Moving Elements in a Numeric Vector to the Front in APL

Moving Elements in a Numeric Vector to the Front in APL#

In this article, we will explore how to move elements of a numeric vector that have a certain value to the front of that vector in APL (A Programming Language).

Traditional Approach#

A traditional way to achieve this in APL is through sorting. However, it’s important to note that we don’t want to sort the vector itself, as that would not yield the desired outcome. Instead, we can sort by a certain criterion.

Using Comparisons and Sorting#

Let’s start with a small function (or Lambda) that compares the left argument to the right argument. This function will help us identify where the target value (in this case, the number 3) is located.

To find the elements that match the target value (3), we can use the equality comparison function. When we perform the comparison, we’ll find that all the numbers corresponding to the target value get rearranged to the front. For instance:

3 {⍺=⍵} 1 2 3 4 1 3 1 4 5

  • Output: 0 0 1 0 0 1 0 0 0

To flip the ones and zeros, we can use the unequal function instead. By doing this, if we sort so that the numbers corresponding to zeros come first, then we can easily identify the position of our target value.

Here’s how we can identify positions where elements are not equal to our target:

3 {⍺≠⍵} 1 2 3 4 1 3 1 4 5

  • Output: 1 1 0 1 1 0 1 1 1

Utilizing the Grade Function#

The grade function in APL provides us with the indices needed to select elements from the vector in such a way that they appear sorted. The output indicates that for the numeric vector to be sorted, we first sort based on the unequal comparison.

This can be done as follows:

3 {⍋⍺≠⍵} 1 2 3 4 1 3 1 4 5

  • Output: 3 6 1 2 4 5 7 8 9

Using these indices, we can reorder our original vector:

3 {⍵[⍋⍺≠⍵]} 1 2 3 4 1 3 1 4 5

  • Output: 3 3 1 2 4 1 1 4 5

It’s important to remember that instead of reordering the Boolean vector, we are reordering the right argument using indexing (with square brackets).

Array Manipulation Using Grading#

Utilizing grading and reordering based on the grade function effectively solves the problem. While this method is neat, there is an even shorter approach that can be employed using set functions.

Set Functions Approach#

Let’s consider the two arguments as sets. The goal is to find the intersection of these sets (the target value) and the set difference (all other elements). If we can combine the intersection and the set difference, we can arrive at a solution.

Implementing Set Difference and Intersection#

To find the set difference in APL, we can utilize the without function, which retrieves the elements on the left that are not in the right. This can be done with the following approach:

3 (~⍨) 1 2 3 4 1 3 1 4 5

  • Output: 1 2 4 1 1 4 5

Simultaneously, we can find the intersection, which will yield the target value as follows:

3 (∩⍨) 1 2 3 4 1 3 1 4 5

  • Output: 3 3

Now, combining both results can be done using concatenation:

3 (∩⍨,~⍨) 1 2 3 4 1 3 1 4 5

  • Output: 3 3 1 2 4 1 1 4 5

Another way to combine the results is by using the union operator instead of concatenation, which would yield:

3 (∩,~)⍨ 1 2 3 4 1 3 1 4 5

  • Output: 3 3 1 2 4 1 1 4 5

This demonstrates how we can resolve the problem using exclusively set functions effectively.

Conclusion#

Using the set functions approach not only provides a concise solution but also showcases the power and elegance of APL in manipulating vectors. Thank you for your attention, and I hope this explanation clarifies how to move elements in a numeric vector effectively!

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