如何将先前的值表示为先前行的乘积

Question

I'm trying to initialize 3 columns of a data.table as a product of previous rows.
i have a formula like this:

tableA[0,] = values  

for each col:

tableA[i, col] = tableA[i - 1, col] * exp(tableB[i -1, col] * tableC[i, col]

(columns 1, 2 and 3 are completely independant)

I feel like i should use cumprod but i really don't see how (of course i can solve the problem with a for loop in 5sec)

Any one could please help me?

---

Also bonus question, would there be a reference site with some examples translating various mathematical formula with sums or prods so that I could familiarize my self with not using for loops.

I have written a little example below.
For ex the expected values for tableA[, .(valA1)] would be :

1.2, 1.620, 3.605, 8.867, 19.734  

assuming no mistake in excel

Thank you!

tableA <- data.table( valA1 = rep(0, 5), valA2 = rep(0, 5), valA3 = rep(0, 5))
tableA[1, valA1 := 1.2]
tableA[1, valA2 := 1.1]
tableA[1, valA3 := 1.3]
    
tableB <- data.table(valB1 = c(0.1, 0.2, 0.3, 0.4, 0.5),
                     valB2 = c(0.2, 0.3, 0.4, 0.5, 0.6),
                     valB3 = c(0.3, 0.4, 0.5, 0.6, 0.7))
tableC <- data.table(valC1 = c(2, 3, 4, 3, 2),
                     valC2 = c(1, 2, 3, 1, 2),
                     valC3 = c(5, 3, 3, 2, 2))
tableA[, ValB1 := shift(tableB$valB1)]
tableA[, ValB2 := shift(tableB$valB2)]
tableA[, ValB3 := shift(tableB$valB3)]
    
tableA[, ValC1 := tableC$valC1]
tableA[, ValC2 := tableC$valC2]
tableA[, ValC3 := tableC$valC3]
    
tableA[, Exp1 := exp(ValB1 * ValC1)]
tableA[, Exp2 := exp(ValB2 * ValC2)]
tableA[, Exp3 := exp(ValB3 * ValC3)]
    
tableA[-1, ValA1 := cumprod something?]

Answer 1

You were nearly there.

How to use cumprod()

According to OP's description, the rule for computing the ValA columns iteratively is

a _i = a _i-1 * f _i

where f _i are constants which are computed from the ValB and ValC columns

So, for i = 1 we get

a ₁ = a ₀ * f ₁ ,

for i = 2, we get

a ₂ = a ₁ * f ₂ = a ₀ * f ₁ * f ₂ ,

and so forth. If this is being continued to i = n we get

a _n = a ₀ * f ₁ * f ₂ * ... * f _n-1 * f _n = a ₀ * cumprod(f)

A fill parameter for shift() is required

Now, we need to consider the value of b ₀ . According to OP's formula

f _n = e ^{b _i-1 c _i}

For i = 1, this becomes

f ₁ = e ^{b ₀ c ₁}

so we need to define b ₀ as 0 in the call to shift()

shift(ValB, fill = 0)

so that f ₁ = 1. Otherwise, f ₁ would be NA and cumprod(f) would become NA as well.

Define a function

Now, this can be wrapped up in a function definition:

myfunc <- function(a, b, c) first(a) * cumprod(exp(shift(b, fill = 0) * c))

When called with the sample datsets

myfunc(tableA$valA1, tableB$valB1, tableC$valC1)

it returns the expected result:

[1] 1.200000 1.619831 3.604999 8.866867 19.733576

Dealing with multiple independent columns

The OP has pointed out that

columns 1, 2 and 3 are completely independent

Coding the same operations for a different set of columns is quite tedious. Therefore, I suggest to reshape and combine the datasets.

A <- melt(tableA, measure.vars = patterns("valA"), value.name = "a")
A[, b := melt(tableB, measure.vars = patterns("valB"))$value]
A[, c := melt(tableC, measure.vars = patterns("valC"))$value]
A[]

 variable abc 1: valA1 1.2 0.1 2 2: valA1 0.0 0.2 3 3: valA1 0.0 0.3 4 4: valA1 0.0 0.4 3 5: valA1 0.0 0.5 2 6: valA2 1.1 0.2 1 7: valA2 0.0 0.3 2 8: valA2 0.0 0.4 3 9: valA2 0.0 0.5 1 10: valA2 0.0 0.6 2 11: valA3 1.3 0.3 5 12: valA3 0.0 0.4 3 13: valA3 0.0 0.5 3 14: valA3 0.0 0.6 2 15: valA3 0.0 0.7 2

Now, all input data are combined in one data.table in long format whereby each independent dataset is identified by the value of variable . myfunc() can be applied on each group:

A[, myfunc(a, b, c), by = variable]

 variable V1 1: valA1 1.200000 2: valA1 1.619831 3: valA1 3.604999 4: valA1 8.866867 5: valA1 19.733576 6: valA2 1.100000 7: valA2 1.641007 8: valA2 4.036226 9: valA2 6.021342 10: valA2 16.367705 11: valA3 1.300000 12: valA3 3.197484 13: valA3 10.616021 14: valA3 28.857337 15: valA3 95.809732

This result can be reshaped to wide format again

dcast(A[, myfunc(a, b, c), by = variable], rowid(variable) ~ variable)[, variable := NULL][]

 valA1 valA2 valA3 1: 1.200000 1.100000 1.300000 2: 1.619831 1.641007 3.197484 3: 3.604999 4.036226 10.616021 4: 8.866867 6.021342 28.857337 5: 19.733576 16.367705 95.809732