如何从C#中的数组中获取随机值
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How to get random values from array in C#
提问by Shebystian
Possible Duplicate:
Access random item in list
可能的重复：
访问列表中的随机项目
I have an array with numbers and I want to get random elements from this array. For example: {0,1,4,6,8,2}. I want to select 6 and put this number in another array, and the new array will have the value {6,....}.
我有一个带数字的数组，我想从这个数组中获取随机元素。例如：{0,1,4,6,8,2}。我想选择 6 并将这个数字放在另一个数组中，新数组的值将是 {6,....}。
I use random.next(0, array.length), but this gives a random number of the length and I need the random array numbers.
我使用 random.next(0, array.length)，但这给出了长度的随机数，我需要随机数组数。
for (int i = 0; i < caminohormiga.Length; i++ )
{
if (caminohormiga[i] == 0)
{
continue;
}
for (int j = 0; j < caminohormiga.Length; j++)
{
if (caminohormiga[j] == caminohormiga[i] && i != j)
{
caminohormiga[j] = 0;
}
}
}
for (int i = 0; i < caminohormiga.Length; i++)
{
int start2 = random.Next(0, caminohormiga.Length);
Console.Write(start2);
}
return caminohormiga;
采纳答案by Tilak
I use the random.next(0, array.length), but this give random number of the length and i need the random array numbers.
我使用 random.next(0, array.length)，但这给出了长度的随机数，我需要随机数组数。
Use the return value from random.next(0, array.length)as index to get value from the array
使用random.next(0, array.length)作为索引的返回值从array
Random random = new Random();
int start2 = random.Next(0, caminohormiga.Length);
Console.Write(caminohormiga[start2]);
回答by faester
You just need to use the random number as a reference to the array:
您只需要使用随机数作为对数组的引用：
var arr1 = new[]{1,2,3,4,5,6}
var rndMember = arr1[random.Next(arr1.Length)];
回答by SergeyS
Console.Write(caminohormiga[start2]);
回答by Murshid Ahmed
Try like this
像这样尝试
int start2 = caminohormiga[ran.Next(0, caminohormiga.Length)];
instead of
代替
int start2 = random.Next(0, caminohormiga.Length);
回答by Seth Flowers
To shuffle
洗牌
int[] numbers = new [] {0, 1, 4, 6, 8, 2};
int[] shuffled = numbers.OrderBy(n => Guid.NewGuid()).ToArray();
回答by Erik
I noticed in the comments you wanted no repeats, so you want the numbers to be 'shuffled' similar to a deck of cards.
我在评论中注意到你不想重复，所以你希望数字像一副纸牌一样“洗牌”。
I would use a List<>for the source items, grab them at randomand pushthem to a Stack<>to create the deck of numbers.
我会用List<>的源项目，抓住他们随意和推动他们去Stack<>创造数字的甲板。
Here is an example:
下面是一个例子：
private static Stack<T> CreateShuffledDeck<T>(IEnumerable<T> values)
{
var rand = new Random();
var list = new List<T>(values);
var stack = new Stack<T>();
while(list.Count > 0)
{
// Get the next item at random.
var index = rand.Next(0, list.Count);
var item = list[index];
// Remove the item from the list and push it to the top of the deck.
list.RemoveAt(index);
stack.Push(item);
}
return stack;
}
So then:
那么：
var numbers = new int[] {0, 1, 4, 6, 8, 2};
var deck = CreateShuffledDeck(numbers);
while(deck.Count > 0)
{
var number = deck.Pop();
Console.WriteLine(number.ToString());
}
