gpt4 book ai didi

arrays - 如何从 AutoHotkey 数组中删除重复项?

转载 作者:行者123 更新时间:2023-12-02 02:43:14 30 4
gpt4 key购买 nike

我有一个array AutoHotkey 中包含重复条目的字符串。

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

我想删除所有重复项,以便仅保留唯一值。

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

理想情况下,我正在寻找类似于 Trim() 的函数这将返回一个修剪后的数组,同时保持原始数组不变。 (即 trimmedArray := RemoveDuplicates(nameArray))

如何从 AutoHotkey 数组中删除重复项?

最佳答案

保持原来的完整,只循环一次,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。这可能被证明更高效、更快,但我将把测试留给您。

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:最初我不打算测试,但我很好奇,也厌倦了等待 OP。结果并不令我惊讶:

enter image description here

我们在这里看到的是 10,000 个测试的平均执行时间,数字越低,任务计算速度越快。明显的赢家是我的没有哈斯基方法的脚本变体,但仅以微小的优势!所有其他方法都注定失败,因为它们不是线性解决方案。

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)
}

关于arrays - 如何从 AutoHotkey 数组中删除重复项?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46432447/

30 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com