reduce

问题 I try to understand how does the reduce() method work in java-8. For example I have this code: public class App { public static void main(String[] args) { String[] arr = {"lorem", "ipsum", "sit", "amet"}; List<String> strs = Arrays.asList(arr); int ijk = strs.stream().reduce(0, (a, b) -> { System.out.println("Accumulator, a = " + a + ", b = " + b); return a + b.length(); }, (a, b) -> { System.out.println("Combiner"); return a * b; }); System.out.println(ijk); } } And the output is this:

问题 I try to understand how does the reduce() method work in java-8. For example I have this code: public class App { public static void main(String[] args) { String[] arr = {"lorem", "ipsum", "sit", "amet"}; List<String> strs = Arrays.asList(arr); int ijk = strs.stream().reduce(0, (a, b) -> { System.out.println("Accumulator, a = " + a + ", b = " + b); return a + b.length(); }, (a, b) -> { System.out.println("Combiner"); return a * b; }); System.out.println(ijk); } } And the output is this:

问题 I have the following data that I want to sort based on the date - not including the timestamp. NOTE: I have access to moment for this task. My data looks like the following: const data = [ { "fixture": "AC v Inter", "kickOffTime": "2018-06-14T15:00:00Z", }, { "fixture": "DC v NYC", "kickOffTime": "2018-06-15T12:00:00Z", }, { "fixture": "AFC v LPC", "kickOffTime": "2018-06-15T15:00:00Z", }, { "fixture": "DTA v MC", "kickOffTime": "2018-06-15T18:00:00Z", }, { "fixture": "LAC v GC", "kickOffTime

问题 Is it possible to leave the return value of a thrust::reduce operation in device-allocated memory? In case it is, is it just as easy as assigning the value to a cudaMalloc'ed area, or should I use a thrust::device_ptr? 回答1: Is it possible to leave the return value of a thrust::reduce operation in device-allocated memory? The short answer is no. thrust reduce returns a quantity, the result of the reduction. This quantity must be deposited in a host resident variable: Take for example reduce,

问题 List<Integer> integers = Arrays.asList(1, 2, 3, 5, 6, 8, 9, 10); integers.stream().filter((integer) -> integer % 2 == 0).collect(Collectors.toList()); As shown above integers is a List, from which we need to filter only even numbers. I can achieve by using .filter() method. But, is there any possibility to achieve the same with .reduce() method. Hope, .reduce() method filtered out all the other elements by performing given BynaryOperation and return reduced list. If my understanding is not

问题 In Java 8, Stream has a method reduce: T reduce(T identity, BinaryOperator<T> accumulator); Is the accumulator operator allowed to modify either of its arguments? I presume not since the JavaDoc says the accumulator should be NonInterfering, though all examples talk of modifying the collection, rather than modifying the elements of the collection. So, for a concrete example, if we have integers.reduce(0, Integer::sum); and suppose for a moment that Integer was mutable, would sum be allowed to

问题 There are many ways to write a Python program that computes a histogram. By histogram, I mean a function that counts the occurrence of objects in an iterable and outputs the counts in a dictionary. For example: >>> L = 'abracadabra' >>> histogram(L) {'a': 5, 'b': 2, 'c': 1, 'd': 1, 'r': 2} One way to write this function is: def histogram(L): d = {} for x in L: if x in d: d[x] += 1 else: d[x] = 1 return d Are there more concise ways of writing this function? If we had dictionary comprehensions

问题 Given n partial sums it's possible to sum all the partial sums in log2 parallel steps. For example assume there are eight threads with eight partial sums: s0, s1, s2, s3, s4, s5, s6, s7 . This could be reduced in log2(8) = 3 sequential steps like this; thread0 thread1 thread2 thread4 s0 += s1 s2 += s3 s4 += s5 s6 +=s7 s0 += s2 s4 += s6 s0 += s4 I would like to do this with OpenMP but I don't want to use OpenMP's reduction clause. I have come up with a solution but I think a better solution

问题 What is the best way to transform an array like this: const arr = [ { name: 'Bob' }, { name: 'Ben' } { name: 'Cole' } { name: 'Mary' } { name: 'Travis' } ] to an object like: const obj = { 'B': ['Bob', 'Ben'], 'C': ['Cole'], 'M': ['Mary'], 'T': ['Travis'] } Using only vanilla JS 回答1: I created an array where the key is the first letter of the name using the reduce function and restructuring the 'name' from the objects. If the key exists in the array the name is pushed (using spread operator).

问题 So I am trying to re-implement the reduce method, so it can add a couple of numbers that normally can be done using reduce , like: (reduce + [1 2 3]) ;; 6 (newRd + [1 2 3]) ;; 6 So I thought maybe it can be done using a recursive function that adds the first element of the vector every time it is called and do it again for the rest of the vector. Something like this: (defn newRd [list] (let [sum 0] (if (not= 0 (count list)) (+ sum (first list)) (newRd (rest list)) ) ) ) I think I am not doing