result[1:length(result) - 1]
df[nrow(df), "result"] <- result[length(result)]
}
df
## Date rain code cumsum diff result
## 1 2009-04-01 0.0 0 0.0 0 NA
## 2 2009-04-02 0.0 0 0.0 0 NA
## 3 2009-04-03 0.0 0 0.0 1 NA
## 4 2009-04-04 0.7 1 0.7 0 NA
## 5 2009-04-05 54.2 1 54.9 -1 54.9
## 6 2009-04-06 0.0 0 54.9 0 NA
## 7 2009-04-07 0.0 0 54.9 0 NA
## 8 2009-04-08 0.0 0 54.9 0 NA
## 9 2009-04-09 0.0 0 54.9 0 NA
## 10 2009-04-10 0.0 0 54.9 0 NA
## 11 2009-04-11 0.0 0 54.9 1 NA
## 12 2009-04-12 5.3 1 60.2 0 NA
## 13 2009-04-13 10.1 1 70.3 0 NA
## 14 2009-04-14 6.0 1 76.3 0 NA
## 15 2009-04-15 8.7 1 85.0 -1 30.1
## 16 2009-04-16 0.0 0 85.0 0 NA
## 17 2009-04-17 0.0 0 85.0 0 NA
## 18 2009-04-18 0.0 0 85.0 0 NA
## 19 2009-04-19 0.0 0 85.0 0 NA
## 20 2009-04-20 0.0 0 85.0 0 NA
## 21 2009-04-21 0.0 0 85.0 0 NA
## 22 2009-04-22 0.0 0 85.0 0 NA
## 23 2009-04-23 0.0 0 85.0 0 NA
## 24 2009-04-24 0.0 0 85.0 1 NA
## 25 2009-04-25 4.3 1 89.3 0 NA
## 26 2009-04-26 42.2 1 131.5 0 NA
## 27 2009-04-27 45.6 1 177.1 0 NA
## 28 2009-04-28 12.6 1 189.7 0 NA
## 29 2009-04-29 6.2 1 195.9 0 NA
## 30 2009-04-30 1.0 1 196.9 0 111.9