ramda.js

Ramda chain usage

偶尔善良 提交于 2019-11-29 02:03:11
From the documentation : var duplicate = n => [n, n]; R.chain(duplicate, [1, 2, 3]); //=> [1, 1, 2, 2, 3, 3] R.chain(R.append, R.head)([1, 2, 3]); //=> [1, 2, 3, 1] The first example is very straight forward, it applies duplicate() to every element in the array and concatenates the result. But I am having trouble understanding the second example. How exactly is it mapping R.append + R.head over the array? Can someone please provide a step by step explanation for the second example ? I am familiar with compose and currying. Thanks The second example is showing how R.chain can be used things

Ramda: How to filter based on a value in a nested array

大城市里の小女人 提交于 2019-11-28 21:43:50
I'm trying to accomplish this in a functional manner (with Ramda). My JSON is structured like this [ {username: 'bob', age: 30, tags: ['work', 'boring']}, {username: 'jim', age: 25, tags: ['home', 'fun']}, {username: 'jane', age: 30, tags: ['vacation', 'fun']} ] and I am trying to filter based on a value in the 'tags' property, but have not been successful. I am able to filter on ints/strings (age and username), but I can't figure out how to do so with values in nested arrays (tags). Any help would be much appreciated. There are many ways you could do this. But I think the cleanest one would

Can't wrap my head around “lift” in Ramda.js

孤街醉人 提交于 2019-11-28 19:20:33
Looking at the source for Ramda.js, specifically at the "lift" function. lift liftN Here's the given example: var madd3 = R.lift(R.curry((a, b, c) => a + b + c)); madd3([1,2,3], [1,2,3], [1]); //=> [3, 4, 5, 4, 5, 6, 5, 6, 7] So the first number of the result is easy, a , b , and c , are all the first elements of each array. The second one isn't as easy for me to understand. Are the arguments the second value of each array (2, 2, undefined) or is it the second value of the first array and the first values of the second and third array? Even disregarding the order of what's happening here, I

Good names for flipped versions of `lt`, `lte`, `gt`, and `gte`?

此生再无相见时 提交于 2019-11-27 20:27:39
问题 I've been working for some time on a Javascript FP library called Ramda , and I'm having a slight problem with naming things. (You've heard the old line, right? "There are only two hard problems in Computer Science: cache invalidation, naming things, and off-by-one errors.") In this library, (almost) every function of more than one parameter is automatically curried. And this works well for most use-cases. But there are some issues with a few functions which are non-commutative binary

Ramda chain usage

放肆的年华 提交于 2019-11-27 16:23:50
问题 From the documentation: var duplicate = n => [n, n]; R.chain(duplicate, [1, 2, 3]); //=> [1, 1, 2, 2, 3, 3] R.chain(R.append, R.head)([1, 2, 3]); //=> [1, 2, 3, 1] The first example is very straight forward, it applies duplicate() to every element in the array and concatenates the result. But I am having trouble understanding the second example. How exactly is it mapping R.append + R.head over the array? Can someone please provide a step by step explanation for the second example ? I am

Ramda: How to filter based on a value in a nested array

假如想象 提交于 2019-11-27 02:02:49
问题 I'm trying to accomplish this in a functional manner (with Ramda). My JSON is structured like this [ {username: 'bob', age: 30, tags: ['work', 'boring']}, {username: 'jim', age: 25, tags: ['home', 'fun']}, {username: 'jane', age: 30, tags: ['vacation', 'fun']} ] and I am trying to filter based on a value in the 'tags' property, but have not been successful. I am able to filter on ints/strings (age and username), but I can't figure out how to do so with values in nested arrays (tags). Any help

Sort an array of objects based on another array of ids

大城市里の小女人 提交于 2019-11-26 22:06:13
问题 I have 2 arrays a = [2,3,1,4] b = [{id: 1}, {id: 2}, {id: 3}, {id: 4}] How do I get b sorted based on a ? My desired output would be c = [{id: 2}, {id: 3}, {id: 1}, {id: 4}] I would prefer to use Ramda or regular JS. 回答1: Ramda really shines for these types of problems. Where the size of the data is small, we can use a simple reduce function, and indexOf helper. // match id of object to required index and insert var sortInsert = function (acc, cur) { var toIdx = R.indexOf(cur.id, a); acc

RxJS Promise Composition (passing data)

别来无恙 提交于 2019-11-26 07:37:21
问题 I\'m brand new to Rx and am finding it difficult to find documentation on composing promises such that data from the first promise is passed into the second and so on. Here\'s three very basic promises, the calculations on the data aren\'t important, just that something async has to be done using data from the previous promise. const p1 = () => Promise.resolve(1); const p2 = x => { const val = x + 1; return Promise.resolve(val); }; const p3 = x => { const isEven = x => x % 2 === 0; return