数据表中的累积和,但每行使用分组

假设我有一个data.table如下(你可以把w看作一个分组变量):

set.seed(1)
prQ = CJ(Q1 = 1:10, Q2=1:10,w=1:2)
prQ[,pQ:=runif(100,0,1)]
prQ[,pQ:=pQ/sum(pQ),by=w]

  > prQ
     Q1 Q2 w          pQ
  1:  1  1 1 0.004889560
  2:  1  1 2 0.007553012
  3:  1  2 1 0.010549565
  4:  1  2 2 0.018433927
  5:  1  3 1 0.003714138
 ---                    
196: 10  8 2 0.016183006
197: 10  9 1 0.008384253
198: 10  9 2 0.008323492
199: 10 10 1 0.014932841
200: 10 10 2 0.012278353

如何为给定的w(称为CDF)计算新列,该列执行以下操作:

例如,假设Q1 = 4且Q2 = 6.定义一个新列,

对于所有Q1 <= 4且Q2 <= 6,CDF = sum(pQ),保持固定. 例如,一行:

CDF0 = sum(prQ[Q1<=4 & Q2<=6 & w==1,pQ])
prQ[Q1==4 & Q2==6,CDF:=CDF0]

对于给定的w,我想对所有行执行此操作.

使用强力完成所需的输出:

for(w0 in 1:2){
  for(j in 1:10){
    for(p in 1:10){
          CDF0 = sum(prQ[Q1<=j & Q2<=p & w==w0,pQ])
          prQ[Q1==j & Q2==p & w==w0,CDF:=CDF0]

    }
  }
}



  > head(prQ)
   Q1 Q2 w          pQ         CDF
1:  1  1 1 0.004889560 0.004889560
2:  1  1 2 0.007553012 0.007553012
3:  1  2 1 0.010549565 0.015439125
4:  1  2 2 0.018433927 0.025986939
5:  1  3 1 0.003714138 0.019153263
6:  1  3 2 0.018234648 0.044221587
最佳答案
根据pQ值构建矩阵内每个可能的子矩阵(行数=唯一Q2的数量和列数=唯一Q1的数量)的可能方法:

#ensure that order is correct as values will be used to generate the matrix 
#so that all elements in the top left sub-matrix will always be 
#smaller than or equal to the bottom right element of this sub-matrix
setorder(prQ, w, Q1, Q2)

#create all possible permutations of row and column indices
subMatIdx <- prQ[, CJ(as.integer(as.factor(Q1)), as.integer(as.factor(Q2)), unique=TRUE)]

#sum every sub matrix
prQ[, CDF :=
    {
        nr <- uniqueN(Q2)

        .(Map(function(i, j) sum(matrix(pQ, nrow=nr)[1L:j, 1L:i]), 
            subMatIdx[["V1"]], subMatIdx[["V2"]]))
    },
    by=.(w)]

输出:

     Q1 Q2 w          pQ        CDF
  1:  1  1 1 0.004889560 0.00488956
  2:  1  2 1 0.010549565 0.01543912
  3:  1  3 1 0.003714138 0.01915326
  4:  1  4 1 0.017396970 0.03655023
  5:  1  5 1 0.011585652 0.04813589
 ---                               
196: 10  6 2 0.001196193  0.5713282
197: 10  7 2 0.017785668  0.6535378
198: 10  8 2 0.016183006  0.7734989
199: 10  9 2 0.008323492   0.871678
200: 10 10 2 0.012278353          1

编辑:
如果Q1和Q2为负数或任何实数怎么办?
subMatIdx上的行应该已经处理好了.

例如.:

set.seed(1)
prQ = CJ(Q1 = -1:10, Q2=-1:10,w=1:2)
prQ[,pQ:=runif(nrow(prQ),0,1)]
prQ[,pQ:=pQ/sum(pQ),by=w]

setorder(prQ, w, Q1, Q2)

#create all possible permutations of row and column indices
subMatIdx <- prQ[, CJ(as.integer(as.factor(Q1)), 
    as.integer(as.factor(Q2)), unique=TRUE)]

prQ[, CDF := {
        nr <- uniqueN(Q2)

        .(Map(function(i, j) sum(matrix(pQ, nrow=nr)[1L:j, 1L:i]), 
            subMatIdx[["V1"]], subMatIdx[["V2"]]))
    },
    by=.(w)]

输出:

     Q1 Q2 w          pQ         CDF
  1: -1 -1 1 0.003607862 0.003607862
  2: -1  0 1 0.007784212  0.01139207
  3: -1  1 1 0.002740553  0.01413263
  4: -1  2 1 0.012836710  0.02696934
  5: -1  3 1 0.008548709  0.03551805
 ---                                
284: 10  6 2 0.011164332   0.6425251
285: 10  7 2 0.007638237   0.7360602
286: 10  8 2 0.005403923   0.8270053
287: 10  9 2 0.002008067   0.9193811
288: 10 10 2 0.002242777           1

转载注明原文:数据表中的累积和,但每行使用分组 - 代码日志