lambda

问题 I usually comment my functions using multi-line docstrings with """, as mentioned in : https://www.python.org/dev/peps/pep-0257/ def func1(x): """ This function does ... """ ... But what is the best way to comment a lambda function ? I hesitate between : # This function does ... func2 = lambda x: ... or : func2 = lambda x: ... """ This function does ... """ or else ? 回答1: tbh, even assigning a lambda to a variable seems unpythonic to me. if it needs a name, define it as a regular function.

问题 This is more a kind of theoretical question. Is it possible in C++11 to combine functions into a new function? For example : auto f = [](int i){return i * 2;}; auto g = [](int i){return i + 10;}; So this works: auto c = f(g(20)); // = 60 But I want an object that stores the combination, like auto c = f(g); std::cout << c(20) << std::endl; //prints 60 Edit: Additionally what i want to create is a function a, which you can give a function b and an int n , and which returns the n'th combination

问题 Look this code: class MyClass_1(): @staticmethod def method_1(func): return func(1, 2, 3) class MyClass_2(): my_func = lambda a,b,c : a*b*c # I need to call this method def method_2(self): result = MyClass_1.method_1(self.my_func) print(result) My error: TypeError: () takes 3 positional arguments but 4 were given I need to call the lambda function my_func in the same way as the code above, but a self is appearing from somewhere I don't know and causing this error. What am I missing? 回答1:

问题 Why am I seeing different result on GPU compare to sequential CPU? import numpy from numba import cuda from functools import reduce A = (numpy.arange(100, dtype=numpy.float64)) + 1 cuda.reduce(lambda a, b: a + b * 20)(A) # result 12952749821.0 reduce(lambda a, b: a + b * 20, A) # result 100981.0 import numba numba.__version__ # '0.34.0+5.g1762237' Similar behavior happens when using Java Stream API to parallelize reduction on CPU: int n = 10; float inputArray[] = new float[n]; ArrayList<Float

问题 Why am I seeing different result on GPU compare to sequential CPU? import numpy from numba import cuda from functools import reduce A = (numpy.arange(100, dtype=numpy.float64)) + 1 cuda.reduce(lambda a, b: a + b * 20)(A) # result 12952749821.0 reduce(lambda a, b: a + b * 20, A) # result 100981.0 import numba numba.__version__ # '0.34.0+5.g1762237' Similar behavior happens when using Java Stream API to parallelize reduction on CPU: int n = 10; float inputArray[] = new float[n]; ArrayList<Float

问题 I was trying to answer this question, but I did not because I don't understand Streams well enough. Please tell me if my explanation is correct or not. My answer : import java.util.Arrays; import java.util.stream.Stream; public class Temp { public static void main(String [] args){ String [] input = {"1,2", "3,4", "5"}; String [] expected = {"1", "2", "3", "4", "5"}; String [] actual = Stream.of(input) .flatMap(s -> Arrays.stream(s.split(","))) .toArray(String [] :: new); //Testing - Runs only

问题 @FunctionalInterface public interface Runnable { public abstract void run(); } public class MethodReference1 { public static String ThreadStatus() { System.out.println( Thread.currentThread().getName() + " is running..."); return "threadname"; } public static void main(String[] args) { Thread t1 = new Thread(() -> ThreadStatus()); t1.start(); } } In the above example using Java 8, ThreadStatus() returns a string but Runnable interface "run()" method doesnt return any value. But it still works

问题 I am building an advanced search screen and am using nHibernate to query the DB. I have already built my DataAccess layer and have built a generic method which works great - I pass in an expression to be used as a predicate and passes back a collection of an object that matches the predicate: public object LoadByPredicate<T>(Expression<Func<T, bool>> predicate) where T : class Example usage is: var items = _query.LoadByPredicate<StaticTypes>(x => x.StaticName == type) as List<StaticTypes>;

问题 Say I have this code: #lang racket (define a ((λ (x) x) ((λ (y) y) (λ (z) ((λ (w) w) z))))) I know intuitively that this lambda expression is (extensionally) equal to (λ (z) z) My question is if there is a way to print out the body of a in case I want to see how much the function got simplified internally by Racket. More information: By default, if I type a into the interpreter, I get #<procedure:y> (This seems to give a hint as to how much evaluation happened). I can change the output style

问题 I have the following class class Book implement Borrowable { @Override public String toString(Function<? extends Borrowable , String> format) { return format.apply(this); } } This gives me an error that i cannot use "apply" on this(Book object). My current formatter is Function<Book, String> REGULAR_FORMAT = book -> "name='" + book.name + '\'' + ", author='" + book.author + '\'' + ", year=" + book.year; I don't want to make the lambda function of the type Function<Borrowable, String> as I