6

使用data.tablein R，我正在尝试对不包括所选元素的子集进行操作。我正在使用by运算符，但我不知道这是否是正确的方法。

这是一个例子。例如，Deltain的IAH:SNA值为 (3+3)/2，它是 in 的平均值，Stops已IAH:SNA被Delta排除。

library(data.table)
s1 <- "Market   Carrier Stops
IAH:SNA     Delta     1
IAH:SNA     Delta     1
IAH:SNA Southwest     3
IAH:SNA Southwest     3
MSP:CLE Southwest     2
MSP:CLE Southwest     2
MSP:CLE  American     2
MSP:CLE   JetBlue     1"

d <- data.table(read.table(textConnection(s1), header=TRUE))

setkey(d, Carrier, Market)

f <- function(x, y){
         subset(d, !(Carrier %in% x) & Market == y, Stops)[, mean(Stops)]}

d[, s := f(.BY[[1]], .BY[[2]]), by=list(Carrier, Market)]

##     Market   Carrier Stops     s
## 1: MSP:CLE  American     2  1.666667
## 2: IAH:SNA     Delta     1  3.000000
## 3: IAH:SNA     Delta     1  3.000000
## 5: IAH:SNA Southwest     3  1.000000
## 6: IAH:SNA Southwest     3  1.000000
## 7: MSP:CLE Southwest     2  1.500000
## 8: MSP:CLE Southwest     2  1.500000

上面的解决方案在大型数据集（它本质上是一个 ）上表现很差，mapply但我不确定如何以类似快速data.table的方式做到这一点。

也许一个人可以（动态地）产生一个这样做的因素？我只是不确定如何。. .

有没有办法改进它？

编辑：只是为了它，这是一种获得上述更大版本的方法

library(data.table)
dl.dta <- function(...){
      ## input years ..
      years <- gsub("\\.", "_", c(...))
      baseurl <- "http://www.transtats.bts.gov/Download/"
      names <- paste("Origin_and_Destination_Survey_DB1BMarket", years, sep="_")
      info <- t(sapply(names, function(x) file.exists(paste(x, c("zip", "csv"), sep="."))))
      to.download <- paste(baseurl, names, ".zip", sep="")[!apply(info, 1, any)]
      if (length(to.download) > 0){
          message("starting download...")
          sapply(to.download,
                 function(x) download.file(x, rev(strsplit(x, "/")[[1]])[1]))}

      to.unzip <- paste(names,  "zip", sep=".")[!info[, 2]]
      if (length(to.unzip > 0)){
          message("starting to unzip...")
          sapply(to.unzip, unzip)}
      paste(names, "csv", sep=".")}

countWords.split <- function(x, s=":"){
    ## Faster on my machine than grep for some reanon
    sapply(strsplit(as.character(x), s), length)}

countWords.grep <- function(x){
    sapply(gregexpr("\\W+", x), length)+1}

fname <- dl.dta(2013.1)
cols <- rep("NULL", 41)
## Columns to keep: 9 is Origin, 18 is Dest, 24 is groups of airports in travel
## 30 is RPcarrier (reporting carrier).  
## For more columns: 35 is market fare and 36 is distance.
cols[9] <- cols[18] <- cols[24] <- cols[30] <- NA
d <- data.table(read.csv(file=fname,  colClasses=cols))
d[, Market := paste(Origin, Dest, sep=":")]
## should probably
d[, Stops := -2 + countWords.split(AirportGroup)]
d[, Carrier := RPCarrier]
d[, c("RPCarrier", "Origin", "Dest", "AirportGroup") := NULL]
4

2 回答 2

4

使用一点初等数学：

d[, c("tmp.mean", "N") := list(mean(Stops), .N), by = Market]
d[, exep.mean := (tmp.mean * N - sum(Stops)) / (N - .N), by = list(Market,Carrier)]

#     Market   Carrier Stops tmp.mean N exep.mean
# 1: IAH:SNA     Delta     1     2.00 4  3.000000
# 2: IAH:SNA     Delta     1     2.00 4  3.000000
# 3: IAH:SNA Southwest     3     2.00 4  1.000000
# 4: IAH:SNA Southwest     3     2.00 4  1.000000
# 5: MSP:CLE Southwest     2     1.75 4  1.500000
# 6: MSP:CLE Southwest     2     1.75 4  1.500000
# 7: MSP:CLE  American     2     1.75 4  1.666667
# 8: MSP:CLE   JetBlue     1     1.75 4  2.000000
于 2013-07-27T09:34:10.480 回答
3

@Roland 的答案适用于某些功能（当它起作用时最好），但不是一般情况下。不幸的是，您不能按原样将拆分-应用-组合策略应用于数据来完成任务，但是如果您使数据更大，则可以。让我们从一个更简单的例子开始：

dt = data.table(a = c(1,1,2,2,3,3), b = c(1:6), key = 'a')

# now let's extend this table the following way
# take the unique a's and construct all the combinations excluding one element
combinations = dt[, combn(unique(a), 2)]

# now combine this into a data.table with the excluded element as the index
# and merge it back into the original data.table
extension = rbindlist(apply(combinations, 2,
                  function(x) data.table(a = x, index = setdiff(c(1,2,3), x))))
setkey(extension, a)

dt.extended = extension[dt, allow.cartesian = TRUE]
dt.extended[order(index)]
#    a index b
# 1: 2     1 3
# 2: 2     1 4
# 3: 3     1 5
# 4: 3     1 6
# 5: 1     2 1
# 6: 1     2 2
# 7: 3     2 5
# 8: 3     2 6
# 9: 1     3 1
#10: 1     3 2
#11: 2     3 3
#12: 2     3 4

# Now we have everything we need:
dt.extended[, mean(b), by = list(a = index)]
#   a  V1
#1: 3 2.5
#2: 2 3.5
#3: 1 4.5

回到原始数据（并做一些稍微不同的操作，以简化表达式）：

extension = d[, {Carrier.uniq = unique(Carrier);
                 .SD[, rbindlist(combn(Carrier.uniq, length(Carrier.uniq)-1,
                          function(x) data.table(Carrier = x,
                                   index = setdiff(Carrier.uniq, x)),
                          simplify = FALSE))]}, by = Market]
setkey(extension, Market, Carrier)

extension[d, allow.cartesian = TRUE][, mean(Stops), by = list(Market, Carrier = index)]
#    Market   Carrier       V1
#1: IAH:SNA Southwest 1.000000
#2: IAH:SNA     Delta 3.000000
#3: MSP:CLE   JetBlue 2.000000
#4: MSP:CLE Southwest 1.500000
#5: MSP:CLE  American 1.666667
于 2013-07-27T18:01:40.063 回答