问题
I have an array which I map
over. I need to compare the current element with the previous. I am detecting if the current element is the same as the previous element by comparing their id
s and doing something different based on this condition. Is there any purely functional way to do it without doing index math?
items.map((item, index) => {
if(item.id === items[index - 1 > 0 ? index - 1 : 0].id) {
// do something
} else {
// do something else
}
})
The code works but I would like to avoid doing math on the index. Is there any way to do it?
回答1:
Are you sure that you want a map? This sounds like an XY problem. If you want to map over adjacent elements of an array then you'd have to define your own function.
const mapAdjacent = (mapping, array) => {
const {length} = array, size = length - 1, result = new Array(size);
for (let i = 0; i < size; i++) result[i] = mapping(array[i], array[i + 1]);
return result;
};
const items = [1, 2, 3, 4, 5];
const result = mapAdjacent((x, y) => [x, y], items);
console.log(result); // [[1, 2], [2, 3], [3, 4], [4, 5]]
Note that this will throw a RangeError
if you give it an empty array as input.
const mapAdjacent = (mapping, array) => {
const {length} = array, size = length - 1, result = new Array(size);
for (let i = 0; i < size; i++) result[i] = mapping(array[i], array[i + 1]);
return result;
};
const items = [];
const result = mapAdjacent((x, y) => [x, y], items); // RangeError: Invalid array length
console.log(result);
I think this is good behaviour because you shouldn't be giving mapAdjacent
an empty array to begin with.
Here's a purely functional implementation of mapAdjacent
which uses reduceRight
. As an added bonus, it works for any iterable object.
const mapAdjacent = (mapping, [head, ...tail]) =>
tail.reduceRight((recur, item) => prev =>
[mapping(prev, item), ...recur(item)]
, _ => [])(head);
const items = "hello";
const result = mapAdjacent((x, y) => [x, y], items);
console.log(result); // [['h', 'e'], ['e', 'l'], ['l', 'l'], ['l', 'o']]
Unlike the iterative version, it returns an empty array instead of throwing an error if you give it an empty array as input.
const mapAdjacent = (mapping, [head, ...tail]) =>
tail.reduceRight((recur, item) => prev =>
[mapping(prev, item), ...recur(item)]
, _ => [])(head);
const items = "";
const result = mapAdjacent((x, y) => [x, y], items);
console.log(result); // []
Note that this is an unintended side effect of array destructuring with rest elements in JavaScript. The equivalent Haskell version does raise an exception.
mapAdjacent :: (a -> a -> b) -> [a] -> [b]
mapAdjacent f (x:xs) = foldr (\y g x -> f x y : g y) (const []) xs x
main :: IO ()
main = do
print $ mapAdjacent (,) "hello" -- [('h','e'),('e','l'),('l','l'),('l','o')]
print $ mapAdjacent (,) "" -- Exception: Non-exhaustive patterns in function mapAdjacent
However, returning an empty array might be desirable for this function. It's equivalent to adding the mapAdjacent f [] = []
case in Haskell.
回答2:
Not a particularly fast implementation, but destructuring assignment makes it particularly elegant -
const None =
Symbol ()
const mapAdjacent = (f, [ a = None, b = None, ...more ] = []) =>
a === None || b === None
? []
: [ f (a, b), ...mapAdjacent (f, [ b, ...more ]) ]
const pair = (a, b) =>
[ a, b ]
console.log(mapAdjacent(pair, [ 1, 2, 3 ]))
// [ [ 1, 2 ], [ 2, 3 ] ]
console.log(mapAdjacent(pair, "hello"))
// [ [ h, e ], [ e, l ], [ l, l ], [ l, o ] ]
console.log(mapAdjacent(pair, [ 1 ]))
// []
console.log(mapAdjacent(pair, []))
// []
Or write it as a generator -
const mapAdjacent = function* (f, iter = [])
{ while (iter.length > 1)
{ yield f (...iter.slice(0,2))
iter = iter.slice(1)
}
}
const pair = (a, b) =>
[ a, b ]
console.log(Array.from(mapAdjacent(pair, [ 1, 2, 3 ])))
// [ [ 1, 2 ], [ 2, 3 ] ]
console.log(Array.from(mapAdjacent(pair, "hello")))
// [ [ h, e ], [ e, l ], [ l, l ], [ l, o ] ]
console.log(Array.from(mapAdjacent(pair, [ 1 ])))
// []
console.log(Array.from(mapAdjacent(pair, [])))
// []
回答3:
The reduce() function provides a functional what you need:
items.reduce((previousValue, currentValue) => {
if(currentValue.id === previousValue.id) {
// do something
} else {
// do something else
}
});
回答4:
As I mentioned in a comment, I would suggest using reduce
. Here is an example:
const input = [
{id: 1, value: "Apple Turnover"},
{id: 1, value: "Apple Turnover"},
{id: 2, value: "Banana Bread"},
{id: 3, value: "Chocolate"},
{id: 3, value: "Chocolate"},
{id: 3, value: "Chocolate"},
{id: 1, value: "Apple"},
{id: 4, value: "Danish"},
];
// Desired output: Array of strings equal to values in the above array,
// but with a prefix string of "New: " or "Repeated: " depending on whether
// the id is repeated or not
const reducer = (accumulator, currentValue) => {
let previousValue, descriptions, isRepeatedFromPrevious;
if (accumulator) {
previousValue = accumulator.previousValue;
descriptions = accumulator.descriptions;
isRepeatedFromPrevious = previousValue.id === currentValue.id;
} else {
descriptions = [];
isRepeatedFromPrevious = false;
}
if (isRepeatedFromPrevious) {
// The following line is not purely functional and performs a mutation,
// but maybe we do not care because the mutated object did not exist
// before this reducer ran.
descriptions.push("Repeated: " + currentValue.value);
} else {
// Again, this line is mutative
descriptions.push("New: " + currentValue.value);
}
return { previousValue: currentValue, descriptions }
};
const output = input.reduce(reducer, null).descriptions;
document.getElementById('output').innerText = JSON.stringify(output);
<output id=output></output>
来源:https://stackoverflow.com/questions/57767346/functional-way-to-get-previous-element-during-map