我有多个地点和年份的一些数据
big.data <- data.frame(loc.id = rep(1:3, each = 10*3),
year = rep(rep(1981:1983, each = 10),times = 3),
day = rep(1:10, times = 3*3),
CN = rep(c(50,55,58), each = 10*3),
top.FC = rep(c(72,76,80),each = 10*3),
DC = rep(c(0.02,0.5,0.8), each = 10*3),
WAT0 = rep(c(20,22,26), each = 10*3),
Precp = sample(1:100,90, replace = T),
ETo = sample(1:10,90, replace = T))
我有一个函数:water.model，它使用内部称为water.update的第二个函数
water.model <- function(dat){
top.FC <- unique(dat$top.FC)
dat$WAT <- -9.9
dat$RO <- -9.9
dat$DR <- -9.9
dat$WAT[1] <- top.FC/2 # WAT.i is a constant
dat$RO[1] <- NA
dat$DR[1] <- NA
for(d in 1:(nrow(dat)-1)){
dat[d + 1,10:12] <- water.update(WAT0 = dat$WAT[d],
RAIN.i = dat$Precp[d + 1],
ETo.i = dat$ETo[d + 1],
CN = unique(dat$CN),
DC = unique(dat$DC),
top.FC = unique(dat$top.FC))
}
return(dat)
}
water.update <- function(WAT0, RAIN.i, ETo.i, CN, DC, top.FC){
S = 25400/CN - 254; IA = 0.2*S
if (RAIN.i > IA) { RO = (RAIN.i - 0.2 * S)^2/(RAIN.i + 0.8 * S)
} else {
RO = 0
}
if (WAT0 + RAIN.i - RO > top.FC) {
DR = DC * (WAT0 + RAIN.i - RO - top.FC)
} else {
DR = 0
}
dWAT = RAIN.i - RO - DR - ETo.i
WAT1 = WAT0 + dWAT
WAT1 <- ifelse(WAT1 < 0, 0, WAT1)
return(list(WAT1,RO,DR))
}
如果我在 X 年的单个位置运行上述函数
big.data.sub <- big.data[big.data$loc.id == 1 & big.data$year == 1981,]
water.model(big.data.sub)
loc.id year day CN top.FC DC WAT0 Precp ETo WAT RO DR
1 1 1981 1 50 72 0.02 20 52 5 36.0000 NA NA
2 1 1981 2 50 72 0.02 20 12 9 39.0000 0.0000000 0.000000
3 1 1981 3 50 72 0.02 20 3 2 40.0000 0.0000000 0.000000
4 1 1981 4 50 72 0.02 20 81 9 107.8750 3.2091485 0.915817
5 1 1981 5 50 72 0.02 20 37 10 133.4175 0.0000000 1.457501
6 1 1981 6 50 72 0.02 20 61 7 184.5833 0.3937926 2.440475
7 1 1981 7 50 72 0.02 20 14 10 186.0516 0.0000000 2.531665
8 1 1981 8 50 72 0.02 20 9 6 186.5906 0.0000000 2.461032
9 1 1981 9 50 72 0.02 20 77 9 248.3579 2.4498216 3.782815
10 1 1981 10 50 72 0.02 20 18 6 256.4708 0.0000000 3.887159
如何针对所有位置和年份运行此操作？
big.data %>% group_by(loc.id, year) %>% # apply my function here.
我的最终数据应类似于上面的内容，其中包含三个名为 WAT、RO 和 DR 的新列，这些列是在函数运行时生成的.
最佳答案
我们可以通过使用map循环遍历list来分割数据并应用water.model
library(tidyverse)
split(big.data, big.data[c('loc.id', 'year')], drop = TRUE) %>%
map_df(water.model)
或者在group_by之后应用do内的函数
big.data %>%
group_by(loc.id, year) %>%
do(data.frame(water.model(.)))
关于r - 使用 dplyr 将函数应用于多个组，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/50393990/