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I am working on dataframe transformations and was working Arun and Ricardo on a previous post

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Arun, suggested a brilliant solution ( matrix multiplication ) to achieve what i was trying to do.

That solution worked for a small data set like what i mentioned in the example, now i am running the same solution on a data frame which has the following sizes:

Total rows: 143345
Total Persons: 98461
Total Items :  30

Now, when i run the following command

 A <- acast(Person~Item+BorS,data=df,fun.aggregate=length,drop=FALSE)

I get this error..

Error: segfault from C stack overflow

Is this because, i dont have enough processing/memory power. My machine has 4 GB RAM, 2.8 GHz i7 processor ( Macbook) ? How do we handle these type of cases ?

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1 回答 1

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一个data.table解决方案。这是通过首先聚合，然后创建新的 data.table 并通过引用填充来实现的

library(data.table)

# some sample data
DT <- data.table(Person = sample(98461, 144000, replace = TRUE), item = sample(c(letters,LETTERS[1:4]), 144000, replace = TRUE), BorS = sample(c('B','S'), 144000, replace = TRUE))
# aggregate to get the number of rows in each subgroup by list item and BorS 
# the `length` of each subgroup
DTl <- DT[,.N , by = list(Person, item, BorS)]
# the columns you want to create
newn <- sort(DT[, do.call(paste0,do.call(expand.grid,list(unique(item),unique(BorS) )))])
# create a column which has this id combination in DTl
DTl[, comnb := paste0(item, BorS)]
# set the key so we can join / subset easily
setkey(DTl, comnb)
# create a data.table that has 1 row for each person, and has  columns for all the combinations
# of item and BorS
DTb <- DTl[, list(Person)][, c(newn) := 0L]
# set the key so we can join / subset easily
setkey(DTb, Person)
# this bit could be far quicker, but I think
# would require a feature request to data.table
for(nn in newn){
  # for each of the cominations extract which persons have
  # this combination >0
  pp <- DTl[list(nn), list(Person,N)]
  # for the people who have N > 0
  # assign the correct numbers in the correct column in DTb
  DTb[list(pp[['Person']]), c(nn) := pp[['N']]]
}

DTb要完成您的初始问题，您可以从矩阵中提取适当的列

A <- DTb[,-1,with = FALSE]

results <- crossprod(A)
于 2013-03-15T02:46:47.653 回答