R：具有 2 个大数据集的模式匹配金融时间序列数据：

Question

我的问题可能比较复杂，请耐心看完

我正在处理以下案例，我有来自 2 个交易所（纽约和伦敦）的金融时间序列的两个时间数据集

两个数据集如下所示：

伦敦数据集：

Date        time.second Price
2015-01-05  32417   238.2
2015-01-05  32418   238.2
2015-01-05  32421   238.2
2015-01-05  32422   238.2
2015-01-05  32423   238.2
2015-01-05  32425   238.2
2015-01-05  32427   238.2
2015-01-05  32431   238.2
2015-01-05  32435   238.47
2015-01-05  32436   238.47

纽约数据集：

NY.Date     Time    Price
2015-01-05  32416   1189.75
2015-01-05  32417   1189.665
2015-01-05  32418   1189.895
2015-01-05  32419   1190.15
2015-01-05  32420   1190.075
2015-01-05  32421   1190.01
2015-01-05  32422   1190.175
2015-01-05  32423   1190.12
2015-01-05  32424   1190.14
2015-01-05  32425   1190.205
2015-01-05  32426   1190.2
2015-01-05  32427   1190.33
2015-01-05  32428   1190.29
2015-01-05  32429   1190.28
2015-01-05  32430   1190.05
2015-01-05  32432   1190.04

可以看到，有3列：日期，时间（秒），价格

我想做的是利用london数据集作为参考，找到最近但更早的数据项在纽约数据集中。

最近但更早是什么意思？我的意思是，例如，

"2015-01-01","21610","15.6871"在伦敦数据集中，我想找到纽约数据集中的数据，在同一天，并且最近但更早或相同的时间，看看我当前的程序会很有帮助]:

# I am trying to avoid using for-loop
for(i in 1:dim(london_data)[1]){ #for each row in london data set
    print(i)
    tempRow<-london_data[i,]
    dateMatch<-(which(NY_data[,1]==tempRow[1])) # select the same date
    dataNeeded<-(london_before[dateMatch,]) # subset the same date data
    # find the nearest but earlier data in NY_data set
    Found<-dataNeeded[which(dataNeeded[,2]<=tempRow[2]),] 
    # Found may be more than one row, each row is of length 3
    if(length(Found)>3)
    {    # Select the data, we only need "time" and "price", 2nd and 3rd  
         # column
         # the data is in the final row of **Found**
         selected<-Found[dim(Found)[1],2:3] 
         if(length(selected)==0) # if nothing selected, just insert 0 and 0
             temp[i,]<-c(0,0)
         else
            temp[i,]<-selected
     }
     else{ # Found may only one row, of length 3
         temp[i,]<-Found[2:3] # just insert what we want
     }
   print(paste("time is", as.numeric(selected[1]))) #Monitor the loop
 }
 res<-cbind(london_data,temp)
 colnames(res)<-c("LondonDate","LondonTime","LondonPrice","NYTime","NYPrice")

上面列出的数据集的正确输出是**（仅部分）**：

      "LondonDate","LondonTime","LondonPrice","NYTime","NYPrice"
 [1,] "2015-01-05" "32417"      "238.2"       "32417"    "1189.665" 
 [2,] "2015-01-05" "32418"      "238.2"       "32418"    "1189.895" 
 [3,] "2015-01-05" "32421"      "238.2"       "32421"    "1190.01"  
 [4,] "2015-01-05" "32422"      "238.2"       "32422"    "1190.175" 
 [5,] "2015-01-05" "32423"      "238.2"       "32423"    "1190.12"  
 [6,] "2015-01-05" "32425"      "238.2"       "32425"    "1190.205" 
 [7,] "2015-01-05" "32427"      "238.2"       "32427"    "1190.33"  
 [8,] "2015-01-05" "32431"      "238.2"       "32430"    "1190.05"  
 [9,] "2015-01-05" "32435"      "238.47"      "32432"    "1190.04"  
 [10,] "2015-01-05" "32436"      "238.47"      "32432"    "1190.04"

我的问题是，伦敦数据集有超过 5,000,000 列，我试图避免 for-loop 但我仍然 至少需要一个，以上程序运行成功，但耗时约24小时。

如何避免使用 for 循环并加速程序？

我们将不胜感激。

Answer 1

在 @Jan Gorecki 评论的基础上使用 data.table 这是解决方案：

library(data.table)

df1 <- data.table(Date=rep("05/01/2015", 10),   
              time.second=c(32417, 32418, 32421, 32422, 32423, 32425, 32427, 32431, 32435, 32436),  
              Price=c(238.2, 238.2, 238.2, 238.2, 238.2, 238.2, 238.2, 238.2, 238.47, 238.47))

df2 <- data.table(NY.Date=rep("05/01/2015", 16),    
              Time=c(32416, 32417, 32418, 32419, 32420, 32421, 32422, 32423, 32424, 32425, 32426, 32427, 32428, 32429, 32430, 32432),   
              Price=c(1189.75, 1189.665, 1189.895, 1190.15, 1190.075, 1190.01, 1190.175, 1190.12, 1190.14, 1190.205, 1190.2, 1190.33, 1190.29, 1190.28, 1190.05, 1190.04))


setnames(df2, c("Date", "time.second", "NYPrice"))

setkey(df1,"Date", "time.second")
setkey(df2,"Date", "time.second")

df2[, NYTime:=time.second]

df3 <- df2[df1, roll=TRUE]
df3

R：具有 2 个大数据集的模式匹配金融时间序列数据：

R: Pattern-matching financial time-series data with 2 large data sets:

r

time-series

apply

lapply

data.table