R：设置初始条件的for循环的dplyr解决方案

时间：2018-03-05 19:51:17

标签： r dplyr data.table rcpp purrr

我有一年中有40天的数据和一些数据

set.seed(123)
df <- data.frame(day = 1:40,rain = runif(40,min = 0, max = 3), petc = runif(40, min = 0.3, max = 8),swc = runif(40, min = 27.01, max = 117.43))

我想计算每天另一个名为aetc的变量，计算方法如下：

SW.ini <- 2 # setting some initial values 
SW.max <- 5
SW.min <- 0

第1天，

1）确定名为PAW(day1) = SW.ini + rain(day1)

的变量

2）如果PAW(day1) >= SWC(day1), aetc(day1) = petc(day1); 

If `PAW(day1) < SWC(day1), aetc(day1) = PAW(day1)/SWC(day1) * petc(day1)`

3）检查是否aetc(day1) > PAW(day1). If yes, aetc(day1) = paw(day1)

4）更新SW(day1) = SW.ini + rain(day1) - aetc(day1)

5）如果SW(day1) > SW.max, SW(day1) = SW.max. Similarly if SW（第1天）＆lt; SW.min，SW（day1）= SW.min`

重复第2天

1）确定PAW(day2) = SW(day1) + rain(day2)
2）如果PAW(day2) >= SWC(day2), aetc(day2) = petc(day2);    如果PAW(day2) < SWC(day2), aetc(day2) = PAW(day2)/SWC(day2) * petc(day2)

3）检查是否aetc(day2) > PAW(day2)。如果是，aetc(day2) = paw(day2)

4）更新SW(day2) = SW(day1) + rain(day2) - aetc(day2)

5）如果SW(day2) > SW.max, SW(day2) = SW.max. Similarly if SW（第2天）＆lt; SW.min，  SW（第2天）= SW.min`

这是我优雅的for循环：

      df$PAW <- NA
      df$aetc <- NA
      df$SW <- NA

      df$PAW[1] <- SW.ini + df$rain[1]

      df$aetc[1] <- ifelse(df$PAW[1] >= df$swc[1], df$petc[1],(df$PAW[1]/df$swc[1])*df$petc[1])
      df$aetc[1] <- ifelse(df$aetc[1] > df$PAW[1], df$PAW[1], df$aetc[1])
      df$SW[1] <- SW.ini + df$rain[1] -  df$aetc[1]
      df$SW[1] <- ifelse(df$SW[1] > SW.max, SW.max, ifelse(df$SW[1] < 0, 0,df$SW[1]))

      for (day in 2:nrow(df)){

        df$PAW[day] <- df$SW[day - 1] + df$rain[day]
        df$aetc[day] <- ifelse(df$PAW[day] >= df$swc[day], df$petc[day], (df$PAW[day]/df$swc[day]) * df$petc[day])
        df$aetc[day] <- ifelse(df$aetc[day] > df$PAW[day], df$PAW[day],df$aetc[day])
        df$SW[day] <- df$SW[day - 1] + df$rain[day] -  df$aetc[day]
        df$SW[day] <- ifelse(df$SW[day] > SW.max,SW.max, ifelse(df$SW[day] < 0, 0,df$SW[day]))
      }

我的问题是，这只是一年的数据，我想要运行多年。 

      set.seed(123)
      df <- data.frame(year = 1980:2015, day = rep(1:40, each = 36),rain = 
      runif(40*36,min = 0, max = 3), petc = runif(40*36, min = 0.3, max = 8),swc = runif(40*36, min = 27.01, max = 117.43))

所以我想做一些像

这样的事情 
                df %>% group_by(year) # and then run the above function for each year.

是否有dplyr或任何其他解决方案？

谢谢

3 个答案:

答案 0 :(得分：5)

  

注意：我最初在你的后续问题R: for loop within a foreach loop上发布了这个答案，但在看到这个问题后，似乎这个答案在这里更为重要。 （我不会在我的回答中解决与并行化相关的任何问题，这是您跟进的主题）。

使用Rcpp和data.table

使用C ++编译逻辑并使用data.table分组操作按组应用它可以从基线开始加速~2,000倍，远远超过您希望通过并行化获得的速度。

在您的原始示例中， 39,420,000行，这将在我的计算机上以 1.883秒执行;在 28,800行的修订后的版本中，执行 0.004秒 

library(data.table)
library(Rcpp)

在R脚本中内联定义并编译C++函数CalcSW()：

一个注意事项：C / C++中的计数从0开始，与R不同，后者从1开始 - ＆＃39 ;为什么这里的指数不同 

Rcpp::cppFunction('
List CalcSW(NumericVector SW_ini,
            NumericVector SW_max,
            NumericVector rain,
            NumericVector swc,
            NumericVector PETc) {

  int n = SW_ini.length();
  NumericVector SW(n);
  NumericVector PAW(n);
  NumericVector aetc(n);

  double SW_ini_glob = SW_ini[0];
  double SW_max_glob = SW_max[0];

  SW[0] = SW_ini_glob;
  PAW[0] = SW[0] + rain[0];

  if (PAW[0] > swc[0]){
    aetc[0] = PETc[0];
  } else {
    aetc[0] = PAW[0]/swc[0]*PETc[0];
  }

  if (aetc[0] > PAW[0]){
    aetc[0] = PAW[0];
  }

  SW[0] = SW[0] + rain[0] - aetc[0];

  if(SW[0] > SW_max_glob){
    SW[0] = SW_max_glob;
  }

  if(SW[0] < 0){
    SW[0] = 0;
  }

  for (int i = 1; i < n; i++) {

    PAW[i] = SW[i-1] + rain[0];

    if (PAW[i] > swc[i]){
      aetc[i] = PETc[i];
    } else {
      aetc[i] = PAW[i]/swc[i]*PETc[i];
    }

    if (aetc[i] > PAW[i]){
      aetc[i] = PAW[i];
    }

    SW[i] = SW[i-1] + rain[i] - aetc[i];

    if(SW[i] > SW_max_glob){
      SW[i] = SW_max_glob;
    }

    if(SW[i] < 0){
     SW[i] = 0;
    }
  }
  return Rcpp::List::create(Rcpp::Named("SW") = SW,
                            Rcpp::Named("PAW") = PAW,
                            Rcpp::Named("aetc") = aetc);
}')

创建data.table 

df <- data.table(loc.id = rep(1:10, each = 80*36), 
                 year = rep(rep(1980:2015, each = 80), times = 10),
                 day = rep(rep(1:80, times = 36),times = 10),
                 rain = runif(10*36*80, min = 0 , max = 5),
                 swc = runif(10*36*80,min = 0, max = 50),
                 SW_max = rep(runif(10, min = 100, max = 200), each = 80*36),
                 SW_ini = runif(10*36*80),
                 PETc = runif(10*36*80, min = 0 , max = 1.3),
                 SW = as.numeric(NA),
                 PAW = as.numeric(NA), 
                 aetc = as.numeric(NA))

setkey(df, loc.id, year, day)

对CalcSW()和df的每个组合在loc.id执行函数year，同时将返回值分配给三列：

system.time({
  df[,  c("SW","PAW","aetc") := CalcSW(SW_ini,
                                       SW_max,
                                       rain,
                                       swc,
                                       PETc), keyby = .(loc.id, year)]
})

... 

   user  system elapsed 
  0.004   0.000   0.004

结果：

head(df)

... 

   loc.id year day       rain       swc   SW_max     SW_ini      PETc       SW      PAW       aetc
1:      1 1980   1 0.35813251 28.360715 177.3943 0.69116310 0.2870478 1.038675 1.049296 0.01062025
2:      1 1980   2 1.10331116 37.013022 177.3943 0.02742273 0.4412420 2.125335 1.396808 0.01665171
3:      1 1980   3 1.76680011 32.509970 177.3943 0.66273062 1.1071233 3.807561 2.483467 0.08457420
4:      1 1980   4 3.20966558  8.252797 177.3943 0.12220454 0.3496968 6.840713 4.165693 0.17651342
5:      1 1980   5 1.32498191 14.784203 177.3943 0.66381497 1.2168838 7.573160 7.198845 0.59253503
6:      1 1980   6 0.02547458 47.903637 177.3943 0.21871598 1.0864713 7.418750 7.931292 0.17988449

我并非100％肯定我完全实现了你的逻辑，但逻辑应该非常简单地调整我可能错过的东西，我以非常类似的方式实现它。 / p>

另一个注意事项：通过自动缩进和代码突出显示（无论您是否使用RStudio或Emacs），您可以更轻松地编写C++您创建一个单独的文件，名称类似TestCode.cpp格式如下。

然后，您可以使用Rcpp::sourceCpp("TestCode.cpp")在R脚本中编译函数，也可以将除前三行之外的所有内容复制并粘贴为Rcpp::cppFunction()的参数。就像我上面做的那样。 

#include <Rcpp.h>
using namespace Rcpp;

// [[Rcpp::export]]
List CalcSW(NumericVector SW_ini,
                     NumericVector SW_max,
                     NumericVector rain,
                     NumericVector swc,
                     NumericVector PETc) {

  int n = SW_ini.length();
  NumericVector SW(n);
  NumericVector PAW(n);
  NumericVector aetc(n);

  double SW_ini_glob = SW_ini[0];
  double SW_max_glob = SW_max[0];

  SW[0] = SW_ini_glob;
  PAW[0] = SW[0] + rain[0];

  if (PAW[0] > swc[0]){
    aetc[0] = PETc[0];
  } else {
    aetc[0] = PAW[0]/swc[0]*PETc[0];
  }

  if (aetc[0] > PAW[0]){
    aetc[0] = PAW[0];
  }

  SW[0] = SW[0] + rain[0] - aetc[0];

  if(SW[0] > SW_max_glob){
    SW[0] = SW_max_glob;
  }

  if(SW[0] < 0){
    SW[0] = 0;
  }

  for (int i = 1; i < n; i++) {

    PAW[i] = SW[i-1] + rain[0];

    if (PAW[i] > swc[i]){
      aetc[i] = PETc[i];
    } else {
      aetc[i] = PAW[i]/swc[i]*PETc[i];
    }

    if (aetc[i] > PAW[i]){
      aetc[i] = PAW[i];
    }

    SW[i] = SW[i-1] + rain[i] - aetc[i];

    if(SW[i] > SW_max_glob){
      SW[i] = SW_max_glob;
    }

    if(SW[i] < 0){
      SW[i] = 0;
    }
  }
  return Rcpp::List::create(Rcpp::Named("SW") = SW,
                            Rcpp::Named("PAW") = PAW,
                            Rcpp::Named("aetc") = aetc);
}

答案 1 :(得分：1)

您可以将代码包装在另一个for循环中，并将每年df保存在列表中：

color

答案 2 :(得分：1)

来自Matt的data.table插图很好地说明了data.table的速度有多快，因为它可以在没有副本和移动数据的情况下进行计算。

但是，要回答关于使用管道的问题的关键，您可以使用group_by和do来完成您的目标（虽然比data.table慢得多）

下面我设置了Matt所做的相同虚拟数据。然后我使用你的函数（但在PETc上修复了案例）。它并不快，但它很容易遵循。 

df <- data.frame(loc.id = rep(1:10, each = 80*36), 
                 year = rep(rep(1980:2015, each = 80), times = 10),
                 day = rep(rep(1:80, times = 36),times = 10),
                 rain = runif(10*36*80, min = 0 , max = 5),
                 swc = runif(10*36*80,min = 0, max = 50),
                 SW_max = rep(runif(10, min = 100, max = 200), each = 80*36),
                 SW_ini = runif(10*36*80),
                 PETc = runif(10*36*80, min = 0 , max = 1.3) 
                 )

my_fun <- function(df){
  SW.ini <- 2 # setting some initial values 
  SW.max <- 5
  SW.min <- 0

  df$PAW <- NA
  df$aetc <- NA
  df$SW <- NA

  df$PAW[1] <- SW.ini + df$rain[1]

  df$aetc[1] <- ifelse(df$PAW[1] >= df$swc[1], df$PETc[1],(df$PAW[1]/df$swc[1])*df$PETc[1])
  df$aetc[1] <- ifelse(df$aetc[1] > df$PAW[1], df$PAW[1], df$aetc[1])
  df$SW[1] <- SW.ini + df$rain[1] -  df$aetc[1]
  df$SW[1] <- ifelse(df$SW[1] > SW.max, SW.max, ifelse(df$SW[1] < 0, 0,df$SW[1]))

  for (day in 2:nrow(df)){

    df$PAW[day] <- df$SW[day - 1] + df$rain[day]
    df$aetc[day] <- ifelse(df$PAW[day] >= df$swc[day], df$PETc[day], (df$PAW[day]/df$swc[day]) * df$PETc[day])
    df$aetc[day] <- ifelse(df$aetc[day] > df$PAW[day], df$PAW[day],df$aetc[day])
    df$SW[day] <- df$SW[day - 1] + df$rain[day] -  df$aetc[day]
    df$SW[day] <- ifelse(df$SW[day] > SW.max,SW.max, ifelse(df$SW[day] < 0, 0,df$SW[day]))
  }
  return(df)
}


library(tictoc)
library(tidyverse)


tic()
df  %>% 
  group_by(year) %>%
  do(my_fun(.)) -> 
  out
toc()
#> 5.075 sec elapsed
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