## PyTorch ## Train-Val-Test划分 主讲人：龙良曲 ## Recap  ![Image](/uploads/documents/8/e/e/3/8ee3008fab4a9fcce57f4bda7ddee796/p2_2 colspan="2">KaggleVal Set ## ● ● ● print('train:', len(train_db), 'test:', len(test_db)) train_db, val_db = torch.utils.data.random_split(train_db 'db2:', len(val_db)) train_loader = torch.utils.data.DataLoader( train_db, batch_size=batch_size, shuffle=True) val_loader = torch.utils.data.DataLoader( val_db, batch_size=batch_size

parametric type, Val{x}, and a constructor Val(x) = Val{x}(), which serves as a customary way to exploit this technique for cases where you don't need a more elaborate hierarchy. Val is defined as: as: julia> struct Val{x} end julia> Val(x) = Val{x}() Val There is no more to the implementation of Val than this. Some functions in Julia's standard library accept Val instances as arguments firstlast(::Val{true}) = "First" firstlast(generic function with 1 method) julia> firstlast(::Val{false}) = "Last" firstlast(generic function with 2 methods) julia> firstlast(Val(true))

'int64' b = (np.arange(15) ** 2).reshape(3, 5) // Kotlin import org.jetbrains.numkt.* fun main() { val a = arrange(15).reshape(3, 5) println(a.shape.contentEquals(intArrayOf(3, 5))) // true println(a java.lang.Integer // create an array of ints, we square each element and the shape to (3, 5) val b = (arrange(15) ** 2).reshape(3, 5) } Unlike Python, Kotlin is a statically typed language. This random.random((3, 3)) b *= a # Success a *= b # TypeError at runtime // Kotlin // ... val a = ones(3, 3) * 3 val b = Random.random(3, 3) b *= a // Success a *= b // Compilation error: // Type mismatch:

asyncOverlay() = async(CommonPool) { // start two async operations val original = asyncLoadImage("original") val overlay = asyncLoadImage("overlay") // and then apply overlay // launches new coroutine in UI context launch(UI) { // wait for async overlay to complete val image = asyncOverlay().await() // and then show it in UI showImage(image) } Here, async { element has been retrieved, and resumed when the next element is requested. Here's an example: val seq = buildSequence { for (i in 1..5) { // yield a square of i yield(i * i)

template void add(CollT& coll, const T& val) { coll.push_back(val); } std::vector coll1; std::set col12; add(col11, 42); // OK add(col12 add(CollT& coll, const T& val) { coll.push_back(val); } template void add(CollT& coll, const T& val) { coll.insert(val); } std::vector coll1; add(CollT& coll, const T& val) { coll.push_back(val); } template void add(CollT& coll, const T& val) { coll.insert(val); } std::vector coll1;

with val • mutable variables with var To assign a value, use the assignment operator =. For example: fun main() { //sampleStart val popcorn = 5 // There are 5 boxes of popcorn val hotdog value can be reassigned after declaration. We recommend that you declare all variables as read-only (val) by default. Declare mutable variables (var) only if necessary. ## String templates It's useful the code within curly braces {} after the dollar sign $. For example: fun main() { //sampleStart val customers = 10 println("There are $customers customers") // There are 10 customers println("There

该问题解法的一个简单命令式版本可以这样编写： 【Kotlin 1.6.0 及更高版本】 fun main() { var n = readln().toInt() // 读取输入的整数 val reached = HashSet() // 可变的哈希 set while (reached.add(n)) n = f(n) // 迭代函数 f println(reached 同样，如果输入不是整数，那么 String.toInt() 函数会抛出异常。 【早期版本】 fun main() { var n = readLine()!!.toInt() // 读取输入的整数 val reached = HashSet() // 可变的哈希 set while (reached.add(n)) n = f(n) // 迭代函数 f println(reached 读取输入 val n = readln().toInt() val s = readln() val fl = readln().split(" ").map { it.toInt() } // 定义局部函数 f fun f(c: Char) = '0' + fl[c - '1'] // 贪婪查找第一个与最后一个索引 val i = s

时已达到的值，该问题解法的一个简单命令式版本可以这样编写： 【Kotlin 1.6.0 及更高版本】 fun main() { var n = readln().toInt() // 读取输入的整数 val reached = HashSet() // 可变的哈希 set while (reached.add(n)) n = f(n) // 迭代函数 f println(reached 数，那么 String.toInt() 函数会抛出异常。 【早期版本】 fun main() { var n = readLine()!.toInt() // 读取输入的整数 val reached = HashSet() // 可变的哈希 set while (reached.add(n)) n = f(n) // 迭代函数 f println(reached B：长数问题用一个简单的贪心算法实现，可以采用这种风格编写而无需任何可变变量： 【Kotlin 1.6.0 及更高版本】 fun main() { // 读取输入 val n = readln().toInt() val s = readln() val fl = readln().split(" ").map { it.toInt() } // 定义局部函数 f fun

for 迴圈為例： /// == File: iteration.kt === /* 蠻層 for 迴圈 */ fun nestedForLoop(n: Int): String { val res = StringBuilder() // 迴圈 i = 1, 2, ..., n-1, n for (i in 1..n) { // 迴圈 j = 1, 2 === /* 遞迴 */ fun recur(n: Int): Int { // 終止條件 if (n == 1) return 1 // 遞：遞迴呼叫 val res = recur(n - 1) } // 迥：返回結果 return n + res } 圖 2-3 展示了該函式的遞迴過程。  = 1 if (n == 1 || n == 2) return n - 1 // 遞迴呼叫 f(n) = f(n - 1) + f(n - 2) val res = fib(n - 1) + fib(n - 2) // 返回結果 f(n) return res } 觀察以上程式碼，我們在函式內遞迴呼叫了兩個函式，這意味著從一個呼叫

for 循环为例： /// == File: iteration.kt === /* 双层 for 循环 */ fun nestedForLoop(n: Int): String { val res = StringBuilder() // 循环 i = 1, 2, ..., n-1, n for (i in 1..n) { // 循环 j = 1, 2 === /* 递归 */ fun recur(n: Int): Int { // 终止条件 if (n == 1) return 1 // 递：递归调用 val res = recur(n - 1) } // 归：返回结果 return n + res } 图 2-3 展示了该函数的递归过程。  return n - 1 // 递归调用 $ f(n) = f(n-1) + f(n-2) $ val res = fib(n - 1) + fib(n - 2) // 返回结果 $ f(n) $ return res } 观察以上代码，我们在函数内递归调用了两个函数，这意味着从一个调用产生了两个调用分支。如图