保存R中嵌套for循环的输出

问题描述 投票:0回答:2

我正在试图弄清楚如何存储我将在多次迭代中运行的模拟输出。我在这里有一个简化的例子,我模拟了50年后冬季温度的变化,我想这样做3次:

# function to draw the winter temperature in a particular year
draw_winter_temperature <- function(){
  intercept <- winterIntercept
  beta1 <- rnorm(1, winterBeta, winterBetaSD)
  val = rnorm(1, intercept + beta1 * years[i], winterSigma)
  return(val)
}

# parameters for winter temperature function
winterIntercept <- -0.633
winterBeta <- 0.08
winterBetaSD <- 0.04
winterSigma <- 1.30 

# number of iterations
iter = seq(1,3,1)

# number of years to run in each simulation
years <- seq(1,50,1)

# matrix to store 50 winter temperatures for each iteration? Not sure best approach
temps <- matrix(NA, nrow = length(years), ncol = length(iter))

# start of nested for loop - loop over each iteration
for (i in seq_along(iter)) {

  # within each iteration, loop over each year
  for (j in seq_along(years)) {

  # draw winter temp. anomaly
  winterTemp <- draw_winter_temperature()

  # store winter temps, not sure what to do here
  temps <- winterTemp

  }
}

我想我有点不确定索引在这里将输出保存在我的空矩阵中(或以另一种格式保存)。我应该说多年来我的循环中还存储了其他函数和数据,所以我常常对每次迭代存储数据的最佳方法感到困惑。

r for-loop
2个回答
1
投票

我完全避免了循环。使用迭代和年份创建数据框:

data1 <- data.frame(iter = rep(1:3, each = 50), 
                    year = rep(1:50, 3))

以稍微不同的方式定义函数,以便提供所有默认参数:

draw_winter_temperature <- function(intercept = -0.633, 
                                    winterbeta = 0.08, 
                                    winterbetasd = 0.04,
                                    wintersigma = 1.30,
                                    year = 1){
  beta1 <- rnorm(1, winterbeta, winterbetasd)
  val   <- rnorm(1, intercept + beta1 * year, wintersigma)
  val
}

使用purrr::map_dbl()dplyr::mutate()创建一个每年采样值的列:

library(dplyr)
library(purrr)

data1 <- data1 %>% 
  mutate(winterTemp  = map_dbl(year, ~draw_winter_temperature(year = .)))

0
投票

我想你需要temps中的[j,i]子集。此外,重新定义函数以跨越年份参数:

> rm(list = ls())
> 
> # parameters for winter temperature function
> winterIntercept <- -0.633
> winterBeta <- 0.08
> winterBetaSD <- 0.04
> winterSigma <- 1.30 
> 
> # number of iterations
> iter = seq(1,3,1)
> 
> # number of years to run in each simulation
> years <- seq(1,50,1)
> 
> # matrix to store 50 winter temperatures for each iteration? Not sure best approach
> temps <- matrix(NA, nrow = length(years), ncol = length(iter))
> 
> draw_winter_temperature <- function(i) {
+   intercept <- winterIntercept
+   beta1 <- rnorm(1, winterBeta, winterBetaSD)
+   val = rnorm(1, intercept + beta1 * years[i], winterSigma)
+   return(val)
+ }
> 
> 
> # start of nested for loop - loop over each iteration
> for (i in seq_along(iter)) {
+   
+   # within each iteration, loop over each year
+   for (j in seq_along(years)) {
+     
+     # draw winter temp. anomaly
+     winterTemp <- draw_winter_temperature(j)
+     
+     # store winter temps, not sure what to do here
+     temps[j, i] <- winterTemp
+     
+   }
+ }
> 
> temps
             [,1]        [,2]        [,3]
 [1,]  0.07484416 -2.09659305  1.36146643
 [2,]  0.91077388  0.91630545  0.15919461
 [3,]  2.07646459  1.47967364 -0.80436293
 [4,] -1.06727442 -1.35760417 -2.23396119
 [5,]  0.67833459 -0.49733006 -0.08458830
 [6,] -2.94754201 -0.83298461 -0.09547269
 [7,] -0.27718212 -1.26709144  1.07166328
 [8,] -2.48913267  1.15023058 -0.45768205
 [9,]  0.81201831  1.73791432 -1.12876304
[10,] -1.26716059 -0.59580422 -1.28796699
[11,]  1.73244202  1.45400888 -0.13374732
[12,]  0.06181381  1.42056333 -0.28493854
[13,]  1.32197105 -0.26822322 -0.64071024
[14,]  0.31681581  2.17959219  1.05139421
[15,] -0.40649165  0.40584711 -1.24801466
[16,] -0.37852770 -0.01912727 -1.57738425
[17,]  1.07342597 -0.37121912 -1.78913387
[18,]  2.37652192  2.34135284  1.57626300
[19,]  2.26526300 -0.22457651  1.22608845
[20,]  0.43476700  1.20513841  3.15556064
[21,]  2.59591314 -0.30772162 -0.18688845
[22,]  1.06465892  1.71525510 -1.96024033
[23,]  0.33484399  1.81225896  2.11832405
[24,]  1.90063594  3.51556209  1.07306117
[25,] -0.01940422  2.28460208  3.21963212
[26,]  3.28816991  2.95440512  3.24107428
[27,]  0.96753190 -0.10513793 -1.21244216
[28,]  2.54233795 -0.71844973  1.81812858
[29,]  2.17954621  2.98536138  3.66748584
[30,]  0.81877427 -1.35965678  2.56338093
[31,]  0.29897273  1.11832993  0.87086081
[32,]  3.41605791  4.72562882  4.07046433
[33,]  3.48441756 -0.09328612  1.64075691
[34,] -1.45902220 -0.38229884  4.62577573
[35,]  0.76568048  3.05618914 -2.04568644
[36,]  0.51939827  0.46442133  2.45079102
[37,]  3.31122478  3.72294771  3.79475270
[38,]  3.22139644  2.33785724 -0.38787472
[39,]  2.95662591  1.14475461  2.97768620
[40,]  0.41263989  3.12365715 -0.05141565
[41,]  1.38138984  3.76526281  1.96114126
[42,] -0.18932163  3.05150559  1.01252868
[43,]  2.60596068  4.58785954 -0.62515705
[44,]  4.11088547  1.61733269  2.51478319
[45,]  2.54634136  5.03258017  3.69739035
[46,]  3.90668226  5.10453707  9.44805817
[47,] -0.01088235 -0.72829163  5.23958084
[48,]  5.69817388  0.12111318  4.43908874
[49,]  0.62863949  1.44337538  6.37749772
[50,]  6.03095905 -1.06661804  2.96625244

它也可以使用嵌套的sapply来完成:

> sapply(seq(1, 3, 1), function(x){
+   sapply(seq(1, 50, 1), function(i) {draw_winter_temperature(i)})
+ })
             [,1]       [,2]        [,3]
 [1,] -0.89724777 -0.3248484 -0.90899374
 [2,] -1.94179327 -1.1282839  0.31149541
 [3,] -0.42025521  0.7724711  0.74888190
 [4,] -1.33109020  1.9129040  0.87848969
 [5,] -0.47958978  0.0847167  0.41101536
 [6,] -0.79725942 -0.3091579 -1.24232376
 [7,]  1.77518276 -0.1922316 -1.47421230
 [8,]  0.69493304 -0.2019549  0.13570145
 [9,] -1.30457205  0.3104541  0.42987422
[10,]  0.17036195 -1.7092747 -0.83657188
[11,]  0.32147101 -0.9007392 -1.19378838
[12,]  0.67253811  1.3546455  0.65988074
[13,]  1.50350007  0.5492711  1.79631547
[14,] -2.51666246 -0.9415391 -0.29810167
[15,] -0.40993616 -0.4815619 -0.64542730
[16,]  1.77118058  0.1315932 -0.75448968
[17,]  0.50048465  2.7956318 -0.87173764
[18,] -3.21569892 -0.3323585 -1.23359412
[19,]  0.31825702  0.3584551  0.06093396
[20,]  3.89144244 -2.4149095  1.78682882
[21,] -0.43390072  0.7793848 -0.10369739
[22,] -2.86730666 -0.7670008  1.52314298
[23,] -1.79228394  0.9572841  0.48042120
[24,] -0.43718475 -0.1275890  1.21647106
[25,] -1.15994958  1.6406028  4.60036997
[26,]  0.01317481  1.4994584  2.66739115
[27,]  4.17871392  0.7907429  2.73681377
[28,]  4.28581141  1.6371291  0.68691671
[29,]  1.93210763  2.9950647  4.21374229
[30,]  2.58606149  1.0610567 -0.85244266
[31,]  1.80716913  4.5783347  2.52572309
[32,] -0.59753770  1.4036211  4.67023114
[33,] -0.62456359  3.8556583  1.45738330
[34,]  3.97709081  1.9414286  2.13243487
[35,] -0.44078059  0.1845302  2.93266443
[36,]  3.95186775  0.6522162  1.33296220
[37,]  2.44364830  2.8547030  1.20660793
[38,]  4.41777409  1.9200504  2.71936027
[39,]  0.28099830  2.7876076  3.92335543
[40,]  3.40916089  0.1663190  0.42249808
[41,]  4.14691320  2.2004945  2.03032191
[42,]  0.61603992  4.6339235  2.82358688
[43,]  3.15565113  4.2119661  5.04334175
[44,]  2.63957253  5.9546210  2.77191218
[45,] -1.50822755  0.9417640  6.71999760
[46,]  1.29101617  3.2551090  1.59456129
[47,]  5.98411236  2.0040345  4.03826529
[48,]  2.33718663 -3.1029647  2.84654953
[49,] -1.77860722  4.4929658  2.93310143
[50,]  7.12512178  5.3284199  2.97240287

希望能帮助到你。

© www.soinside.com 2019 - 2024. All rights reserved.