问题描述
我有以下数据：
dat <- structure(list(year = c(1979L,1979L,1980L,1981L,1982L,1983L,1984L,1985L,1986L,1987L,1988L,1989L,1990L,1991L,1992L,1993L,1994L,1995L,1996L,1996L),mon = c(5L,5L,7L,6L,7L),day = c(16L,17L,18L,19L,20L,21L,22L,23L,24L,25L,26L,4L,8L,9L,10L,11L,12L,13L,14L,15L,16L,27L,28L,29L,30L,1L,2L,3L,31L,2L),phase = c(6L,1L),Rainfall = c(23.2,35.575,37.4,6.425,10.275,3.05,50.075,23.05,2,1.4,3.325,12.84,0.68,7.78,12.88,91.48,41.08,4.48,0.26,2.32,13.25,64.5,21.55,82.175,33.725,48.95,3.8,16.875,4.7,7.7,48.7,25.275,3.625,0.075,2.5,3.525,0.725,0.2,0.625,0.25,2.85,6.15,10.675,41.975,24.975,127.775,86.225,19.95,1.725,11.125,1.775,5.825,5.975,18.125,3.725,11.75,13.975,0.1,1,4.775,0.225,2.625,0.575,13.375,0.825,0.45,2.2,4.5,0.05,0.975,5.375,9.1,27.3,47.7,31.475,4.8,11.45,3.15,2.3,14.975,77.25,112.225,69.675,27.625,43.65,34.85,47.325,65.725,83.825,29.525,29.95,12.575,3.2,30.95,26.25,4.15,0.025,0.5,0.375,5.15,0.55,2.025,10.525,0.4,9.225,5.25,1.3,3.175,7.825,1.15,3.475,4.275,3.45,3.95,23.525,6.2,5.7,6.1,4.975,2.7,0.95,1.55,37.525,53.8,26.275,101.25,81.825,26.05,6.4,6.75,0.65,2.475,1.45,0.775,5.8,0.36,0.02,0.8,2.64,3.44,26.8,17.98,3.88,33.48,8.08,15.8,11.52,21.44,31.18,13.06,12.92,0.24,9.4,4.24,4.36,2.34,5.72,16.56,10.96,24.12,2.96,28.48,14.72,6.32,0.3,3.46,0.62,0.76,0.46,17.22,10.92,1.96,2.92,3.86,2.88,0.72,0.06,1.62,28.74,0.64,1.18,0.42,5.46,3.56,0.44,0.48,4.9,1.48,19.94,7.28,29.56,8.72,1.5,2.42,4.62,1.2,13.88,9.76,26.32,11,23.8,10.08,17.04,47.6,15.22,4.06,60.3,71.2,16.54,0.88,0.04,0.34,0.36)),row.names = c(NA,266L),class = "data.frame")
此数据中有四列：年，月，日，阶段，降雨。
我想：
（1）计算连续至少三天（长度> = 3）每天降雨量低于5毫米的情况
（2）和连续几天的开始（首次出现）应该具有等于1的阶段。
dat2<-subset(dat,phase == 1)
countruns = function(x){
rainfall_rle_df <- data.frame(unclass(rle(x$Rainfall < 5)))
nrow(subset(rainfall_rle_df,lengths >= 3 & values == TRUE))
}
countruns(dat2)
此脚本给我的值为5，但这是不正确的。整个数据的正确答案应该只有4。
所以问题在于此脚本忽略了连续几天的开始阶段。
例如，上面数据中的1996年具有以下值：
year month day phase Rainfall
1996 6 28 8 0.100
1996 6 29 1 0.040
1996 6 30 1 0.340
1996 7 1 1 0.000
1996 7 2 1 0.360
开始阶段是阶段8，因此不应计入该阶段。
满足上述条件的年份应该是1983、1984、1991和1993。
关于如何在R中执行此操作的任何建议？还是在R中有更好的方法？
我必须将脚本放入函数countruns（）中。
我将不胜感激！
解决方法
在按“年”分组后，我们用rleid（来自data.table）创建分组列，然后保留{Rainfall”的all值小于5，{{ 1}}组，其中“ {phase}的filter值为1，并在first中进行频率计数（n()）
summarise如果我们要将其包装在函数中，library(dplyr)
library(data.table)
dat %>%
# // create a Date column as it is easier to check for consecutive days
mutate(Date = as.Date(paste(year,mon,day,sep="-"))) %>%
# // create a group with year
group_by(year) %>%
# // add more groups with rleid on logical vector
group_by(grp = rleid(Rainfall < 5),# // checks for difference between adjacent Date
# // if the difference is greater than 1,cumsum increments by 1
grp2 = cumsum(c(0,abs(diff(Date))) > 1),.add = TRUE) %>%
# // filter groups where all Rainfall is less than 5 and number of rows >= 3
filter(all(Rainfall < 5),n() >= 3) %>%
# // filter the groups where the first value of phase is 1
filter(first(phase) == 1) %>%
# // get the frequency count
summarise(n = n(),.groups = 'drop') %>%
# // remove the columns that are not needed
select(-grp,-grp2)
# A tibble: 4 x 2
# year n
# <int> <int>
#1 1983 8
#2 1984 11
#3 1991 6
#4 1993 5
会执行{{}} + enquo（当我们将未加引号的变量名作为参数传递给函数esp的列名时） / p>
!!-应用功能
f1 <- function(data,year,month,rainfall) {
data %>%
mutate(Date = as.Date(str_c({{year}},{{month}},{{day}},sep="-"))) %>%
group_by({{year}}) %>%
group_by(grp = rleid({{rainfall}} < 5),grp2 = cumsum(c(0,.add = TRUE) %>%
filter(all({{rainfall}} < 5),n() >=3 ) %>%
filter(first(phase) == 1) %>%
summarise(n = n(),.groups = 'drop') %>%
select(-grp,-grp2)
}