Remove Elements from a Slice Iteratively

When iterating over a slice, removing an element within the loop can be tricky due to the shifting of subsequent elements. A common incorrect approach is using append to remove an element, as seen in the example below:

a := []string{"abc", "bbc", "aaa", "aoi", "ccc"}
for i := range a { // BAD
    if conditionMeets(a[i]) {
        a = append(a[:i], a[i 1:]...)
    }
}

This method does not work correctly because the loop does not account for the shifted elements. To properly remove elements while iterating, you can either use a downward loop or employ an alternative method that avoids constant copy operations.

Downward Loop

A downward loop iterates in reverse order, starting from the last element. This approach allows you to remove elements without having to manually decrement loop variables:

a := []string{"abc", "bbc", "aaa", "aoi", "ccc"}
for i := len(a) - 1; i >= 0; i-- {
    if conditionMeets(a[i]) {
        a = append(a[:i], a[i 1:]...)
    }
}

Alternate Method for Many Removals

If you need to remove a large number of elements, using append can be inefficient due to excessive copying. An alternative approach is to create a new slice and copy only the non-removable elements:

a := []string{"abc", "bbc", "aaa", "aoi", "ccc"}
b := make([]string, len(a))
copied := 0
for _, s := range(a) {
    if !conditionMeets(s) {
        b[copied] = s
        copied  
    }
}
b = b[:copied]

In-Place Removal for Many Removals (General Purpose)

An in-place removal technique involves maintaining two indices and assigning non-removable elements in the same slice while zeroing out removed element locations:

a := []string{"abc", "bbc", "aaa", "aoi", "ccc"}
copied := 0
for i := 0; i