JavaScript函数式编程 - 链接函数和使用匿名函数 [英] JavaScript Functional Programming - Chaining Functions and using an anonymous function
问题描述
所以我做了一个
em> pass()函数接受函数作为参数,以便我可以使用匿名函数来操作该值并返回它。因此,在这种情况下,它从reduce中获取值并将其传递给函数,以便它可以操纵该值然后返回它。
它 WORKS 但是我真的不想在Object原型中添加一个方法!
我怎么能以另一种方式做这件事,同时仍然保持功能的链接?
简单示例
Object.prototype.pass = function(fn){
return fn(this);
};
var value = 1;
var new_value = value.pass(function(num){
return num + 1;
});
console.log(value,new_value); //输出:1 2
CodeWars上下文的问题
Object.prototype.pass = function(fn){return fn(this)}; function digPow(n,p){返回n .toString().split('').reduce(函数(total,num,i){return total + Math.pow(parseInt(num),(p + i))},0).pass(function (总数){return(total%n == 0)?Math.floor(total / n):-1;});} // digPow(89,1)应该返回1,因为8¹+9²= 89 = 89 * 1console.log(Test Case 1 returns(,digPow(89,1),)should return 1)// digPow(92,1)应该返回-1,因为没有k,例如9¹+2²等于92 * kconsole.log(测试用例2返回(,digPow(92,1),)应该返回-1)// digPow(695,2)应该返回2,因为6²+9³+ 5 4 = 1390 = 695 * 2console.log(测试用例3返回(,digPo (测试用例4返回(w(695,2),)应该返回2)// digPow(46288,3)应该返回51,因为4 3 + 6 4 + 2 5 + 8 8 + = 2360688 = 46288 * 51console.log ,digPow(46288,3),)应该返回51)
b
$ b Code Wars Instructions
有些数字具有有趣的属性。例如:
89 - >8¹+9²= 89 * 1
695 - >6²+ 9 3 + 5 4 = 1390 = 695 * 2
46288 - > 4 3 + 6 4 + 2 5 5 + 8 8 + 8 8 = 2360688 = 46288 * 51给定
positive整数n写成abcd ...(a,b,c,d ...是数字)和
a正整数p我们希望找到一个正整数k,如果
存在,比如对于p的连续
次幂取n的数位之和等于k * n。换句话说:
是否有整数k,例如:(a ^ p + b ^(p + 1)+ c ^(p + 2)+ d ^
(p + 3)+ ...)= n * k如果是这种情况,我们将返回k,如果没有返回
-1。
注意:n,p将始终以严格正整数形式给出。
解决方案只是不使用方法链 - 你想应用于结果的功能不是它的一种方法。有各种各样的方法: 一个简单的变量: IIFE调用(此方法对静态函数效果更好): 实验绑定运算符(使用静态方法更好):
函数digPow(n,p){
const total = n
.toString()
.split('')
.reduce(函数(total,num,i){
return total + Math.pow(parseInt(num),(p + i))
},0);
返回(总共%n == 0)? Math.floor(total / n):-1;
}
函数digPow(n,p){
return(函数(总计){
return(total %n == 0)?Math.floor(total / n):-1;
}(n
.toString()
.split('')
.reduce (函数(total,num,i){
return total + Math.pow(parseInt(num),(p + i))
},0));
}
function digPow(n,p){
return n
.toString()
.split('')
.reduce(function(total,num,i){
return total + Math.pow(parseInt(num),(p + i))
},0)
:: function(){
return(this%n == 0)?Math.floor(this / n):-1;
}();
}
您也可以在传递
方法中使用任何方法。
I was doing a question on CodeWars and practicing some functional programming when I encountered a problem while trying to apply a function to a value.
So I made a pass() function that accepts a function as an argument so that I could use an anonymous function to manipulate that value and then return it. So, in this case, it takes the value from reduce and passes it to a function so it can manipulate that value then return it.
It WORKS but I really don't want to add a method to the Object prototype!
How can I do this another way while still maintaining the chaining of functions?
Simple Example
Object.prototype.pass = function(fn) {
return fn(this);
};
var value = 1;
var new_value = value.pass(function(num){
return num + 1;
});
console.log(value, new_value); // Outputs: 1 2
CodeWars Problem for context
Object.prototype.pass = function(fn) {
return fn(this)
};
function digPow(n, p) {
return n
.toString()
.split('')
.reduce(function(total, num, i) {
return total + Math.pow(parseInt(num), (p + i))
}, 0)
.pass(function(total) {
return (total % n == 0) ? Math.floor(total / n) : -1;
});
}
//digPow(89, 1) should return 1 since 8¹ + 9² = 89 = 89 * 1
console.log("Test Case 1 returns (", digPow(89, 1), ") should return 1")
//digPow(92, 1) should return -1 since there is no k such as 9¹ + 2² equals 92 * k
console.log("Test Case 2 returns (", digPow(92, 1), ") should return -1")
//digPow(695, 2) should return 2 since 6² + 9³ + 5⁴= 1390 = 695 * 2
console.log("Test Case 3 returns (", digPow(695, 2), ") should return 2")
//digPow(46288, 3) should return 51 since 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51
console.log("Test Case 4 returns (", digPow(46288, 3), ") should return 51")
Code Wars Instructions
Some numbers have funny properties. For example:
89 --> 8¹ + 9² = 89 * 1
695 --> 6² + 9³ + 5⁴= 1390 = 695 * 2
46288 --> 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51 Given a positive integer n written as abcd... (a, b, c, d... being digits) and a positive integer p we want to find a positive integer k, if it exists, such as the sum of the digits of n taken to the successive powers of p is equal to k * n. In other words:
Is there an integer k such as : (a ^ p + b ^ (p+1) + c ^(p+2) + d ^ (p+3) + ...) = n * k If it is the case we will return k, if not return -1.
Note: n, p will always be given as strictly positive integers.
The solution is just to not use method chaining - the functionality you want to apply to the result is not a method of it. There are various ways around:
A simple variable:
function digPow(n, p) { const total = n .toString() .split('') .reduce(function(total, num, i) { return total + Math.pow(parseInt(num), (p + i)) }, 0); return (total % n == 0) ? Math.floor(total / n) : -1; }
An IIFE call (this approach works better with static functions):
function digPow(n, p) { return (function(total) { return (total % n == 0) ? Math.floor(total / n) : -1; }(n .toString() .split('') .reduce(function(total, num, i) { return total + Math.pow(parseInt(num), (p + i)) }, 0)); }
The experimental bind operator (works better with static "methods" as well):
function digPow(n, p) { return n .toString() .split('') .reduce(function(total, num, i) { return total + Math.pow(parseInt(num), (p + i)) }, 0) :: function() { return (this % n == 0) ? Math.floor(this / n) : -1; }(); }
You also could use any of the approaches with your pass
"method".
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