更新：

Question

我有length(Date_List)天数，我有length(ISIN_Table$ID)个项目的信息。对于每一天（j中的循环），我创建了一个零的数据框，可以容纳所有项目（length(ISIN_Table$ID)）和一些列（4）。

每个项目在每个矩阵中都是一行，但根据日期会有不同的填充。

#create list that will hold matrices
df.list<-vector("list", length(Dates_List))
for (j in 1:(length(Dates_List))){
  df.list[[j]] <- data.frame(matrix(0, nrow = length(ISIN_Table$ID),ncol=4))
}

#Loop over number of days
for (j in 1:(length(Dates_List))){
  date<-Dates_List[j]
  #create empty dataframe 
  df.list[[j]] <- data.frame(matrix(0, nrow=length(ISIN_Table$ID), ncol=4))

  #loop over every item
  for (i in 1:(length(ISIN_Table$ID))){
    #check whether item is known at date
    if (nrow(data.raw[data.raw$ID==i & data.raw$Date==date,]) < 1){
      ID<-i
      df.list[[j]][i,1]<-date
      df.list[[j]][i,2]<-ID     #fill up the row
    }
    else{
      #fill up the row
      df.list[[j]][i,]<-c(
        as.character(data.raw[data.raw$ID==i & data.raw$Date==date,"Date"]),
        (data.raw[data.raw$ID==i & data.raw$Date==date,"ID"]),
        (data.raw[data.raw$ID==i & data.raw$Date==date,"Bid.Price"]),
        (data.raw[data.raw$ID==i & data.raw$Date==date,"Ask.Price"]))
    }
  }
}

代码给了我想要的确切输出，然而它却非常慢。我很感激有关如何提高速度的任何意见，目前的版本是不可行的。

更新：

# create dummy data:

Dates_List<-c("2007-01-02", "2007-01-03")
ISIN_Table<-data.frame(c(1,2,3))
colnames(ISIN_Table)<-"ID"
ID<-rep(1:2, len=2, each=1)
Date<-c("2007-01-02","2007-01-02","2007-01-03", "2007-01-03")
Bid.Price<-rep(100,4)
Ask.Price<-rep(100,4)
data.raw<-data.frame(ID, Date, Bid.Price, Ask.Price)

要求df.list [[1]]返回：

          X1 X2  X3  X4
1 2007-01-02  1 100 100
2 2007-01-02  2 100 100
3 2007-01-02  3   0   0

Answer 1

<强>更新根据@ Arun的建议，您可以在拆分之前添加缺失的行并完全避免使用mapply

Dates_List <- c("2007-01-02", "2007-01-03")
ISIN_Table <- data.frame(c(1, 2, 3))
colnames(ISIN_Table) <- "ID"
ID <- rep(1:2, len = 2, each = 1)
Date <- c("2007-01-02", "2007-01-02", "2007-01-03", "2007-01-03")
Bid.Price <- rep(100, 4)
Ask.Price <- rep(100, 4)
data.raw <- data.frame(ID, Date, Bid.Price, Ask.Price)

temp <- expand.grid(Dates_List, ISIN_Table$ID)
names(temp) <- c("Date", "ID")

data.raw <- merge(temp, data.raw, all.x = TRUE)
data.raw[is.na(data.raw)] <- 0
data.raw
##         Date ID Bid.Price Ask.Price
## 1 2007-01-02  1       100       100
## 2 2007-01-02  2       100       100
## 3 2007-01-02  3         0         0
## 4 2007-01-03  1       100       100
## 5 2007-01-03  2       100       100
## 6 2007-01-03  3         0         0


splitdata <- split(data.raw, data.raw$Date)

splitdata
## $`2007-01-02`
##         Date ID Bid.Price Ask.Price
## 1 2007-01-02  1       100       100
## 2 2007-01-02  2       100       100
## 3 2007-01-02  3         0         0
## 
## $`2007-01-03`
##         Date ID Bid.Price Ask.Price
## 4 2007-01-03  1       100       100
## 5 2007-01-03  2       100       100
## 6 2007-01-03  3         0         0

OLD ANSWER

您可以使用split按日期拆分数据，然后使用mapply和merge来获取甚至在给定日期没有任何数据的ID的行。

Dates_List <- c("2007-01-02", "2007-01-03")
ISIN_Table <- data.frame(c(1, 2, 3))
colnames(ISIN_Table) <- "ID"
ID <- rep(1:2, len = 2, each = 1)
Date <- c("2007-01-02", "2007-01-02", "2007-01-03", "2007-01-03")
Bid.Price <- rep(100, 4)
Ask.Price <- rep(100, 4)
data.raw <- data.frame(ID, Date, Bid.Price, Ask.Price)

splitdata <- split(data.raw, data.raw$Date)

mapply(FUN = function(x, date) merge(x, 
                          data.frame(ID = ISIN_Table$ID, 
                                     Date = rep(date, length(ISIN_Table$ID))), 
                                 all.y = TRUE), 
       splitdata, t(names(splitdata)), SIMPLIFY = FALSE)

## $`2007-01-02`
##   ID       Date Bid.Price Ask.Price
## 1  1 2007-01-02       100       100
## 2  2 2007-01-02       100       100
## 3  3 2007-01-02        NA        NA
## 
## $`2007-01-03`
##   ID       Date Bid.Price Ask.Price
## 1  1 2007-01-03       100       100
## 2  2 2007-01-03       100       100
## 3  3 2007-01-03        NA        NA

数据填充速度慢

更新：

1 个答案: