Question

我真的很难创建一个运行模型的函数，其中包含所有变量a，b，d，g＆amp; N有多个版本，如下面的data.table所示，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'
)

我要做的是运行下面的函数来计算lnN的值，并将其放入标题中与输入到模型中的变量具有相同编号的列中。即使用a1，b1，d1，g1＆amp; 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)]

}

我还没有找到一个关于如何实现这个目标的例子。我已经尝试使用mapply，但我看不到如何迭代每个变量的所有迭代。谢谢你的帮助！

Answer 1

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，只选择包含数字的列（以及rain），重命名列以删除数字，应用公式，重命名结果列以附加数字，然后cbind将结果放到原始表格上。

Answer 2

所以过去看了上面的评论，并意识到在尝试将迭代次数变为数千次时可能存在问题，因为Frank和Weihuang都建议我重新考虑如何构建数据。

我所做的是将随机生成的变量矩阵作为单独的数据帧。 e包含a，b，g＆amp;的多变量随机值。在d中N Nit {现在称为rnm}。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{{1}的列}。然后将bob合并到我在函数（bob）之前创建的空数据框中。这将循环，直到达到所需的循环次数。

然后我使用data.y将两者合并在一起，然后使用存储在数据框cbind中的名称依次重命名所有bob列，该数据框包含编号为{的列标题列表从我在Excel中生成的bobs文件中读取的bob.1到bob.n。

Answer 3

此处＆＃39; sa melt - 和 - dcast（recast）版本利用melt的{{1}}功能patterns来利用提供的名称。有关说明，请参阅newly implemented，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']

R Loop用于匹配的data.table列

3 个答案: