如何从AutoHotkey数组中删除重复项？

保持原样，仅循环一次，保留顺序：

nameArray := ["Chris","Joe","Marcy","Chris","Elina","Timothy","Joe"]

trimmedArray := trimArray(nameArray)

trimArray(arr) { ; Hash O(n) 

    hash := {}, newArr := []

    for e, v in arr
        if (!hash.Haskey(v))
            hash[(v)] := 1, newArr.push(v)

    return newArr
}

使用haskey方法的替代方法是检查哈希对象中的值。这可能会被证明更有效，更快捷，但是我将把测试交给您。

trimArray(arr) { ; Hash O(n) 

    hash := {}, newArr := []

    for e, v in arr
        if (!hash[v])
            hash[(v)] := 1, newArr.push(v)

    return newArr
}

编辑：起初我不打算测试，但我既好奇又厌倦了等待OP。结果并不令我感到惊讶：

我们在这里看到的是10,000个测试的平均执行时间，该数字越小，任务的计算速度就越快。明显的赢家是我的脚本版本没有Haskey方法，但是只有很少的利润！所有其他方法都注定要失败，因为它们不是线性解。

测试代码在这里：

setbatchlines -1 

tests := {test1:[], test2:[], test3:[], test4:[]}

Loop % 10000 {
    nameArray := ["Chris","Joe","Marcy","Chris","Elina","Timothy","Joe"]

    QPC(1)

    jimU(nameArray)

    test1 := QPC(0), QPC(1)

    AbdullaNilam(nameArray)

    test2 := QPC(0), QPC(1)

    ahkcoderVer1(nameArray)

    test3 := QPC(0), QPC(1)

    ahkcoderVer2(nameArray)

    test4 := QPC(0)

    tests["test1"].push(test1), tests["test2"].push(test2)
    , tests["test3"].push(test3), tests["test4"].push(test4)
}

scripts := ["Jim U         ", "Abdulla Nilam  "
            , "ahkcoder HasKey", "ahkcoder Bool  " ]

for e, testNums in tests ; Averages Results
    r .= "Test Script " scripts[A_index] "`t:`t" sum(testNums) / 10000 "`n"


msgbox % r

AbdullaNilam(names) {

    for i, namearray in names
        for j, inner_namearray in names
            if (A_Index > i && namearray = inner_namearray)
                names.Remove(A_Index)
    return names
}

JimU(nameArray) {
  hash := {}
  for i, name in nameArray
    hash[name] := null

  trimmedArray := []
  for name, dummy in hash
    trimmedArray.Insert(name)

  return trimmedArray
}

ahkcoderVer1(arr) { ; Hash O(n) - Linear

    hash := {}, newArr := []

    for e, v in arr
        if (!hash.Haskey(v))
            hash[(v)] := 1, newArr.push(v)

    return newArr
}

ahkcoderVer2(arr) { ; Hash O(n) - Linear

    hash := {}, newArr := []

    for e, v in arr
        if (!hash[v])
            hash[(v)] := 1, newArr.push(v)

    return newArr
}

sum(arr) {
    r := 0
    for e, v in arr
        r += v
    return r
}

QPC(R := 0) ; https://autohotkey.com/boards/viewtopic.php?t=6413
{
    static P := 0, F := 0, Q := DllCall("QueryPerformanceFrequency", "Int64P", F)
    return ! DllCall("QueryPerformanceCounter", "Int64P", Q) + (R ? (P := Q) / F : (Q - P) / F) 
}