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如何在Dart中合并两个列表?

flutter dart

Jordan Davies · 技术社区 · 7 年前

我正在尝试合并2个列表,因此输出如下:

List<String> array1 = ["John", "Bob", "Fred", "June", "Tom"];
List<String> array2 = ["House", "Flat", "Bungalow"];

List<String> output = //["John", "House", "Bob", "Flat", "Fred", "Bungalow", "June", "Tom"]

我怎样才能做到这一点?我已经看了这本书 zip

:为了澄清,我基本上想做与zip函数相同的事情,但是如果一个数组比另一个数组长,我只想将剩余的项附加到末尾。

4 回复 | 直到 7 年前

Aly Yasser Pranajaya 6 年前

我认为这也应该奏效

void main() {
  var list1 = ["John", "Bob", "Fred", "June", "Tom"];
  var list2 = ["House", "Flat", "Bungalow"];
  var list3 = [...list1, ...list2];

  print(list1);
  print(list2);
  print(list3);
}

输出应如下所示(在中测试 DartPad )

[John, Bob, Fred, June, Tom]
[House, Flat, Bungalow]
[John, Bob, Fred, June, Tom, House, Flat, Bungalow]

Günter Zöchbauer 7 年前

我不知道任何可以使用的东西,但这应该可以满足您的需要:

import 'dart:math' as math;

void main() {
  List<String> array1 = ["John", "Bob", "Fred", "June", "Tom"];
  List<String> array2 = ["House", "Flat", "Bungalow"];

  List<String> output = List<String>(array1.length + array2.length);
  int i = 0;
  for (i; i < math.min(array1.length, array2.length); i++) {
    output[i * 2] = array1[i];
    output[i * 2 + 1] = array2[i];
  }
  print(output);
  if (array1.length != array2.length) {
    if (array1.length > array2.length) {
      output.setRange(i * 2, output.length, array1.sublist(i));
    } else {
      output.setRange(i * 2, output.length, array2.sublist(i));
    }
  }
  print(output);
}

SpencerPark 7 年前

另一个解决方案是下面的解决方案,它的效率可能不如Gunter的解决方案,但在本质上更具功能性,我喜欢从多个角度来解决问题。如果数组像这些一样短,那么连接上的优化可能无论如何都不是问题。

import 'dart:math' as math;

main() {
  List<String> array1 = ["John", "Bob", "Fred", "June", "Tom"];
  List<String> array2 = ["House", "Flat", "Bungalow"];

  //["John", "House", "Bob", "Flat", "Fred", "Bungalow", "June", "Tom"]
  List<String> output =
      Iterable.generate(math.max(array1.length, array2.length))
          .expand((i) sync* {
    if (i < array1.length) yield array1[i];
    if (i < array2.length) yield array2[i];
  }).toList();

  print(output);
}

pbn 7 年前

merge 功能:

import 'dart:math';

List<String> merge(List<String> a, List<String> b) {
    List<String> output = [];

    var min_length = min(a.length, b.length);
    var max_length = max(a.length, b.length);

    for(var i = 0; i < min_length; i++) {
      output.add(a[i]);
      output.add(b[i]);
    }

    List<String> longer = a.length > b.length ? a : b;

    for(var i = min_length; i < max_length; i++) {
      output.add(longer[i]);
    }

    return output;
  }

void main() {
    merge(
      ["John", "Bob", "Fred", "June", "Tom"],
      ["House", "Flat", "Bungalow"]
    ).forEach((e) => print(e));

    print('------');

    merge(
      ["John", "Bob"],
      ["House", "Flat", "Bungalow"]
    ).forEach((e) => print(e));
}

Chris Joseph 6 年前

你可以用 addAll()

List<int> a =[1,2];  List<int> b =[1,2];
List<int> c =[];
c.addAll(a);
c.addAll(b);

或 insertAll(index,Iterateable<>)

List<int> a =[1,2];  
List<int> b =[1,2];
a.insertAll(a.length,b);

Dharman vijay 5 年前

我也找不到解决方案,如果可能的话,我不想使用列表转换或数学函数,所以我写了以下内容:

extension IterableExtensions<T> on Iterable<T> {
  Iterable<T> merge(Iterable<T> other) sync* {
    
    Iterator<T> iter1 = this.iterator;
    Iterator<T> iter2 = other.iterator;

    bool has1 = iter1.moveNext();
    bool has2 = iter2.moveNext();
    while(has1 && has2) {
      yield iter1.current;
      yield iter2.current;
      has1 = iter1.moveNext();
      has2 = iter2.moveNext();
    }
    while(has1) {
      yield iter1.current;
      has1 = iter1.moveNext();
    }
    while(has2) {
      yield iter2.current;
      has2 = iter2.moveNext();
    }
  }
}

也许可以稍微整理一下,但现在可以用了。

例子 here .

Cristiana Pereira 5 年前

void main() {
  List<String> myList = ['a', 'b'];
  List<String> myList2 = ['a2', 'b2'];
  List<String> myList3 = [myList, myList2].expand((x) => x).toList();
  print(myList);  // [a, b]
  print(myList2); // [a, b]
  print(myList3); // [a, b, a2, b2]
}