Question

我对R和编程很新。我想知道是否有人可以帮助我以下内容：我在R中有一个for循环，每次迭代取决于之前的所有。我也有一个200k行的数据集，所以这个for循环需要永远执行。有没有更快的方法来做到这一点？

这是我的代码：

library(data.table)

for (i in 2:nrow(DT)) {
    if (DT$price2[i]%in%DT$price2[1:i-1]) {   #if price occurred before
      DT$dummyid[i] = DT$id[max(which(DT$price2[i] == DT$price2[1:i-1]))]  #record the latest id of this price's order
      if (sum(DT$size[(which(DT$price2[i] == DT$price2[1:i-1]))]) == 0) #if cumulative sum == 0 (excl. current order)
      {DT$id[i] = i} #means order just got cancelled, so assign new id
      else {DT$id[i] = DT$dummyid[i]}  #assign the latest id
    }
    else {DT$id[i] = i} #price never occurred, so new id

    if (DT$id[i]%in%DT$id[1:i-1]) {    #if id occurred before
      if (sum(DT$size[(which(DT$id[i] == DT$id[1:i]))]) == 0) #and cumulative sum == 0 (incl. current order)
        {DT$arc[i] = "C"}  #cancel the order
        else {
          DT$arc[i] = "R" #replace the order
          DT$price3[i] = sum(DT$size[(which(DT$id[i] == DT$id[1:i]))])  #with order of this size
      }
    }
    else {
      DT$arc[i] = "A"   #add new order
    }
  }
}

这里我正在尝试从TAQ NYSE OpenBook重新格式化原始数据，以便订单包可以读取它。如果有人知道其他任何方式，我会非常感激。

基本上，原始数据（最左边的4列）是特定股票当日的买入/卖出订单的价格变化。

我在这里手动编辑了第一行。此处键入0 = BID，1 = ASK，price2 = price + type * 0.000001以区分出价/要求之间的相同价格。尺寸并不总是正面的。

我正在尝试使用代码：

我正在尝试以与一个订单相同的价格处理所有订单，直到它的累积尺寸== 0，这对我来说是可以的。 arc列应为A = add（如果价格之前从未发生过，或者此价格的订单刚被取消），C =取消（如果累计大小== 0），R =替换（如果价格发生之前）和累积大小＆lt;＆gt; 0）。

我还需要id来指示哪个订单被取消，添加或替换。

编辑：

很抱歉不包含示例数据集。这是：

> head(DT, n = 20)
    time type price size price2 dummyid id arc price3
 1:    0    0 20.00  200  20.00       0  1   A     NA
 2:    0    0 24.41  200  24.41      NA NA  NA     NA
 3:    0    0 32.50  200  32.50      NA NA  NA     NA
 4:    0    0 38.40 1000  38.40      NA NA  NA     NA
 5:    0    0 40.50 1700  40.50      NA NA  NA     NA
 6:    0    0 41.50  100  41.50      NA NA  NA     NA
 7:    0    0 41.69  100  41.69      NA NA  NA     NA
 8:    0    0 42.28  100  42.28      NA NA  NA     NA
 9:    0    0 43.00  100  43.00      NA NA  NA     NA
10:    0    0 45.00 1700  45.00      NA NA  NA     NA
11:    0    0 45.12  300  45.12      NA NA  NA     NA
12:    0    0 45.76  200  45.76      NA NA  NA     NA
13:    0    0 46.00  100  46.00      NA NA  NA     NA
14:    0    0 46.76  200  46.76      NA NA  NA     NA
15:    0    0 47.00  200  47.00      NA NA  NA     NA
16:    0    0 48.00  500  48.00      NA NA  NA     NA
17:    0    0 48.10  100  48.10      NA NA  NA     NA
18:    0    0 48.25  400  48.25      NA NA  NA     NA
19:    0    0 48.71  300  48.71      NA NA  NA     NA
20:    0    0 49.05  500  49.05      NA NA  NA     NA

> dput(head(DT,20))
structure(list(time = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0), type = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), price = c(20, 
24.41, 32.5, 38.4, 40.5, 41.5, 41.69, 42.28, 43, 45, 45.12, 45.76, 
46, 46.76, 47, 48, 48.1, 48.25, 48.71, 49.05), size = c(200L, 
200L, 200L, 1000L, 1700L, 100L, 100L, 100L, 100L, 1700L, 300L, 
200L, 100L, 200L, 200L, 500L, 100L, 400L, 300L, 500L), price2 = c(20, 
24.41, 32.5, 38.4, 40.5, 41.5, 41.69, 42.28, 43, 45, 45.12, 45.76, 
46, 46.76, 47, 48, 48.1, 48.25, 48.71, 49.05), dummyid = c("0", 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA), id = c("1", NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), arc = c("A", NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA), price3 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA)), .Names = c("time", "type", 
"price", "size", "price2", "dummyid", "id", "arc", "price3"), class = c("data.table", 
"data.frame"), row.names = c(NA, -20L))

谢谢！

for循环取决于之前的所有迭代

0 个答案: