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Home/ Questions/Q 6866573
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T03:12:06+00:00 2026-05-27T03:12:06+00:00

I would like to make the code more efficient. The example creates a vector

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I would like to make the code more efficient.

The example creates a vector (called ‘new_vector’). The values of this ‘new_vector’ are changed based on if/else-conditions that refer to the values of three other vectors of the same length.

If the conditions are fulfilled, the corresponding elements of the ‘new_vector’ are updated using values from one of the other vectors (in the example elements of M_date are written into new_vector).

Here is the example code:

new_vector<-c(9,9,9)
S_date<-c(1,1,as.Date('2010/08/01'))
V_date<-c(1,as.Date('2010/09/01'),1)
M_date<-c(2,as.Date('2010/07/01'),1)

for (i in 1:3) {
    if ( (S_date[i]==1) & (V_date[i]==1 | M_date[i] < V_date[i]) ) {
    new_vector[i]<-M_date[i]
    }
}

The result of the example is:

> new_vector
[1]     2 14791     9

The example is simplified and in reality the vectors are larger and there are additional if/else-conditions.

How can I avoid the loop and use implicit methods for vector operations instead?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T03:12:07+00:00Added an answer on May 27, 2026 at 3:12 am

    If you write the expression without the [i] bits you get a vector True/False result:

    > S_date==1 & (V_date==1 | M_date < V_date)
    [1]  TRUE  TRUE FALSE
    

    assign that to a vector, and replace in new_vector by that result:

    > result = S_date==1 & (V_date==1 | M_date < V_date)
    > new_vector[result]=M_date[result]
    > new_vector
    [1]     2 14791     9
    

    Its a fairly general pattern. Compute a boolean vector, then replace those matching values with the corresponding values from another vector.

    It works because the FALSE value in the third element of result means that new_vector[3] doesn’t get touched.

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