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Linq学习笔记--聚合函数/Aggregator

发布时间:2020-05-22 21:01:22 所属栏目:程序设计 来源:互联网
导读:以下代码均来自微软官网 /// summary /// 去掉重复项之后,获得个数 /// /summary public void Linq1() { int[] factorsOf300 = { 2, 2, 3, 5, 5 }; int uniqueFactors = factorsOf300.Distinct().Count(); Console.WriteLi

以下代码均来自微软官网


/// <summary>
/// 去掉重复项之后,获得个数
/// </summary>
public void Linq1()
{
int[] factorsOf300 = { 2,2,3,5,5 };

int uniqueFactors = factorsOf300.Distinct().Count();

Console.WriteLine("There are {0} unique factors of 300.",uniqueFactors);
}

---结果 3

/// <summary>
/// 带有Lambda表达式的、
/// 获得所有的奇数的个数
/// </summary>
public void Linq2()
{
int[] numbers = { 5,4,1,9,8,6,7,0 };

int oddNumbers = numbers.Count(n => n % 2 == 1);

Console.WriteLine("There are {0} odd numbers in the list.",oddNumbers);
}

---结果 5

/// <summary>
/// 带有Lambda表达式的、
/// 获得所有的奇数的个数
/// </summary>

public void Linq3()
{
List<Product> products = GetProductList(); //获得一个集合、GetProductList()代码省略....

var categoryCounts =
from p in products
group p by p.Category into g //按每个对象的"分类"进行分组,并给每个组起名为g(个人理解..)
select new { Category = g.Key,ProductCount = g.Count() };
}

/// <summary>
/// 求和
/// </summary>
public void Linq4()
{
int[] numbers = { 5,0 };

double numSum = numbers.Sum();

Console.WriteLine("The sum of the numbers is {0}.",numSum);
}

---结果 The sum of the numbers is 45.

/// <summary>
/// 求:所有字母长度的和
/// </summary>
public void Linq5()
{
string[] words = { "cherry","apple","blueberry" };

double totalChars = words.Sum(w => w.Length);

Console.WriteLine("There are a total of {0} characters in these words.",totalChars);
}

---结果 There are a total of 20 characters in these words.

/// <summary>
/// 求最小值
/// </summary>
public void Linq6()
{
int[] numbers = { 5,0 };

int minNum = numbers.Min();

Console.WriteLine("The minimum number is {0}.",minNum);
}

---结果 The minimum number is 0.


/// <summary>
/// 求最短的单词长度
/// </summary>
public void Linq7()
{
string[] words = { "cherry","blueberry" };

int shortestWord = words.Min(w => w.Length);

Console.WriteLine("The shortest word is {0} characters long.",shortestWord);
}

---结果 The shortest word is 5 characters long.

/// <summary>
/// 求大值
/// </summary>
public void Linq8()
{
int[] numbers = { 5,0 };

int maxNum = numbers.Max();

Console.WriteLine("The maximum number is {0}.",maxNum);
}

---结果 The maximum number is 9.

/// <summary>
/// 求平均数
/// </summary>
public void Linq9()
{
int[] numbers = { 5,0 };

double averageNum = numbers.Average();

Console.WriteLine("The average number is {0}.",averageNum);
}

---结果 The average number is 4.5.

===============================华丽的分割线===============================

/// <summary>
/// 累计求乘积
/// </summary>
public void Linq10()
{
double[] doubles = { 1.7,2.3,1.9,4.1,2.9 };

//累计的一个循环相乘的过程,返回最后相乘后的结果
double product = doubles.Aggregate((runningProduct,nextFactor) => runningProduct * nextFactor);

Console.WriteLine("Total product of all numbers: {0}",product);
}

---结果 Total product of all numbers: 88.33081

/// <summary>
/// 我也不知道这个例子是到底要干什么.可能只是为了讲述语法吧...
/// </summary>
public void Linq11()
{
double startBalance = 100.0;

int[] attemptedWithdrawals = { 20,10,40,50,70,30 };

//乍一看有些复杂,让人找不着北,其实慢慢细看,其实也就是Lambda里嵌套一个Lambda,
//我们可以把代码拆分为AAAA() :下面....
double endBalance =
attemptedWithdrawals.Aggregate(startBalance,
(balance,nextWithdrawal) =>
((nextWithdrawal <= balance) ? (balance - nextWithdrawal) : balance));

Console.WriteLine("Ending balance: {0}",endBalance);
}


/// <summary>
/// 这是上面拆分来的代码~
/// </summary>
public void AAAA()
{
double startBalance = 100.0;

int[] attemptedWithdrawals = { 20,30 };

double endBalance =
attemptedWithdrawals.Aggregate(startBalance,(balance,nextWithdrawal) =>aaa(balance,nextWithdrawal));

Console.WriteLine("Ending balance: {0}",endBalance);
}


double aaa(double a,double b)
{
return (b <= a) ? (a - b) : a;
}

---结果 DE>Ending balance: 20DE>

/// <summary>
/// let关键字的用法。传统的子查询看起来始终有些不太直观
/// 两种写法效果一样,并且let更加直观
/// (个人感觉就像一个给子查询的别名)
/// </summary>
public void Linq12()
{
int[] numbers = new[] { 1,9 };
//传统下的子查询做法
var query = from num in numbers
select num * (from n in numbers where n % 2 == 0 select n).Count();

//使用LET关键字的做法
//var query = from num in numbers
//let evenNumbers = from n in numbers //给查询出来的能整除2的数字一个"名字"
//where n % 2 == 0
//select n
//select num * evenNumbers.Count();
foreach (var item in query)
{
Console.WriteLine(item);
}
}

---结果 4 8 12 16 20 24 28 32 36

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