匹配 data.table 列的 R 循环

R Loop for matched data.table columns

我真的很难创建一个 运行 模型的函数,其中所有变量 abdg & N 有多个版本,如 data.table 所示,下面我将其命名为 crm:

crm = data.table(
  East = 26500,
  North = c(115000, 120000, 125000, 130000, 135000, 140000), 
  rain = c(1049.61, 1114.31, 1361.61, 1407.2, 1499.56, 1654.13), 
  crop = 'Wheat', area = c(0.1718, 0.1629, 0.1082, 0.0494, 0.02, 0.004), 
  rn = c("10007", "10018", "10023", "10024", "10025", "10026"), 
  N1 = 184.262648839489, N2 = 180.312874871521, N3 = 178.615847839997,
  N4 = 182.531626054579, a1 = 0.186117715072018, a2 = -0.0232731908915799,
  a3 = 0.227017532149122, a4 = 0.162943230565506, b1 = 0.000478900233700419,
  b2 = 0.000787931973696371, b3 = 0.000458478256537521, b4 = 0.000517304324750896,
  d1 = -0.000328164576390286, d2 = -0.000112122093240884, d3 = 0.000112702113716146,
  d4 = 7.40875908059628e-05, g1 = 4.04709473710477e-06, g2 = 3.68724096485995e-06,
  g3 = 3.47214450131546e-06, g4 = 3.55825543257538e-06, key = 'rn'
)

我想要做的是 运行 下面的函数计算 lnN 的值并将其放入标题中与输入的变量具有相同数字的列中该模型。 IE。使用 a1b1d1g1 & N1 将为所有 2s、3s 和 4s 生成列 lnN1 等等.

n <- 1:4
cols <- paste0("lnN",n)
for(i in 1:length(n)){
crm[,(cols) := lapply(.SD ,function (x) {
  N = crm[,7+i]
  a = crm[,11+i]
  b = crm[,15+i]
  d = crm[,19+i]
  g = crm[,23+i]
a + (b*crm[,rain]) + (g*N) + (d*crm[,rain]*N)}), .SDcols = paste0("N",n)]

}

我还没有在任何地方找到有关如何完成此操作的示例。我试过使用 mapply 但我看不到如何在每个变量的所有迭代中迭代 mapply。感谢您的帮助!

怎么样:

library(dplyr)
cbind(crm, do.call(cbind, 
  lapply(1:4, function(x) {
    select(crm, c(contains(as.character(x)), rain)) %>% 
      setnames(gsub("[0-9]", "", names(.))) %>%
      transmute(lnN = a + (b*rain) + (g*N) + (d*rain*N)) %>%
      setnames(paste0("lnN", x))
  })
))

主要思想是,对于每个数字,select 只有包含数字的列(以及 rain),重命名列以删除数字,应用公式,重命名结果列追加数字,然后 cbind 结果到原始 table.

所以看了上面的评论并意识到在尝试将迭代次数更改为数千时可能会出现问题,因为 Frank 和 Weihuang 都建议我重新考虑我的数据结构。

我所做的是将随机生成的变量矩阵保留为单独的数据帧。 e 包含 abgd 的多元随机值,其中 N(现在称为 Nit) =18=]。所以 crm 现在只有前六列。代码如下所示:

for(i in 1:n){
  a = e[1,i]
  b = e[2,i]
  g = e[3,i]
  d = e[4,i]
Nit = rnm[i]
bob = a + (b*crm$rain) + (g*Nit) + (d*crm$rain*Nit)
data.y <- cbind(data.y, bob)
}
crm <- cbind(crm, data.y)
names(crm)[c(7:n)] = names(bobs)

对于 1:n 的每次迭代,它读取每个参数的 i 值(所有 1、所有 2 等)并将其放入模型中并创建一个名为的列bobbob 然后合并到我在函数 (data.y) 之前创建的空数据框中。循环直到达到所需的循环次数。

然后我使用 cbind 将两者合并在一起,然后使用存储在包含列标题列表的数据框 bobs 中的名称按顺序重命名所有 bob 列从我在 Excel 中生成的 .csv 文件中读取的编号 bob.1bob.n

这里有一个 melt-和-dcast (recast) 版本利用 newly implemented functionality of melt for patterns to utilize provided names. See the Installation wiki 作为说明,因为这是目前正在开发的功能。

library(data.table) 1.10.5+
# create character version of 1:N (Number of output columns)
N = paste0(seq_len(length(grep('^b', names(crm)))))
# join crm to a melt & recast version of itself using rain as 
#   the join key (note this will fail if the amount of rain may
#   not be unique -- in this case, we should include some ID in
#   id.vars, like rn, and adjust accordingly)
crm = crm[crm[ , melt(.SD, id.vars = 'rain', 
                      measure.vars = patterns(N = '^N[0-9]', a = '^a[0-9]', 
                                              b = '^b', d = '^d', g = '^g'))
               # use the formula to generate ln
               ][ , ln := a + b*rain + g * N + d * rain * N
                  # reshape wide
                  ][ , dcast(.SD, rain ~ variable, value.var = 'ln')
                     # rename the columns here
                     ][ , setnames(.SD, N, paste0('ln', N))],
          on = 'rain']

# by-reference version
crm[crm[ , melt(.SD, id.vars = 'rain', 
                measure.vars = patterns(N = '^N[0-9]', a = '^a[0-9]', 
                                        b = '^b', d = '^d', g = '^g'))
         # use the formula to generate ln
         ][ , ln := a + b*rain + g * N + d * rain * N
            # reshape wide
            ][ , dcast(.SD, rain ~ variable, value.var = 'ln')],
    # mget tends to be sort of slow, which is why I used the
    #   assign-by-copy approach first above; in larger examples,
    #   this slow-down may be outweighed by the cost of copying
    paste0('ln', N) := mget(N), on = 'rain']