Apply same arguments to Lodash flow callbacks - javascript

The Lodash flow() function works, as shown in this question, by giving it a bunch of functions in an array. The result of the previous function will be applied to the next one.
And this is exactly what is problematic for my use case, as it throws away any arguments provided. I need to pass them on to the next function in line, instead of just the return value *).
*) Yes, the return value should get returned modified and be passed on the next function
Using flow() is just what I came up with. Any other solution achieving the goal is ok as well.
What it does:
_.flow( [ firstFn, secondFn ] )
( first, second, third ) => firstFn( first, second, third ) => return first * first
( first ) => secondFn( first )
What it should do:
_.flow( [ firstFn, secondFn ] )
( first, second, third ) => firstFn( first, second, third ) => return first * first
( first, second, third ) => secondFn( resultFromFirstFn, second, third )
Here's some code to demo it:
const checkLog = ( val, attr, group ) => {
console.log( val, attr, group )
return val
}
// This basically is "easy to use"-API and what will be mostly used.
const Callbacks = {
demo: [ _.trim, _.toLower, checkLog ],
}
// _.cond() is just a switch
// [ <condition-to-exec-cb>, <cb-if-cond-true> ]
const cond = _.cond( _.map(
[
[ _.stubTrue, Callbacks.demo ],
],
// Monkey patching _.flow() to each array of cond/cb above
cb => _.set( cb, '1', _.flow( _.last( cb ) ) )
) )
const res = _.map( {
" city": " Sample City",
"street": "Sample Street",
}, cond )
console.log( res )
Note: In case you wonder about why I write it like this: There's a single line comment that points to which part will get altered and extended a lot in the future and this must, due to the target group working there, be simple. So it's just an array of stacked callbacks.

The problem with _.flow() is that it expects everything after the first argument to be a unary function so it can pass the result through all of them. This presents a challenge when you want to apply the same set of arguments to all function and only alter the first one. There are a few ways you could go about this.
Using Lodash
You could leverage _.partialRight to do a partial allocation of every function. partialRight will apply arguments from right to left depending on how many arguments the function takes.
const fn = (a, b, c, d) => console.log(a, b, c, d);
// behaves similar to (a, b) => console.log(a, b, "yak", "zebra")
const newFn = _.partialRight(fn, "yak", "zebra");
newFn("alpaca", "beaver");
<script src="https://cdn.jsdelivr.net/npm/lodash#4.17.19/lodash.min.js"></script>
However you can still pass arguments left to right and this would push the partially applied arguments to right:
const fn = (a, b, c, d) => console.log(a, b, c, d);
// behaves similar to (a, b) => console.log(a, b, "yak", "zebra")
const newFn = _.partialRight(fn, "yak", "zebra");
newFn("alpaca", "beaver", "cat");
<script src="https://cdn.jsdelivr.net/npm/lodash#4.17.19/lodash.min.js"></script>
Assuming all of the function will have at most take n amount of arguments we can:
Get an array of n arguments to be applied to all functions.
Get all functions.
Do partialRight on all functions using all but the first argument. Since we've applied n-1 arguments to each, now all of them can be called as if they are unary.
Use flow on the new functions from step 3.
Start the chain using the first argument.
This will then call all functions such that the first argument is the last result and the second and onwards arguments would be based on the initial ones thus always the same for all functions
function nAryFlow(argsArray, ...fns) { //1. and 2. - get arguments and functions
const start = argsArray[0];
const rest = argsArray.slice(1);
const convertedFns = fns.map(f => _.partialRight(f, ...rest)) //3. Turn all functions into unary ones
return _.flow( //4. `flow` through the functions
...convertedFns
)(start); //5. use the initial fist argument to kick things off
}
const fn1 = (who1, who2, who3) => `${who1}, ${who2}, and ${who3} are best friends.`;
const fn2 = (friends, who2, who3) => `${friends} Especially ${who2} and ${who3}.`;
const fn3 = (story, who2) => `${story} They all met at a party thrown by ${who2}`;
const args = ["Alice", "Bob", "Carol"];
const result = nAryFlow(args, fn1, fn2, fn3);
console.log(result)
<script src="https://cdn.jsdelivr.net/npm/lodash#4.17.19/lodash.min.js"></script>
Without using Lodash
All of the above can very easily be done without Lodash. Instead, we can use Array#reduce to go through all of the functions and apply the arguments. This time, we're directly applying them instead of pre-processing the functions but the overall operation and effect is the same:
The first function takes all the arguments.
Any further function takes the result of the last function and the second argument onwards from the start:
function nAryFlow(argsArray, ...fns) {
const start = argsArray[0];
const rest = argsArray.slice(1);
return fns.reduce((last, f) => f(last, ...rest), start);
}
const fn1 = (who1, who2, who3) => `${who1}, ${who2}, and ${who3} are best friends.`;
const fn2 = (friends, who2, who3) => `${friends} Especially ${who2} and ${who3}.`;
const fn3 = (story, who2) => `${story} They all met at a party thrown by ${who2}`;
const args = ["Alice", "Bob", "Carol"];
const result = nAryFlow(args, fn1, fn2, fn3);
console.log(result)
Variation using more higher order functions
Just as a variation, this could be split up into multiple higher order functions which might produce a nicer syntax for some situations nAryFlow(f1, f2, f3, fN)(arg1, arg2, arg3, argN):
function nAryFlow(...fns) {
return function (...args) {
const start = args[0];
const rest = args.slice(1);
return fns.reduce((last, f) => f(last, ...rest), start);
}
}
const fn1 = (who1, who2, who3) => `${who1}, ${who2}, and ${who3} are best friends.`;
const fn2 = (friends, who2, who3) => `${friends} Especially ${who2} and ${who3}.`;
const fn3 = (story, who2) => `${story} They all met at a party thrown by ${who2}`;
const chain = nAryFlow(fn1, fn2, fn3);
const result1 = chain("Alice", "Bob", "Carol");
const result2 = chain("Rachel Green", "Monica Geller", "Chandler Bing");
console.log(result1);
console.log(result2);
Using a Functor
For a slightly different take here, you can use an algebraic structure called Functor to ease the syntax. The essential thing about functors is that they have a .map() method that accepts a function. If you're reminded of Array#map then you're not wrong.
The basic idea is that the functor holds a value and allows it to be modified with functions via .map(). It can then dictate how the function is applied to its value. The result of .map is always the same type of functor as what was already mapped, so you can always continue mapping and be sure that the application would be the same every time. With arrays you always get a new array of the same length with every member transformed. Other functors can apply the function given to .map() different to what an array does but it would always be consistent.
So, background done, here is how this functor can look like:
class NAryFlow {
constructor(argsArray) {
this.value = argsArray[0];
this.rest = argsArray.slice(1);
}
static of(argsArray) {
return new NAryFlow(argsArray);
}
map(fn) {
return NAryFlow.of(
[ fn(this.value, ...this.rest), ...this.rest ]
);
}
}
const fn1 = (who1, who2, who3) => `${who1}, ${who2}, and ${who3} are best friends.`;
const fn2 = (friends, who2, who3) => `${friends} Especially ${who2} and ${who3}.`;
const fn3 = (story, who2) => `${story} They all met at a party thrown by ${who2}`;
const result = NAryFlow.of(["Alice", "Bob", "Carol"])
.map(fn1)
.map(fn2)
.map(fn3)
.value;
console.log(result)
A similar idea as the two others above - we take the arguments, and apply them all to each function we give to .map(). Every next time we call .map() the first argument would be the last result.
And here is a slight variation using ES6 getters. I think it has a slightly better syntax but wanted to keep the previous implementation simpler.
class NAryFlow {
constructor(argsArray) {
this.args = argsArray;
}
static of(argsArray) {
return new NAryFlow(argsArray);
}
map(fn) {
return NAryFlow.of(
[ fn(...this.args), ...this.rest ]
);
}
get value() {
return this.args[0];
}
get rest() {
return this.args.slice(1);
}
}
const fn1 = (who1, who2, who3) => `${who1}, ${who2}, and ${who3} are best friends.`;
const fn2 = (friends, who2, who3) => `${friends} Especially ${who2} and ${who3}.`;
const fn3 = (story, who2) => `${story} They all met at a party thrown by ${who2}`;
const result = NAryFlow.of(["Alice", "Bob", "Carol"])
.map(fn1)
.map(fn2)
.map(fn3)
.value;
console.log(result)

flow doesn't throw any arguments, it's just how javascript functions work, they don't respect the a particular signature, you can call a function that only takes 1 parameter with 100 parameters and that won't throw an error.
Also flow is defined such that it passes the initial parameters all to the first function regardless of whether it's going to use them all or not.
You can create your own flow function(since it is a simple one) that only passes the exact amount of parameters a function needs and keep the rest for the rest of functions. To know how many parameters a function takes, you can check its length property like so:
function flow(funcs) {
const length = funcs.length;
for(let func of funcs) {
if (typeof func !== 'function') {
throw new TypeError('Expected a function');
}
}
return function(...args) {
let result = args[0];
for(let func of funcs) {
result = func.apply(this, args.slice(0, func.length)); // call this function with only as many args as it needs
args = [result, ...args.slice(func.length)]; // args becomes the result + the rest of args after the previous call
}
return result;
}
}
Note: This won't work if one of the functions uses the rest parameters, in that case length will return 0. You can't know how many parameters that functions takes. If you wan to call that function with all the available parameters, then just change the for loop above to:
for(let func of funcs) {
result = func.apply(this, args.slice(0, func.length || +Infinity)); // if 'func.length' is 0, '+Infinity' is used instead which will use all the available args
args = [result, ...args.slice(func.length || +Infinity)]; // same goes here, '.slice(+Infinity)' will result in an empty array. 'result' will always be present which is the expected behavior
}
Demo:
function flow(funcs) {
const length = funcs.length;
for(let func of funcs) {
if (typeof func !== 'function') {
throw new TypeError('Expected a function');
}
}
return function(...args) {
let result = args[0];
for(let func of funcs) {
result = func.apply(this, args.slice(0, func.length)); // call this function with only as many args as it needs
args = [result, ...args.slice(func.length)]; // args becomes the result + the rest of args after the previous call
}
return result;
}
}
function square(n) {
return n * n;
}
function add(a, b, c) {
return a + b + c;
}
console.log(flow([square, add])(2, 3, 4));

Related

tried to recreate the _.once function but only works upon at least 2 iterations not one

onceCopy function (testFunc) {
const copyFunc = (a) => {
const copyFunc2 = (b) => {
return testFunc(a);
};
return copyFunc2;
};
return copyFunc;
};
So the function returns the inner function upon first invocation.
Then returns the inner function of the inner function of the second invocation.
Then the second inner function (third invocation) actually returns the passed argument in the parent function and only invokes it with the character we gave it on the second invocation.
Ideally I want to achieve what I'm achieving over many invocations after only the first one if that makes sense.
Edit: Yes sorry, _.once.
Edit: so first invocation onceCopy SHOULD hold a copy of the Func passed
Second invocation SHOULD trigger the copy and gives an ouput
Third invocation SHOULD give the result of the second invocation so should the fourth, fifth, sixth and so on...
My function does do this, but on the second invocation it stores a function (copyFunc2) again, but I just made that because I need somewhere to store "a".
so like we have
function multiplyBy3 (a) {return a*3}
and then once copy stores a copy of multiplyBy3
const actualFunction = onceCopy(multiplyBy3)
then upon second and third invocation what I want
actualFunction(1) = 3
actualFunction(66) = 3
so the passed function ONLY RUNS ONCE
Cant explain more than this, its in the lodash docs.
I'm not familiar with the function you're trying to reimplement, so feel free to correct me if I misunderstood. To wrap a function and ensure it's only called once you don't need multiple nested wrappings, only one with some state.
You need to keep track of whether you already have a result to return (hasResult) and if so, what that result is (result). Keeping these two variables separate allows you to cover the case when result is undefined while keeping the code easy to read and understand.
function once(wrappedFunction) {
let hasResult = false;
let result;
return (...args) => {
if (hasResult) {
return result;
}
result = wrappedFunction.apply(this, args);
hasResult = true;
return result;
}
}
// An example function to wrap
function multiply(a, b) {
return a * b;
}
const demoFunction = once(multiply)
console.log(demoFunction(1, 2)); // 2
console.log(demoFunction(3, 4)); // 2
console.log(demoFunction(5, 6)); // 2
Is this what you were looking for?
The initial function return is kept on subsequent calls. I used a second variable called in case the first call returns undefined, which should also be returned on subsequent calls.
const once = (onceFn) => {
let called;
let value;
return (...args) => {
if (called) return value;
called = true;
return (value = onceFn(...args));
};
};
function multiplyBy3(a) {
return a * 3;
}
const fn = once(multiplyBy3);
console.log(fn(3)); // 9
console.log(fn(66)); // 9
After calling the function for the 1st time, and getting the result, create a new function that returns the result, and use it whenever the wrapped function is called:
const once = fn => {
let func
return (...args) => {
if(func) return func()
const result = fn(...args)
func = () => result
return result
}
}
const multiply = (a, b) => a * b
const demoFunction = once(multiply)
console.log(demoFunction(1, 2)); // 2
console.log(demoFunction(3, 4)); // 2
console.log(demoFunction(5, 6)); // 2

Memoization: can the arguments be used as key in the cache object?

I have this solution for a memoization function.
const slice = Array.prototype.slice
function memoize(fn){
const cache = {}
return (...args) => {
const params = slice.call(args)
console.log(params)
if(cache[params]){
console.log('cached')
return cache[params]
} else{
let result = fn(...args)
cache[params] = result
console.log('not cached')
return result
}
}
}
cache[params] is the point. cache is an object and params is an array. Will this always work? After some researching I am still not confident this code is correct.
Such memoization can work for very simple cases, but it does not work in many other cases, for instance when:
the arguments are objects. They will often stringify to "[object Object]", and so different objects are treated as if they are the same.
the arguments are strings but cannot be distinguished because of the poor way they are separated when the arguments array is stringified (comma delimiter)
Some demos of failures:
Arguments are Objects
const slice = Array.prototype.slice
function memoize(fn){
const cache = {}
return (...args) => {
const params = slice.call(args)
if(cache[params]){
return cache[params]
} else{
let result = fn(...args)
cache[params] = result
return result
}
}
}
// The function we will test with:
function sum(a) {
let total = 0;
for (let value of a) total += value;
return total;
}
function test() {
console.log(sum(new Set([1,2,3]))); // should be 6
console.log(sum(new Set([2,4,6]))); // should be 12
}
console.log("Run test without memoization...");
test(); // Perform test without memoization
sum = memoize(sum); // Memoize the function
console.log("Run test WITH memoization...");
test(); // Test again, but now with memoization
Arguments are strings that have commas:
const slice = Array.prototype.slice
function memoize(fn){
const cache = {}
return (...args) => {
const params = slice.call(args)
if(cache[params]){
return cache[params]
} else{
let result = fn(...args)
cache[params] = result
return result
}
}
}
// The function we will test with:
function compareString(a, b) {
return a.localeCompare(b); // returns -1, 0 or 1.
}
function test() {
console.log(compareString("b,a", "c")); // should be -1
// Change the arguments such that the concatenation with comma remains the same
console.log(compareString("b", "a,c")); // should be 1
}
console.log("Run test without memoization...");
test(); // Perform test without memoization
compareString = memoize(compareString); // Memoize the function
console.log("Run test WITH memoization...");
test(); // Test again, but now with memoization
Other remarks on the code
Calling slice is useless, as args is already a new array.
if(cache[params]) will evaluate to false when the already cached value is a falsy value, like 0, "", false. Doing if (params in cache) would avoid that issue
params will be stringified, which (as shown above) is not guaranteed to uniquely identify an array of arguments.
Improvement
If we can require that the arguments passed to our function are immutable, then we can use these immutable values or references as keys in a Map.
This Map would become a tree of Maps when there are multiple arguments, so that when a lookup is made for the first argument in the main Map, it returns a nested Map, and then in that Map the next argument would be used as key, ...etc, until the deep Map is found for the last argument. In that final Map, a reserved key is used to retrieve the cached value. This reserved key can be Symbol that is only known by the memoize function, so that it can never collide with an argument value.
Disclaimer: This will not work when objects can mutate between calls.
Here is how that looks:
function memoize(fn){
const end = Symbol("end"); // A unique reference, only known here
const cache = new Map;
return (...args) => {
let node = args.reduce((node, arg) => {
if (!node.has(arg)) node.set(arg, new Map);
return node.get(arg);
}, cache);
if (!node.has(end)) node.set(end, fn(...args));
return node.get(end);
}
}
// The function we will test with:
let numCalls = 0;
function length(a) { // Length of a linked list
numCalls++; // Keep track of the number of calls made
return a ? length(a.next) + 1 : 0;
}
function test() {
numCalls = 0; // Reset the number of calls
let head = { next: { next: { next: {}}}}; // Linked list with 4 nodes
console.log(length(head)); // should be 4
// Now exclude the head node:
console.log(length(head.next)); // should be 3
console.log("number of calls: ", numCalls);
}
console.log("Run test without memoization...");
test(); // Perform test without memoization
length = memoize(length); // Memoize the function
console.log("Run test WITH memoization...");
test(); // Test again, but now with memoization
Again, this cannot be used when objects mutate between calls. But for all other scenarios it should work fine.
Object keys are supposed to be string/symbols.
Depending on how you input array affects the output, you can .join() the array and use it as your cache key:
const slice = Array.prototype.slice
function memoize(fn){
const cache = {}
return (...args) => {
const params = slice.call(args)
let paramsString = params.sort().join('');
console.log(paramsString)
if(cache[paramsString]){
console.log('cached')
return cache[paramsString]
} else{
let result = fn(...args)
cache[paramsString] = result
console.log('not cached')
return result
}
}
}
If the order does not matter then you can .sort(). If order is important then you can remove the sort and just join. This is not perfect but works for simple cases.

Array.reduce somewhat complex statement?

export default function compose(...funcs) {
if (funcs.length === 0) {
return arg => arg
}
if (funcs.length === 1) {
return funcs[0]
}
// what does this line do???
return funcs.reduce((a, b) => (...args) => a(b(...args)))
}
I'm reading compose.js source code from redux module. I know basics of reduce function, it just calls the reducer in every element in the collection and returns a single output. It has an accumulator to store the intermediary results and to pass it in each call of the reducer. But reduce in this code is somewhat vague for me, can someone explain it in layman terms?
(a, b) can be written as (previousComposition, currectFunc). The result of the reducing is a function taking (...args) - none or more arguments.
Every "loop" of reducing returns a function which composes previous composition with the current function a(b(...args)) - it can be understood as (a . b)(x) = a(b(x)).
Only a function is returned, it means the functions are not executed yet. It happens first, when the result composition function is called: var res = compose(a, b)(x)
Example:
var plus1 = x => x + 1;
var plus2 = x => x + 2;
var mult2 = x => x * 2;
var composedFn = compose(mult2, plus2, plus1);
// (...x) => mult2(plus2(plus1(...x)))
var res = composedFn(3);
console.log(res); // 12
As you can see in the example, first plus1 was applied, then plus2 and finally mult2.

What is the correct arrow function syntax?

Arrow functions can be written in a lot of different ways, are there any way that is "more correct"?
let fun1 = (a) => { return a; }
let fun2 = a => a;
Just like those cases above, is fun2 faster then fun1? What's the diference between them?
Arrow functions can be written in some diferent ways, like below:
// No params, just returns a string
const noParamsOnlyRet = () => 'lul';
// No params, no return
const noParamsNoRet = () => { console.log('lul'); };
// One param, it retuns what recives
const oneParamOnlyRet = a => a;
// Makes the same thing that the one above
const oneParamOnlyRet2 = a => { return a; };
// Makes the same thing that the one above
const oneParamOnlyRet3 = (a) => a;
/* Makes the same thing that the one above, parentheses in return are optional,
* only needed if the return have more then one line, highly recomended put
* objects into parentheses.
*/
const oneParamOnlyRet4 = (a) => (a);
// Mult params, it returns the sum
const multParamsOnlyRet = (a, b) => a + b;
// Makes the same thing that the one above
const multParamsOnlyRet2 = (a, b) => { return a + b; }
// Curly brackets are needed in multiple line arrow functions
const multParamsMultLines = (a, b) => {
let c = a + b;
return c;
}
So we have some rules:
Parentheses around the params are needed when the function has none or more than one param, if it have just one, parentheses can be omitted.
If the function just have a return the curly brackets and the keyword return can be omitted, if it fits in one line
Curly brackets are needed if the function have more than one line or if it have no return.
As you can see, all of this are valid syntax for arrow functions, none of them is faster than the other, which one you use depends of the way that you code.
From
function(a){return a}
remove function and add => between the () and the {}:
(a) => {return a}
If you only have one argument, you can remove the brackets:
a => {return a}
if everything in the {} is a return statement, you can remove the {return ...} and leave just the statement:
a => a

Curry function without partial application?

I am following a tutorial on firebase authentication in ReactJS. In the code, I stumbled across such function:
const byPropKey = (propertyName, value) => () => ({
[propertyName]: value,
});
I suppose this function is used to set state of a React component. It is used like so:
<input
value={username}
onChange={event => this.setState(byPropKey('username', event.target.value))}
type="text"
placeholder="Full Name"
/>
I am a newcomer to React, JavaScript, ES6 and functional programming in general, so this confuses the hell out of me. I know that the double arrow notation indicates a curried function, which enables partial application of the function. In this case, however, I can not see how this gets used in such scenario.
Here is the tutorial I am referring to:
tutorial
I can not see how this gets used in such scenario.
The tutorial author is currying that function because setState can take either an object or a function. A function gives you access to the previous state and props. Here is a sandbox example.
In the example you provided, the tutorial author is not using the partial application, and thus not leveraging the access to previous state and props. So in this specific case, a non curried implementation (passing an object to setState) would do just fine.
However, if you were to add to this tutorial codebase, the future code might require byPropKey to access the prevState and props:
const byPropKey = (propertyName, value) => (prevState, props) => {
//... do something with prevState and props
return {
[propertyName]: value,
}
};
... which is probably why the tutorial author wrote the function curried.
byPropKey
function uses as helper to map state and returns result like:
{ "statePiceName": value }
to set your state.
Same behavior without this function:
{event => this.setState({ username: event.target.value })}
Also you can just console.log(byPropKey('key', 'customValue')) and see how its works.
You misunderstand the double arrow. This just creates a function, not necessarily a curried or partial function.
For example:
// just a function
let doSomething = () => console.log("did something")
//call it
doSomething()
You can return another function from a function (aka a higher-order function), which is what's going on in your example:
// this will return the function () => console.log("said " + something)
let saySomething = (something) => () => console.log("said " + something)
// call the above function, which returns a different function
let sayHello = saySomething("Hello")
// now call that function
sayHello()
You can make partial functions with bind:
function add(a, b) {console.log(a + b)}
let addFive = add.bind(null, 5)
addFive(3)
addFive(10)
To take it one more step, you can make a generic function to add whatever:
// just a regular function
let add = (a,b) => console.log(a + b)
// a function to create a partial of the above based on passed in argument
let addThis = (a) => add.bind(null, a)
// make an add-10 function
let addTen = addThis(10)
// call it
addTen(12)
// make an add-33 function
let add33 = addThis(33)
add33(100)
EDIT: In response to the comments
Curried functions are those the break down larger functions that take multiple arguments into smaller functions that take fewer or one argument.
For example this simple function:
let add = (a, b, c) => a + b + c
can be curried into three functions that each take a single argument like this:
let addCurried = (a) => (b) => (c) => a + b + c
You can call this with some or all the arguments to get the result or a partial:
// non curried: F(a, b, c) -> sum
// curried: F():a -> (b -> (c -> sum))
let addCurried = (a) => (b) => (c) => a + b + c
// add everything
let total = addCurried(1)(2)(3) // -> 6
// make a partial
let addTwoAndThree = addCurried(2)(3)
// call it
let total2 = addTwoAndThree(100) //-> 105
console.log(total, total2)
It's hard to see given the definition of currying how a function that accepts two arguments and returns function the accepts none is considered currying that function.

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