reduce

Consider the following array -of strings-: let arrayStrings = ["H", "e", "l", "l", "o"] For combining its elements (to get "Hello" as single String), we could: reduce it: let reducedString = arrayStrings.reduce("", { $0 + $1 }) // "Hello" Or join it: let joinedString = arrayStrings.joined() // "Hello" Both would return "Hello" String as output. However, what is the logic to keep in mind to determine what is the better choice for such a process? What is the difference when comparing based on the performance? There are two reasons why joined is a better choice than reduce : Readability If you

Problem 你想要遍历有序集合的所有元素，并且随着你对集合元素的遍历，对比两个相邻的元素 Solution 使用 reduceLeft, foldLeft, reduceRight, foldRight来遍历集合的元素，你的方法作用在相邻的两个元素上，从第一次要遍历的两个相邻元素开始，把你的方法作用在这两个元素上得到返回值，然后把你的方法继续作用在返回值和集合中第三要遍历的元素得到的返回值，再继续和第四个要遍历的元素作用。。。直到遍历完最后一个元素为止： scala> val a = Array(12, 6, 15, 2, 20, 9) a: Array[Int] = Array(12, 6, 15, 2, 20, 9) scala> a.reduceLeft(_ + _) res32: Int = 64 这个例子是这样的：12+6=18, 18+15=33, 33+2=35, 35+20=55, 55+9=64，就是对集合的所有元素求和 接下来你会看到如何使用reduceLeft来计算集合元素的乘积和求最大最小值： scala> a.reduceLeft(_ * _) res33: Int = 388800 scala> a.reduceLeft(_ min _) res34: Int = 2 scala> a.reduceLeft(_ max _) res35: Int =

I am trying to sum up values of a collection of maps by their common keys. I have this snippet: (def data [{:a 1 :b 2 :c 3} {:a 1 :b 2 :c 3}] (for [xs data] (map xs [:a :b])) ((1 2) (1 2)) Final result should be ==> (2 4) Basically, I have a list of maps. Then I perform a list of comprehension to take only the keys I need. My question now is how can I now sum up those values? I tried to use "reduce" but it works only over sequences, not over collections. Thanks. ===EDIT==== Using the suggestion from Joost I came out with this: (apply merge-with + (for [x data] (select-keys x [:col0 :col1 :col2

I've been reading up on reduce and have just found out that there is a 3 argument version that can essentially perform a map reduce like this: String[] strarr = {"abc", "defg", "vwxyz"}; System.out.println(Arrays.stream(strarr).reduce(0, (l, s) -> l + s.length(), (s1, s2) -> s1 + s2)); However I can't see the advantage of this over a mapToInt with a reduce. System.out.println(Arrays.stream(strarr).mapToInt(s -> s.length()).reduce(0, (s1, s2) -> s1 + s2)); Both produce the correct answer of 12, and both appear to work fine in parallel. Is one better than the other, and if so, why? Is one better

In this blog entry, "CSP and transducers in JavaScript" , the author states: First, we have to realise that many array (or other collection) operations like map , filter and reverse can be defined in terms of a reduce . So then we see a number of implementations of this in Clojure aren't lazy, they are eager: user> (defn eager-map [f coll] (reduce (fn [acc v] (conj acc (f v))) [] coll)) #'user/eager-map user> (eager-map inc (range 10)) [1 2 3 4 5 6 7 8 9 10] My question is, are Clojure transducers eager? Beyamor Transducers are very simple functions - they don't have a notion of laziness or,

How can I implement version 7 of the code given in the following link: http://www.cuvilib.com/Reduction.pdf for an input array whose size is an arbitrary number, in other words, not a power of 2? Version 7 already handles an arbitrary number of elements. Perhaps instead of referring to the cuvilib link, you should look at the link to the relevant NVIDIA CUDA reduction sample . It includes essentially the pdf file you are using, but also sample codes that implement reductions 1 through 7 (labelled reduce0 through reduce6 ) If you study the description of the reduction 7 in the document, you'll

Alright, so I had some help a couple months ago with coming up with a solution to keep count of the elements in an array: Loop through multiple array and keep count of each element This solution worked perfectly for me until I realized that it's using ES6 which is not supported by IE 11 . I've tried to convert it to using functions instead of the arrow function so that it'll work across all browsers but am having some issues. Here is the current code that does not work in IE: var b = data.reduce((acc, cur) => { cur.ProductHandlingTypes.map(({ Name }) => Name).forEach(n => acc[n] = (acc[n] || 0