List<String> listA = new List<string> { "A1", "A2" };
List<String> listB = new List<string> { "B1", "B2" };
for(int i = 0; i < listA.Count; i++)
{
text += listA[i] + " and " + listB[i];
}
如何使用foreach
循环执行此操作?
答案 0 :(得分:12)
您可以使用Linq和Zip
方法:
List<String> listA = new List<string> { "A1", "A2" };
List<String> listB = new List<string> { "B1", "B2" };
foreach (var pair in listA.Zip(listB, (a,b) => new {A = a, B = b}))
{
text += pair.A + " and " + pair.B;
}
答案 1 :(得分:4)
你不能使用foreach
做得比使用for
做得更好 - foreach
实际上只有在只有一个要枚举的序列时才能正常工作
但是,您可以使用LINQ非常方便地完成:
text = string.Join("", listA.Zip(listB, (a, b) => a + " and " + b));
这需要.NET 4同时适用于Zip
和specific overload of string.Join
。
答案 2 :(得分:4)
另一种方法是使用简单的Enumerator:
IEnumerator<string> enumerator = listB.GetEnumerator();
enumerator.MoveNext();
foreach(var stra in listA) {
text += stra + " and " + enumerator.Current.ToString() + ", ";
enumerator.MoveNext();
}
答案 3 :(得分:1)
使用LINQ
string text = listA.Zip(listB, (a, b) => new {A = a, B = b}).Aggregate("", (current, pair) => current + (pair.A + " and " + pair.B));
答案 4 :(得分:1)
如果您不想使用LINQ并且希望它们并行迭代,那么您可以选择几个选项 - 如下所示使用新类等,或者您可以使用 foreach ,但仅限于其中一个列表,像这样:
List<String> listA = new List<string> { "A1", "A2" };
List<String> listB = new List<string> { "B1", "B2" };
string text = "";
int i = 0;
foreach (string s in listA) {
text += s + " and " + listB [i++] + "\n";
}
Console.WriteLine (text);
或使用GetEnumerator使其更好一点:
List<String> listA = new List<string> { "A1", "A2" };
List<String> listB = new List<string> { "B1", "B2" };
string text = "";
List<string>.Enumerator e = listB.GetEnumerator ();
foreach (string s in listA) {
e.MoveNext ();
text += s + " and " + e.Current + "\n";
}
Console.WriteLine (text);
你也可以创建一个Enumberable metacollection,它将返回总是一个简单的字符串数组 - 为此你需要创建一个Enumerator和一个来自IEnumerable的类:
首先是枚举器:
private class DoubleStringEnumerator : IEnumerator
{
private DoubleString _elemList;
private int _index;
public DoubleStringEnumerator(DoubleString doubleStringObjt)
{
_elemList = doubleStringObjt;
_index = -1;
}
public void Reset()
{
_index = -1;
}
public object Current {
get {
return _elemList.getNext();
}
}
public bool MoveNext ()
{
_index++;
if (_index >= _elemList.Length)
return false;
else
return true;
}
}
当前方法并没有真正反映出给定示例中的名称 - 但它是出于学习目的。
现在上课:
public class DoubleString : IEnumerable
{
public int Length;
List<String> listA;
List<String> listB;
List<string>.Enumerator eA,eB;
public DoubleString(List<String> newA,List<String> newB)
{
if(newA.Count != newB.Count) {
throw new Exception("Lists lengths must be the same");
}
listA = newA;
listB = newB;
eA = listA.GetEnumerator ();
eB = listB.GetEnumerator ();
Length = listA.Count;
}
IEnumerator IEnumerable.GetEnumerator ()
{
return (IEnumerator)new DoubleStringEnumerator(this);
}
public string[] getNext ()
{
eA.MoveNext ();
eB.MoveNext ();
return new string[] {eA.Current ,eB.Current };
}
}
代码本身:
List<String> listA = new List<string> { "A1", "A2" };
List<String> listB = new List<string> { "B1", "B2" };
DoubleString newDoubleString = new DoubleString (listA, listB);
string text = "";
foreach (string[] s in newDoubleString) {
text += s[0] + " and " + s[1] + "\n";
}
Console.WriteLine (text);
当然最好还是使用LINQ。代码没有被激活,但我没有编译器,所以写下我的头 - 希望它能澄清一些事情。随意提问。
答案 5 :(得分:-2)
List<int> = new [] { 1, 2, 3, 4 };
List<String> words = new [] { "one", "two", "three" };
var numbersAndWords = numbers.Zip(words, (n, w) => new { Number = n, Word = w });
foreach(var nw in numbersAndWords)
{
Console.WriteLine(nw.Number + nw.Word);
}
答案 6 :(得分:-3)
以下是使用foreach的解决方案:
string text = null;
int cnt = 0;
List<String> listA = new List<string> { "A1", "A2" };
List<String> listB = new List<string> { "B1", "B2" };
foreach (string i in listA)
{
text += listA[cnt] + " and " + listB[cnt];
cnt++;
}
谢谢&amp;问候,
Subhankar 强>