CODEWITHSUNDEEP
r
Akshay Geedan
Akshay Geedan

Reputation: 51

Can some one help in finding efficient way to solve the mathematical problem

I have 5 parameters with some values in a named vector.

Vector 1 =("param1" = 10,"param2"=20,param3=30,param4=20,param5=60)

Param1 has some multiplier value which is 3(param1 *3) param2 has some multiplier value which is 2 (Param2 *2)

For param3,param4,param5 we have to take mean if any those three parameters presen and its multiplier values is one (average of three params * 1)

For example if the values are present in Vector 1: Note : average of three params is param3,param4,param5

result = param1 * 3 + param2 *2 + average of three params * 1 / 3 + 2 + 1

If any parameter in the vector1 is not present the code should produce output for

example: if param1 not present in Vector 1

result = param2 * 2 + average of three params * 1 / 2 + 1

the expression should be dynamic based on parameters present in Vector 1

Example

Input = c("param1" = 10,"param2" = 20, "param3" = 40, "param4"= 60, "param5 = 20")

If all the parameters are present in the Input vec the expression should be:

expression = 10*3 + 20*2 + (40+60+20/3) * 1 / 3+2+1

If param2 is not present in the Input vec the expression should be:

expression = 10*3 + (40+60+20/3) * 1 / 3 + 1

If param4 is not present the expression should be :

expression = 10*3 + 20 * 2 + (40+20/2) * 1 / 3 + 2 + 1

If param3, param4, param5 not present

expression = 10 * 3 + 20 * 2 / 3 + 2

Upvotes: 0

Views: 73

Answers (2)

user2332849
user2332849

Reputation: 1450

You can benefit from several facts:

1) When you index a named vector by a name that does not exist, it returns NA.

2) A mathematical operation involving NA will also return NA.

3) The sum() and mean() functions have an optional parameter na.rm, that lets you remove NA values from the computation automatically.

4) A mathematical operation involving a logical condition will treat the logical value (TRUE or FALSE) as numerical 1 or 0.

So your code could be like this:

Vector1 <- c(param1=10, param2=20, param3=30, param4=20, param5=60)

expression <- sum(
  Vector1["param1"] * 3,
  Vector1["param2"] * 2,
  mean(c(Vector1["param3"], Vector1["param4"], Vector1["param5"]), na.rm = TRUE),
  na.rm = TRUE
) / sum(
  3 * !is.na(Vector1["param1"]),
  2 * !is.na(Vector1["param2"]),
  1 * !is.nan(mean(c(Vector1["param3"], Vector1["param4"], Vector1["param5"]), na.rm = TRUE))
)

Upvotes: 2

ThomasIsCoding
ThomasIsCoding

Reputation: 101848

Maybe you can try the code like below

wts <- c(3,2,1)
lst <- list("param1","param2",c("param3","param4","param5"))
u <- unlist(Map(function(v) mean(Input[names(Input)%in% v]), lst))
result <- sum(u*wts,na.rm = TRUE)/sum(wts[!is.na(u)])

Example

> Input
param1 param2 param3 param4 param5 
    10     20     40     60     20 
> result
[1] 18.33333


> Input
param1 param3 param4 param5 
    10     40     60     20 
> result
[1] 17.5


> Input
param1 param2 
    10     20 
> result
[1] 14


> Input
param3 param4 param5 
    40     60     20 
> result
[1] 40

Upvotes: 1

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