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How to Efficiently Remove Elements from a Slice in a Loop in Go?

Barbara Streisand
Barbara StreisandOriginal
2024-10-31 06:21:02584browse

How to Efficiently Remove Elements from a Slice in a Loop in Go?

In-Place Element Removal Within a Slice Loop

Problem:

Accessing slice elements from i within a range for loop and removing them using append() is problematic. Incrementing the loop variable (i) may skip subsequent elements after removal, leading to incomplete processing.

Best Practices:

1. Manual Loop Variable Decrementation

Use a regular for loop with manual loop variable (i) decrementing when an element is removed:

<code class="go">for i := 0; i < len(a); i++ {
    if conditionMeets(a[i]) {
        a = append(a[:i], a[i+1:]...)
        i--
    }
}

2. Downward Loop

Alternatively, use a downward loop to avoid manual decrementation:

<code class="go">for i := len(a) - 1; i >= 0; i-- {
    if conditionMeets(a[i]) {
        a = append(a[:i], a[i+1:]...)
    }
}</code>

3. Non-Removable Element Copying

If numerous elements need to be removed, consider copying non-removable elements to a new slice to improve efficiency:

<code class="go">b := make([]string, len(a))
copied := 0
for _, s := range(a) {
    if !conditionMeets(s) {
        b[copied] = s
        copied++
    }
}
b = b[:copied]</code>

4. In-Place Copying and Zeroing

For general-purpose in-place removal, maintain two indices and assign non-removable elements while zeroing removed element spaces:

<code class="go">copied := 0
for i := 0; i < len(a); i++ {
    if !conditionMeets(a[i]) {
        a[copied] = a[i]
        copied++
    }
}
for i := copied; i < len(a); i++ {
    a[i] = "" // Zero places of removed elements
}
a = a[:copied]</code>

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