数组 – 如何使数组不同

发布时间：2020-12-15 09:07:20 所属栏目：大数据来源：网络整理

导读：我有一个整数和字符串字段数组.为了使它不同,我目前逐行复制到新数组中,并且每条记录检查记录是否已经存在于new中,如果不是我复制它.最后,我将新数组复制回原始数据. 它有效,但速度很慢.有没有更快,更简单的方法来做到这一点？ TArrayMixed = record Field1:

我有一个整数和字符串字段数组.为了使它不同,我目前逐行复制到新数组中,并且每条记录检查记录是否已经存在于new中,如果不是我复制它.最后,我将新数组复制回原始数据.

它有效,但速度很慢.有没有更快,更简单的方法来做到这一点？

TArrayMixed = record
    Field1: integer;
    Field2: integer;
    Field3: integer;
    Field4: string;
    Field5: string;
    Field6: string;
 end;

procedure TForm1.Button10Click(Sender: TObject);
var
  ArrayMixed,ArrayMixed_tmp: array of TArrayMixed;
  i,j,vIdx: integer;
  vExists: boolean;
begin
  SetLength(ArrayMixed,100000);
  for i := 0 to 99999 do
  begin
    ArrayMixed[i].Field1 := 1 + Random(5);
    ArrayMixed[i].Field2 := 1 + Random(5);
    ArrayMixed[i].Field3 := 1 + Random(5);
    ArrayMixed[i].Field4 := 'String';
    ArrayMixed[i].Field5 := 'Another string';
    ArrayMixed[i].Field6 := 'New string';
  end;

  // Sort
  TArray.Sort<TArrayMixed > (ArrayMixed,TComparer<TArrayMixed > .Construct(function(const Left,Right: TArrayMixed): Integer
    begin
      Result := MyCompareAMixed(Left,Right);
    end
    ));

  // Distinct
  SetLength(ArrayMixed_tmp,Length(ArrayMixed));
  vIdx := 0;
  for i := Low(ArrayMixed) to High(ArrayMixed) do
  begin
    vExists := False;
    for j := Low(ArrayMixed_tmp) to vIdx - 1 do
      if (ArrayMixed_tmp[j].Field1 = ArrayMixed[i].Field1) and
        (ArrayMixed_tmp[j].Field2 = ArrayMixed[i].Field2) and
        (ArrayMixed_tmp[j].Field3 = ArrayMixed[i].Field3) and
        (ArrayMixed_tmp[j].Field4 = ArrayMixed[i].Field4) and
        (ArrayMixed_tmp[j].Field5 = ArrayMixed[i].Field5) and
        (ArrayMixed_tmp[j].Field6 = ArrayMixed[i].Field6) then
      begin
        vExists := True;
        Break;
      end;

    if not vExists then
    begin
      ArrayMixed_tmp[vIdx] := ArrayMixed[i];
      Inc(vIdx);
    end;
  end;
  SetLength(ArrayMixed_tmp,vIdx);

  // now copy back to original array
  SetLength(ArrayMixed,0);
  SetLength(ArrayMixed,Length(ArrayMixed_tmp));
  for i := Low(ArrayMixed_tmp) to High(ArrayMixed_tmp) do
    ArrayMixed[i] := ArrayMixed_tmp[i];

  sleep(0);

end;

编辑：

因为在实际数据中字符串并不完全相同,所以当原始数组被填充时,它使得不同数组的部分变慢：

编辑#2 :(在编辑#1中复制了错误的代码)

for i := 0 to 999999 do
  begin
    ArrayMixed[i].Field1 := 1 + Random(5);
    ArrayMixed[i].Field2 := 1 + Random(5);
    ArrayMixed[i].Field3 := 1 + Random(5);
    ArrayMixed[i].Field4 := 'String'+IntToStr(i mod 5);
    ArrayMixed[i].Field5 := 'Another string'+IntToStr(i mod 5);
    ArrayMixed[i].Field6 := 'New string'+IntToStr( i mod 5);
  end;

编辑#3：发布用于排序的代码 – 只排序前3个字段！

TMyArray3 = array[1..3] of Integer;

    function CompareIntegerArray3(const lhs,rhs: TMyArray3): Integer;
    var
      i: Integer;
    begin
      Assert(Length(lhs) = Length(rhs));
      for i := low(lhs) to high(lhs) do
        if lhs[i] < rhs[i] then
          exit(-1)
        else if lhs[i] > rhs[i] then
          exit(1);

      exit(0);
    end;

    function GetMyArrayAMixed(const Value: TArrayMixed): TMyArray3;
    begin
      Result[1] := Value.Field1;
      Result[2] := Value.Field2;
      Result[3] := Value.Field3;
    end;

    function MyCompareAMixed(const lhs,rhs: TArrayMixed): Integer;
    begin
      Result := CompareIntegerArray3(GetMyArrayAMixed(lhs),GetMyArrayAMixed(rhs));
    end;

解决方法

只需检查先前索引旁边的重复项,因为数组已排序.这里也是重复使用的排序比较器.

function RemoveDuplicates(const anArray: array of TArrayMixed): TArray<TArrayMixed>;
var
  j,vIdx: integer;
begin
  // Sort
  TArray.Sort<TArrayMixed > (anArray,TComparer<TArrayMixed >.Construct(function(const Left,Right);
    end
    ));

  // Distinct
  SetLength(Result,Length(anArray));
  vIdx := 0;
  j := 0;
  while (j <= High(anArray) do
  begin
    Result[vIdx] := anArray[j];
    Inc(j);
    While (j <= High(anArray)) and (MyCompareAMixed(Result[vIdx],anArray[j]) = 0) do
      Inc(j);
   Inc(vIdx);
  end;
  SetLength(Result,vIdx);
end;

更新：

在对问题的更新中,声明该数组仅部分排序.减少重复次数的迭代次数的一种方法是：

>查找共享第一个排序条件的项目的开始和停止索引.>在它们之间进行迭代以解决重复问题.

（编辑：李大同）

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