在运行时在数据表中创建列名进行计算

问题描述 投票：0回答：1

我在通过从其他作为变量的列名称计算它们的值并同时满足某些条件的情况下获取数据表来创建28个新列时遇到问题。

我试图实现的是区分四种情况：

  1. 字段在任者（ic）和新来者（nc）== 0 =>期望的结果= NA：这是第一个if子句，并且有效。
  2. 字段ic为> 0，但不存在nc（0）=>所需结果= ic_col的值（即新来者的数目）：这是第二个if子句，它不起作用。
  3. 字段ic == 0，但是存在nc（> 0）=>期望的结果= 0：这是第三个if子句，它可以工作。
  4. 字段ic> 1和nc> 1（=都存在）=>所需结果=两个数字之比，即ic_col / nc_col。这是第四个if子句，不起作用。

您可以看到我尝试了许多不同的变体（如果子句2尝试以一种DT方式进行操作），if子句4试图事先获取值（当然，这是行不通的，因为my_ic_value具有所有其中包含数千个值，而不是只有一行）。我尝试失败的其他方法是

all <- all[, ratio_col:= get(ic_col) / get(nc_col)), by=.(m_island_id_1, m_owner_id_1)]

也仅使用列名不起作用，因为它们是字符串，不能被分割：

all <- all[, ratio_col:= (ic_col / nc_col), by=.(m_island_id_1, m_owner_id_1)]

我希望问题已经很清楚了，我期待所有data.table专家向我展示我的代码有多么无希望的混乱，以及它能做多短而快！ ;-）

谢谢您的帮助！

这里有一些数据（一小部分）：

structure(list(m_island_id_1 = c("020d49b9580f071075cfb6f9da7a426669a56d2d", 
"020d49b9580f071075cfb6f9da7a426669a56d2d", "020d49b9580f071075cfb6f9da7a426669a56d2d"
), m_owner_id_1 = c("01d96c16a2caf720617b266ae5d053243cef0920", 
"4e25ef4cf04f023d96134856098b8104fa6a6add", "7818c5081257a837e7cfad2c1aba4d90dcd18a69"
), Sum_nc = c(2L, 2L, 2L), Sum_ic = c(4L, 4L, 4L), Sum_ic_2 = c(0L, 
0L, 0L), Sum_ic_3 = c(0L, 0L, 0L), Sum_ic_4 = c(0L, 0L, 0L), 
    Sum_ic_5 = c(0L, 0L, 0L), Sum_ic_6 = c(0L, 0L, 0L), Sum_ic_7 = c(0L, 
    0L, 0L), Sum_ic_8 = c(0L, 0L, 0L), Sum_ic_9 = c(0L, 0L, 0L
    ), Sum_ic_10 = c(0L, 0L, 0L), Sum_ic_11 = c(0L, 0L, 0L), 
    Sum_ic_12 = c(0L, 0L, 0L), Sum_ic_13 = c(0L, 0L, 0L), Sum_ic_14 = c(0L, 
    0L, 0L), Sum_ic_15 = c(0L, 0L, 0L), Sum_ic_16 = c(0L, 0L, 
    0L), Sum_ic_17 = c(1L, 1L, 1L), Sum_ic_18 = c(1L, 1L, 1L), 
    Sum_ic_19 = c(2L, 2L, 2L), Sum_ic_20 = c(0L, 0L, 0L), Sum_ic_21 = c(0L, 
    0L, 0L), Sum_ic_22 = c(0L, 0L, 0L), Sum_ic_23 = c(0L, 0L, 
    0L), Sum_ic_24 = c(0L, 0L, 0L), Sum_ic_25 = c(0L, 0L, 0L), 
    Sum_ic_26 = c(0L, 0L, 0L), Sum_ic_27 = c(0L, 0L, 0L), Sum_ic_28 = c(0L, 
    0L, 0L), Sum_nc_2 = c(0L, 0L, 0L), Sum_nc_3 = c(0L, 0L, 0L
    ), Sum_nc_4 = c(0L, 0L, 0L), Sum_nc_5 = c(0L, 0L, 0L), Sum_nc_6 = c(0L, 
    0L, 0L), Sum_nc_7 = c(0L, 0L, 0L), Sum_nc_8 = c(0L, 0L, 0L
    ), Sum_nc_9 = c(0L, 0L, 0L), Sum_nc_10 = c(0L, 0L, 0L), Sum_nc_11 = c(0L, 
    0L, 0L), Sum_nc_12 = c(0L, 0L, 0L), Sum_nc_13 = c(0L, 0L, 
    0L), Sum_nc_14 = c(0L, 0L, 0L), Sum_nc_15 = c(0L, 0L, 0L), 
    Sum_nc_16 = c(1L, 1L, 1L), Sum_nc_17 = c(0L, 0L, 0L), Sum_nc_18 = c(1L, 
    1L, 1L), Sum_nc_19 = c(0L, 0L, 0L), Sum_nc_20 = c(0L, 0L, 
    0L), Sum_nc_21 = c(0L, 0L, 0L), Sum_nc_22 = c(0L, 0L, 0L), 
    Sum_nc_23 = c(0L, 0L, 0L), Sum_nc_24 = c(0L, 0L, 0L), Sum_nc_25 = c(0L, 
    0L, 0L), Sum_nc_26 = c(0L, 0L, 0L), Sum_nc_27 = c(0L, 0L, 
    0L), Sum_nc_28 = c(0L, 0L, 0L), inhabited = c(TRUE, TRUE, 
    FALSE), Sum_time_ic = c(3L, 1L, 0L), groupsize = c(2L, 2L, 
    2L), Ratio_nc_to_ic_16 = c(0, 0, 0), Ratio_nc_to_ic_17 = c(TRUE, 
    TRUE, TRUE), Ratio_nc_to_ic_18 = c(1, 1, 1), Ratio_nc_to_ic_19 = c(TRUE, 
    TRUE, TRUE), ratio_col = c(NaN, NaN, NaN), Ratio_nc_to_ic = c(NA, 
    NA, NA), Ratio_nc_to_ic_ = c(2, 2, 2), Ratio_nc_to_ic_2 = c(NaN, 
    NaN, NaN), Ratio_nc_to_ic_3 = c(NaN, NaN, NaN), Ratio_nc_to_ic_4 = c(NaN, 
    NaN, NaN), Ratio_nc_to_ic_5 = c(NaN, NaN, NaN), Ratio_nc_to_ic_6 = c(NaN, 
    NaN, NaN), Ratio_nc_to_ic_7 = c(NaN, NaN, NaN), Ratio_nc_to_ic_8 = c(NaN, 
    NaN, NaN), Ratio_nc_to_ic_9 = c(NaN, NaN, NaN), Ratio_nc_to_ic_10 = c(NaN, 
    NaN, NaN), Ratio_nc_to_ic_11 = c(NaN, NaN, NaN), Ratio_nc_to_ic_12 = c(NaN, 
    NaN, NaN), Ratio_nc_to_ic_13 = c(NaN, NaN, NaN), Ratio_nc_to_ic_14 = c(NaN, 
    NaN, NaN), Ratio_nc_to_ic_15 = c(NaN, NaN, NaN), Ratio_nc_to_ic_20 = c(NaN, 
    NaN, NaN), Ratio_nc_to_ic_21 = c(NaN, NaN, NaN), Ratio_nc_to_ic_22 = c(NaN, 
    NaN, NaN), Ratio_nc_to_ic_23 = c(NaN, NaN, NaN), Ratio_nc_to_ic_24 = c(NaN, 
    NaN, NaN), Ratio_nc_to_ic_25 = c(NaN, NaN, NaN), Ratio_nc_to_ic_26 = c(NaN, 
    NaN, NaN), Ratio_nc_to_ic_27 = c(NaN, NaN, NaN), Ratio_nc_to_ic_28 = c(NaN, 
    NaN, NaN)), row.names = c(NA, -3L), class = c("data.table", 
"data.frame"), .internal.selfref = <pointer: 0x000001f487d51ef0>)

这是无效代码：

for (j in 2:28){
  ic_col <- paste0("Sum_ic_", j)
  nc_col <- paste0("Sum_nc_", j)
  ratio_col <- paste0("Ratio_nc_to_ic_", j)
  my_ic_value <- all[,get(ic_col)]
  my_nc_value <- all[,get(nc_col)]

  if ((ic_col == 0) & (nc_col == 0)) {
    all <- all[, ratio_col:=  NA, by=.(m_island_id_1, m_owner_id_1)] # no incumbents, no newcomers
  }
  if ((ic_col > 0) & (nc_col == 0)) {
    all <- all[, .(ratio_col=  ic_col), by=.(m_island_id_1, m_owner_id_1)] # Incumbents without newcomers
  }
  if ((ic_col == 0) & (nc_col > 0)) {
    all <- all[, ratio_col:=  0, by=.(m_island_id_1, m_owner_id_1)] # Newcomers without incumbents
  }
  if ((ic_col >= 1) & (nc_col >= 1)) {
    all <- all[, ratio_col:= (my_ic_value / my_nc_value), by=.(m_island_id_1, m_owner_id_1)] # Newcomers with incumbents
  }
}
r data.table calculated-columns
1个回答
0
投票

似乎是按行执行计算，因此您可以删除by（如果我错了，请纠正我）。使用矩阵运算可以覆盖最后两种情况。然后使用一些索引来处理情况1和2，如下所示：

DT[, (ratio_cols) := {
    icmat <- as.matrix(.SD[, mget(ic_cols)])
    ncmat <- as.matrix(.SD[, mget(nc_cols)])

    m <- icmat / ncmat

    #when nc_col==0
    m[is.infinite(m)] <- icmat[is.infinite(m)]

    #when ic_col==0 & nc_col==0
    m[icmat==0 & ncmat==0] <- NA_real_

    as.data.table(m)
}]

数据：

#where dat is your sample data and ratio cols are removed
DT <- dat[, .SD, .SDcols=m_island_id_1:groupsize]
rename <- c("Sum_nc","Sum_ic")
setnames(DT, rename, paste0(rename, "_1"))

jcols <- 1:28
ic_cols <- paste0("Sum_ic_", jcols)
nc_cols <- paste0("Sum_nc_", jcols)
ratio_cols <- paste0("Ratio_nc_to_ic_", jcols)
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