英文:
C# can cast LINQ Aggregate to double to output factorial, this works for integer up to 12
问题
Sure, here's the translation of the code part:
我想使用LINQ填充列表的阶乘。
我有这个，它可以计算到12。我认为这是因为整数范围的限制。但我不能将其重新转换为双精度。是否可以轻松修改（重新转换）？
var listOfFactorials = from fact in Enumerable.Range(0, 171).Reverse().ToList() select new { Number = fact, Factorial = (fact == 0) ? 1d : Enumerable.Range(1, fact).Aggregate((i, j) => (i * j))};
Is there anything else you would like to know or translate?
英文:
I want to fill the list with factorials using linq.
I got this, this works up to 12. I assume it's because of the integer range. But I couldn't retype it to double. Can this be easily modified (retyped)?
var listOfFactorials = from fact in Enumerable.Range(0, 171).Reverse().ToList() select new { Number = fact, Factorial = (fact == 0) ? 1d : Enumerable.Range(1, fact).Aggregate((i, j) => (i * j))};
I want to have the list filled with factorials up to the number 170.
I don't need it for anything, just playing for fun 
答案1
得分: 0
You don't want double - it only gives you 15 digits of precision in exchange for the larger range. You want to use either long or BigInteger.
You can either convert the integers returned from Range before calling Aggregate:
Enumerable.Range(1, fact)
.Select(i => new BigInteger(i))
.Aggregate((i, j) => (i * j))
or use the more verbose version of Aggregate where you specify the types separately:
Enumerable.Range(1, fact)
.Aggregate<int, BigInteger>(new BigInteger(1), (i, j) => (i * j))
Note that long will overflow at 67! while BigInteger is unbounded in theory.
英文:
You don't want double - it only gives you 15 digits of precision in exchange for the larger range. You want to use either long or BigInteger.
You can either convert the integers returned from Range before calling Aggregate:
Enumerable.Range(1, fact)
.Select(i => new BigInteger(i))
.Aggregate((i, j) => (i * j))};
or use the more verbose version of Aggregate where you specify the types separately:
Enumerable.Range(1, fact)
.Aggregate<int, BigInteger>(new BigInteger(1), (i, j) => (i * j))
Note that Long will overflow at 67! while BigInteger is unbounded in theory
答案2
得分: 0
你必须使用 BigInteger:
var listOfFactorials = from fact in Enumerable.Range(0, 200).Reverse().ToList()
select new
{
Number = fact,
Factorial = (fact == 0) ? BigInteger.One : Enumerable.Range(1, fact)
.Select(i => new BigInteger(i))
.Aggregate((i, j) => (i * j))
};
英文:
You have to use BigInteger:
var listOfFactorials = from fact in Enumerable.Range(0, 200).Reverse().ToList()
select new
{
Number = fact,
Factorial = (fact == 0) ? BigInteger.One : Enumerable.Range(1, fact)
.Select( i => new BigInteger(i))
.Aggregate((i, j) => (i * j))
};
答案3
得分: 0
请注意，您的“Aggregate”方法效率不高，因为它每次都会从1到n重新计算，如果您从1开始，您只需为每个后续值执行一次乘法操作：
var listOfFactorials = Enumerable.Range(1, 170)
.Aggregate(new[] { new { Number = 0, Factorial = BigInteger.One } }.ToList(),
(ans, n) => { ans.Add(new { Number = n, Factorial = ans.Last().Factorial * n }); return ans; })
.AsEnumerable()
.Reverse();
由于序列是以“0”值初始化的，范围必须从“1”开始，以减少一个值。
一个更简单（可能更容易理解）的方法是使用“for”循环：
var ans = new[] { new { Number = 0, Factorial = BigInteger.One } }.ToList();
var Factorial = BigInteger.One;
for (int Number = 1; Number < 171; ++Number) {
Factorial *= Number;
ans.Add(new { Number, Factorial });
}
ans.Reverse();
希望这有帮助。
英文:
Note that your Aggregate is not very efficient since it recalculates from 1 to n every time, if you start with 1, you can just do one multiple for each subsequent value:
var listOfFactorials = Enumerable.Range(1, 170)
.Aggregate(new[] { new { Number = 0, Factorial = BigInteger.One } }.ToList(),
(ans, n) => { ans.Add(new { Number = n, Factorial = ans.Last().Factorial * n }); return ans; })
.AsEnumerable()
.Reverse();
Since the sequence is initialized with the 0 value, the range must start at 1 for one fewer values.
A simpler (and probably more understandable) method would be to just use a for loop:
var ans = new[] { new { Number = 0, Factorial = BigInteger.One } }.ToList();
var Factorial = BigInteger.One;
for (int Number = 1; Number < 171; ++Number) {
Factorial *= Number;
ans.Add(new { Number, Factorial });
}
ans.Reverse();
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