sharp - 1 year ago 50

R Question

I would like to find the preceding cumsum (i.e. cumsum minus the present row) for those rows of column B until the sum of the previous rows of column A including present row is <= 7.

I was able to find an answer using a traditional for loop. A vectorized implementation would be very helpful as I need to run it on a large dataset. Sharing my simple code in case it helps.

`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

}

}

}

Here are the 3 conditions to be satisfied:

- If A > 7, then Ans resets to 0
- If cumsum(A)<=7, then Ans is cumsum() of lagB
- If cumsum(A) > 7, then Ans is cumsum() of lagB for the range of previous rows of A for which the sum is <=7

Here is a simplified version of the data (Column A and B) and the desired output is the Column Ans:

`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.

Any help on how can I code this in R?

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Answer Source

**Please see the edit below** which tries to answer the updated question.

If I have understood OP's intention right, then there are 3 rules:

- if
`A`

is greater 7 then`Ans`

is zero and grouping is restarted - if
`cumsum(A)`

within the group is less or equal 7 then`Ans`

is the`cumsum()`

of lagged`B`

- if
`cumsum(A)`

within the group is greater 7 then`Ans`

is lagged`B`

The code below produces the expected result for the given sample data set:

```
# 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)`

creates unique group numbers for all consecutive streaks of `A`

values not greater or greater of 7, resp. The `ifelse()`

clause implements rules 2 and 3 within the grouping. By multiplying the result with `(A <= 7)`

, rule 1 is implemented., thereby using the trick that `as.numeric(TRUE)`

is 1 and `as.numeric(FALSE)`

is 0. Finally, the helper column is removed.

With the additional information provided by the OP, I believe there is *only one* rule left:

- for each row find a window extending backwards which contains as many rows as
`sum(A)`

does not exceed 7. The answer is the sum of lagged`B`

in the same window. - For clarification, if the window has zero length because
`A`

in the inital row already exceeds 7, then the answer is zero.

The variable length of the sliding window is the tricky part here:

```
# 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`

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