Unexpected collection behaviour

1

I am getting an unexpected result to the following

a= Array.rand(20,1,14)

a.do({|item,i| if (item < 5, {a.removeAt(i)})})

I would expect that all values lower then 5 would be expunged from a. that is almost what i get
before
→ [ 6, 7, 3, 1, 13, 14, 9, 9, 14, 2, 7, 2, 8, 3, 8, 3, 7, 6, 14, 9 ]
after
→ [ 6, 7, 1, 13, 14, 9, 9, 14, 7, 8, 8, 7, 6, 14, 9 ]
I suspect the 1 is still there because it appears after the 3 but I don’t understand why.
I am aware that select would work better in this case. However, in the actual case I will have two corresponding lists a and b and I will need to expunge values lower then x from a and the corresponding (i.e. in the same position) values from b. So the real code will also remove an item from b.

2

When you remove an item from the array, all subsequent items “shift left”. The loop continues with the following index, potentially skipping an element.

  1. I would create a new array instead of an in-place operation (I prefer immutability when possible) and use a higher-order function instead of looping.

  2. Another option: iterate from the end of the array to the beginning.

If you choose 1: no side effects, more accessible to reason about and debug, (in parallel systems, it would have a few other “pros”)

If you choose 2, it’s the opposite of what was said about 1, but it’s more memory efficient and faster for very large data. I would reserve this for critical performance situations with large amounts of data. And write good tests.

3

General principle is don’t modify a collection that you are iterating over. A copy or reverse iteration would work yes, but I think reject is what you’re looking for. It returns a new collection.

4

I would put that option in my item 1. a new array with higher-order functions (e.g.,filter, select, etc), SC has a few

5

thanks.
iterate from the end doesn’t work because it ends up trying to access an index that does not exist. Can you elaborate on option 1 - not sure what higher-order function would be.
As I mentioned, I am aware that select or reject would work, but I don’t see how I can combine them with removing the corresponding elements of the corresponding array. In other words, what I would like to do is

a.do({arg item, i; if (item <5, {a.removeAt(i); b.removeAt(i)})})

where a and b have the same length.
copying doesn’t really work either. with c a copy of a, replacing a.do with c.do in the above throws a predictable error a and b become shorter but not c so the index exceeds the size of the array.
thanks

1 Like
6 
~testFunc = { |num| num > 3 };
~numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
~new_numbers = ~numbers.select(~testFunc);

in short, higher-order functions like select, filter, etc., use boolean-returning functions to make decisions about elements. These predicates define a condition or test that each element must pass.

wikipedia

7

I think it works:

~removeItemsInPlace = { |arr, conditionToRemove|
    var i = arr.size - 1;
    while { i >= 0 } {
        if (conditionToRemove.(arr[i])) {
            arr.removeAt(i);
        };
        i = i - 1;
    };
    arr; 
};

~arrayToModify = (1..20);

~conditionToRemove = { |item| item.odd };

~removeItemsInPlace.(~arrayToModify, ~conditionToRemove);

~arrayToModify
// -> -> [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]

But again, it is not safe. Prefer the first option.

8

thanks. I opted for a refinement of this approach by first collecting indexes to be removed into a Set . And in the next step removing those from both lists. The advantage there is the option of using more then one condition for removal.