查找一列的累积和，直到满足另一列的条件和

Question

我想为 B 列的那些行找到前面的 cumsum（即 cumsum 减去当前行），直到 A 列的前几行（包括当前行）的总和 <= 7。

我能够使用传统的 for 循环找到答案。矢量化实现会非常有用，因为我需要在大型数据集上运行它。分享我的简单代码以防有帮助。

dt <- data.frame(A = c(0, 2, 3, 5, 8, 90, 8, 2, 4, 1, 2),
                 B = c(1, 0, 4, 2, 3, 4, 2, 1, 2, 3, 1),
                 Ans = c(0, 1, 1, 4, 0, 0, 0, 2, 3, 5, 6),
                 new=rep(0,11))



dt3 <- dt
for (i in 2:nrow(dt3)){
  set<-0
  count<-0
  k=i-1
  for (j in k:1){
    count=count+dt3$A[j+1]
    if(count<=7){ 
      set<-set+dt3$B[j]
      if(j==1){
        dt3$new[i]=set
      }
    }
    else{
      dt3$new[i]=set
    }
  }
}

满足以下3个条件：

如果 A > 7，则 Ans 重置为 0
如果cumsum(A)<=7，那么Ans就是lagB的cumsum()
如果 cumsum(A) > 7，则 Ans 是 lagB 的 cumsum() 对于 A 的前几行的范围，其总和 <=7

这是数据的简化版本（A 和 B 列），所需的输出是 Ans 列：

dt <- data.frame(A = c(0, 2, 3, 5, 8, 90, 8, 2, 4, 1, 2),
                 B = c(1, 0, 4, 2, 3, 4, 2, 1, 2, 3, 1),
                 Ans = c(0, 1, 1, 4, 0, 0, 0, 2, 3, 5, 6))

dt
    A B Ans   Reason for value in Ans:
1   0 1   0       There are no preceeding rows in B so Ans is 0
2   2 0   1       Sum of value of A from row 2 to 1 is 2 <=7. So Ans is the value of B from first row = 1
3   3 4   1       Sum of value of A from row 3,2 and 1 is 5 <=7. So Ans is the sum of value of B in row 1 and 2, which is 1. 
4   5 2   4       Value of A from row 4 is 5 which is <=7. So Ans is value of B from row 3, which is 4
5   8 3   0       Value of A in row 5 is 8 which is >7. So Ans is 0 (Value of Ans resets to 0 when A > 7).
6  90 4   0
7   8 2   0
8   2 1   2        Value of A in row 8 is 2 which <=7, so Ans is value of B in row 7 which is 2
9   4 2   3        Sum of value of A from row 9 and 8 is 6<=7, so Ans is sum of value of B in row 8 and 7 = 3
10  1 3   5        Sum of value of A from row 10,9 and 8 is 7<=7, so Ans is sum of value of B in row 9,8 and 7 =5.
11  2 1   6        Sum of value of A from row 11,10 and 9 is 7<=7, so Ans is sum of value of B in row 10,9 and 8 =6.

关于如何在 R 中编写此代码的任何帮助？

Answer 1

请参阅下面的编辑，它试图回答更新后的问题。

如果我理解了OP的意图，那么有3条规则：

如果 A 大于 7 则 Ans 为零并重新开始分组
如果组内的cumsum(A)小于或等于7则Ans是滞后B

cumsum()

如果组内 cumsum(A) 大于 7，则 Ans 滞后 B

下面的代码为给定的示例数据集生成预期结果：

# create sample data set
DF <- data.frame(A = c(0, 2, 3, 5, 8, 90, 8, 2, 4, 1),
                 B = c(1, 0, 4, 2, 3, 4, 2, 1, 2, 3),
                 Ans = c(0, 1, 1, 4, 0, 0, 0, 2, 3, 5))
# load data.table, CRAN version 1.10.4 used
library(data.table)
# coerce to data.table
DT <- data.table(DF)
# create helper column with lagged values of
DT[, lagB := shift(B, fill = 0)][]
# create new answer
DT[, new := (A <= 7) * ifelse(cumsum(A) <= 7, cumsum(lagB), lagB), by = rleid(A <= 7)][
  , lagB := NULL][]

     A B Ans new
 1:  0 1   0   0
 2:  2 0   1   1
 3:  3 4   1   1
 4:  5 2   4   4
 5:  8 3   0   0
 6: 90 4   0   0
 7:  8 2   0   0
 8:  2 1   2   2
 9:  4 2   3   3
10:  1 3   5   5

rleid(A <= 7) 为所有连续的 A 值不大于或大于 7 的连续条纹创建唯一的组编号。 ifelse() 子句在分组中实现规则 2 和 3。通过将结果与(A <= 7)相乘，实现了规则1，从而利用了as.numeric(TRUE)为1，as.numeric(FALSE)为0的技巧，最后去掉了helper列。

编辑

根据 OP 提供的附加信息，我相信 只剩下一个 规则：

对每一行求一个向后延伸的window，其中包含的行数不超过sum(A)，不超过7。答案是同一个[=中滞后的B的总和68=].
澄清一下，如果 window 的长度为零，因为初始行中的 A 已经超过 7，则答案为零。

滑动的可变长度window是这里棘手的部分：

# sample data set consists of 11 rows after OP's edit
DF <- data.frame(A = c(0, 2, 3, 5, 8, 90, 8, 2, 4, 1, 2),
                 B = c(1, 0, 4, 2, 3, 4, 2, 1, 2, 3, 1),
                 Ans = c(0, 1, 1, 4, 0, 0, 0, 2, 3, 5, 6))
DT <- data.table(DF) 
DT[, lagB := shift(B, fill = 0)][]

# find window lengths
DT[, wl := DT[, Reduce(`+`, shift(A, 0:6, fill = 0), accumulate = TRUE)][, rn := .I][
  , Position(function(x) x <= 7, right = TRUE, unlist(.SD)), by = rn]$V1][]

# sum lagged B in respective window
DT[, new := DT[, Reduce(`+`, shift(lagB, 0:6, fill = 0), accumulate = TRUE)][
  , rn := .I][, wl := DT$wl][, ifelse(is.na(wl), 0, unlist(.SD)[wl]), by = rn]$V1][]

     A B Ans lagB wl new
 1:  0 1   0    0  7   0
 2:  2 0   1    1  7   1
 3:  3 4   1    0  7   1
 4:  5 2   4    4  1   4
 5:  8 3   0    2 NA   0
 6: 90 4   0    3 NA   0
 7:  8 2   0    4 NA   0
 8:  2 1   2    2  1   2
 9:  4 2   3    1  2   3
10:  1 3   5    2  3   5
11:  2 1   6    3  3   6

查找一列的累积和，直到满足另一列的条件和

Find a cumulative sum of one column until a conditional sum on another column is met

r

cumsum

编辑