段错误

Question

我的工作在数据帧的转换，并在一前一后

Previous Post

阿伦工作阿伦和里卡多，提出了一个辉煌的解决方案（矩阵乘法）来实现我试图做。

这解决方案为小数据集像我在例子中提到，现在我运行它具有以下尺寸的数据帧在同一个解决方案：

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

现在，当我运行以下命令

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

我得到这个错误..

Error: segfault from C stack overflow 

这是becau se，我没有足够的处理/记忆能力。我的机器有4 GB RAM，2.8 GHz i7处理器（Macbook）？我们如何处理这些类型的案件？

Answer 1

A data.table解决方案。这是通过聚合，然后再创建新的data.table，并通过引用填充

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']]] 
} 

要完成你initital问题，您可以提取DTb相应列的矩阵

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

results <- crossprod(A)