Reduce a sequence of items provided by a generator in JavaScript
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3
Say I have a sequence of items and I want to perform a reduce operation via myReducer function (whatever it is). If my items are in an array (say myArray), it's easy:
myArray.reduce(myReducer);
What if, however, my sequence is quite large and I don't want to allocate an array of all of it, only to immediately reduce it item after item? I can create a generator function for my sequence, that part is clear. Is there a straightforward way of how to then perform the reduction? I mean apart from writing the reduce functionality for a generator myself.
javascript generator reduce
asked Nov 23 '18 at 16:19
OndraOndra
556
add a comment |
3
Say I have a sequence of items and I want to perform a reduce operation via myReducer function (whatever it is). If my items are in an array (say myArray), it's easy:
myArray.reduce(myReducer);
What if, however, my sequence is quite large and I don't want to allocate an array of all of it, only to immediately reduce it item after item? I can create a generator function for my sequence, that part is clear. Is there a straightforward way of how to then perform the reduction? I mean apart from writing the reduce functionality for a generator myself.
javascript generator reduce
asked Nov 23 '18 at 16:19
OndraOndra
556
You need to create a lazy evaluation array object/class with a lazy evaluation reduce.
– Dominique Fortin
Nov 23 '18 at 16:26
add a comment |
3
3
3
1
Say I have a sequence of items and I want to perform a reduce operation via myReducer function (whatever it is). If my items are in an array (say myArray), it's easy:
myArray.reduce(myReducer);
What if, however, my sequence is quite large and I don't want to allocate an array of all of it, only to immediately reduce it item after item? I can create a generator function for my sequence, that part is clear. Is there a straightforward way of how to then perform the reduction? I mean apart from writing the reduce functionality for a generator myself.
javascript generator reduce
asked Nov 23 '18 at 16:19
OndraOndra
556
Say I have a sequence of items and I want to perform a reduce operation via myReducer function (whatever it is). If my items are in an array (say myArray), it's easy:
myArray.reduce(myReducer);
What if, however, my sequence is quite large and I don't want to allocate an array of all of it, only to immediately reduce it item after item? I can create a generator function for my sequence, that part is clear. Is there a straightforward way of how to then perform the reduction? I mean apart from writing the reduce functionality for a generator myself.
javascript generator reduce
javascript generator reduce
asked Nov 23 '18 at 16:19
OndraOndra
556
asked Nov 23 '18 at 16:19
OndraOndra
556
asked Nov 23 '18 at 16:19
OndraOndra
556
asked Nov 23 '18 at 16:19
OndraOndra
556
asked Nov 23 '18 at 16:19
OndraOndra
556
556
You need to create a lazy evaluation array object/class with a lazy evaluation reduce.
– Dominique Fortin
Nov 23 '18 at 16:26
add a comment |
You need to create a lazy evaluation array object/class with a lazy evaluation reduce.
– Dominique Fortin
Nov 23 '18 at 16:26
You need to create a lazy evaluation array object/class with a lazy evaluation reduce.
– Dominique Fortin
Nov 23 '18 at 16:26
You need to create a lazy evaluation array object/class with a lazy evaluation reduce.
– Dominique Fortin
Nov 23 '18 at 16:26
add a comment |
1 Answer
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1
For now, ECMA-Script standard provides functions like reduce for arrays, so you're out of luck: you need to implement your own reduce for iterables:
const reduce = (f, i, it) => {
let o = i
for (let x of it)
o = f (o, x)
return o
}
const xs = [1, 2, 3]
const xs_ = {
[Symbol.iterator]: function* () {
yield 1
yield 2
yield 3
}
}
const output1 = reduce ((o, x) => o + x, 10, xs)
const output2 = reduce ((o, x) => o + x, 10, xs_)
console.log ('output1:', output1)
console.log ('output2:', output2)answered Nov 23 '18 at 16:29
Matías FidemraizerMatías Fidemraizer
49.8k1285149
What's the problem with my answer? The answer is there's no current native way of reducing, mapping, filtering, ... iterables.
– Matías Fidemraizer
Nov 23 '18 at 16:50
it´s a nice solution! +1
– Luke
Nov 23 '18 at 16:56
@MatíasFidemraizer: Thanks, that's what I thought. I'm just not as up-to-date with the current state of EcmaScript, so I asked to be sure that I'm not reinventing the wheel.
– Ondra
Nov 23 '18 at 16:58
I tried myself the Symbol.iterator way, that is giving you what you wanted, unfortunatelly I was not able to transform it so it would work with the native reduce.
– Luke
Nov 23 '18 at 17:30
@MatíasFidemraizer I don't see any difference withxs.reduce((o, x) => (o + x), 10)or withArray.prototype.call(xs, (o, x) => (o + x), 10)if it is array like. That doesn't seem to answer the question.
– Dominique Fortin
Nov 23 '18 at 17:35
|
show 5 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
For now, ECMA-Script standard provides functions like reduce for arrays, so you're out of luck: you need to implement your own reduce for iterables:
const reduce = (f, i, it) => {
let o = i
for (let x of it)
o = f (o, x)
return o
}
const xs = [1, 2, 3]
const xs_ = {
[Symbol.iterator]: function* () {
yield 1
yield 2
yield 3
}
}
const output1 = reduce ((o, x) => o + x, 10, xs)
const output2 = reduce ((o, x) => o + x, 10, xs_)
console.log ('output1:', output1)
console.log ('output2:', output2)answered Nov 23 '18 at 16:29
Matías FidemraizerMatías Fidemraizer
49.8k1285149
What's the problem with my answer? The answer is there's no current native way of reducing, mapping, filtering, ... iterables.
– Matías Fidemraizer
Nov 23 '18 at 16:50
it´s a nice solution! +1
– Luke
Nov 23 '18 at 16:56
@MatíasFidemraizer: Thanks, that's what I thought. I'm just not as up-to-date with the current state of EcmaScript, so I asked to be sure that I'm not reinventing the wheel.
– Ondra
Nov 23 '18 at 16:58
I tried myself the Symbol.iterator way, that is giving you what you wanted, unfortunatelly I was not able to transform it so it would work with the native reduce.
– Luke
Nov 23 '18 at 17:30
@MatíasFidemraizer I don't see any difference withxs.reduce((o, x) => (o + x), 10)or withArray.prototype.call(xs, (o, x) => (o + x), 10)if it is array like. That doesn't seem to answer the question.
– Dominique Fortin
Nov 23 '18 at 17:35
|
show 5 more comments
1
For now, ECMA-Script standard provides functions like reduce for arrays, so you're out of luck: you need to implement your own reduce for iterables:
const reduce = (f, i, it) => {
let o = i
for (let x of it)
o = f (o, x)
return o
}
const xs = [1, 2, 3]
const xs_ = {
[Symbol.iterator]: function* () {
yield 1
yield 2
yield 3
}
}
const output1 = reduce ((o, x) => o + x, 10, xs)
const output2 = reduce ((o, x) => o + x, 10, xs_)
console.log ('output1:', output1)
console.log ('output2:', output2)answered Nov 23 '18 at 16:29
Matías FidemraizerMatías Fidemraizer
49.8k1285149
What's the problem with my answer? The answer is there's no current native way of reducing, mapping, filtering, ... iterables.
– Matías Fidemraizer
Nov 23 '18 at 16:50
it´s a nice solution! +1
– Luke
Nov 23 '18 at 16:56
@MatíasFidemraizer: Thanks, that's what I thought. I'm just not as up-to-date with the current state of EcmaScript, so I asked to be sure that I'm not reinventing the wheel.
– Ondra
Nov 23 '18 at 16:58
I tried myself the Symbol.iterator way, that is giving you what you wanted, unfortunatelly I was not able to transform it so it would work with the native reduce.
– Luke
Nov 23 '18 at 17:30
@MatíasFidemraizer I don't see any difference withxs.reduce((o, x) => (o + x), 10)or withArray.prototype.call(xs, (o, x) => (o + x), 10)if it is array like. That doesn't seem to answer the question.
– Dominique Fortin
Nov 23 '18 at 17:35
|
show 5 more comments
1
1
1
For now, ECMA-Script standard provides functions like reduce for arrays, so you're out of luck: you need to implement your own reduce for iterables:
const reduce = (f, i, it) => {
let o = i
for (let x of it)
o = f (o, x)
return o
}
const xs = [1, 2, 3]
const xs_ = {
[Symbol.iterator]: function* () {
yield 1
yield 2
yield 3
}
}
const output1 = reduce ((o, x) => o + x, 10, xs)
const output2 = reduce ((o, x) => o + x, 10, xs_)
console.log ('output1:', output1)
console.log ('output2:', output2)answered Nov 23 '18 at 16:29
Matías FidemraizerMatías Fidemraizer
49.8k1285149
For now, ECMA-Script standard provides functions like reduce for arrays, so you're out of luck: you need to implement your own reduce for iterables:
const reduce = (f, i, it) => {
let o = i
for (let x of it)
o = f (o, x)
return o
}
const xs = [1, 2, 3]
const xs_ = {
[Symbol.iterator]: function* () {
yield 1
yield 2
yield 3
}
}
const output1 = reduce ((o, x) => o + x, 10, xs)
const output2 = reduce ((o, x) => o + x, 10, xs_)
console.log ('output1:', output1)
console.log ('output2:', output2)const reduce = (f, i, it) => {
let o = i
for (let x of it)
o = f (o, x)
return o
}
const xs = [1, 2, 3]
const xs_ = {
[Symbol.iterator]: function* () {
yield 1
yield 2
yield 3
}
}
const output1 = reduce ((o, x) => o + x, 10, xs)
const output2 = reduce ((o, x) => o + x, 10, xs_)
console.log ('output1:', output1)
console.log ('output2:', output2)const reduce = (f, i, it) => {
let o = i
for (let x of it)
o = f (o, x)
return o
}
const xs = [1, 2, 3]
const xs_ = {
[Symbol.iterator]: function* () {
yield 1
yield 2
yield 3
}
}
const output1 = reduce ((o, x) => o + x, 10, xs)
const output2 = reduce ((o, x) => o + x, 10, xs_)
console.log ('output1:', output1)
console.log ('output2:', output2)answered Nov 23 '18 at 16:29
Matías FidemraizerMatías Fidemraizer
49.8k1285149
edited Nov 23 '18 at 17:47
answered Nov 23 '18 at 16:29
Matías FidemraizerMatías Fidemraizer
49.8k1285149
answered Nov 23 '18 at 16:29
Matías FidemraizerMatías Fidemraizer
49.8k1285149
answered Nov 23 '18 at 16:29
Matías FidemraizerMatías Fidemraizer
49.8k1285149
49.8k1285149
What's the problem with my answer? The answer is there's no current native way of reducing, mapping, filtering, ... iterables.
– Matías Fidemraizer
Nov 23 '18 at 16:50
it´s a nice solution! +1
– Luke
Nov 23 '18 at 16:56
@MatíasFidemraizer: Thanks, that's what I thought. I'm just not as up-to-date with the current state of EcmaScript, so I asked to be sure that I'm not reinventing the wheel.
– Ondra
Nov 23 '18 at 16:58
I tried myself the Symbol.iterator way, that is giving you what you wanted, unfortunatelly I was not able to transform it so it would work with the native reduce.
– Luke
Nov 23 '18 at 17:30
@MatíasFidemraizer I don't see any difference withxs.reduce((o, x) => (o + x), 10)or withArray.prototype.call(xs, (o, x) => (o + x), 10)if it is array like. That doesn't seem to answer the question.
– Dominique Fortin
Nov 23 '18 at 17:35
|
show 5 more comments
What's the problem with my answer? The answer is there's no current native way of reducing, mapping, filtering, ... iterables.
– Matías Fidemraizer
Nov 23 '18 at 16:50
it´s a nice solution! +1
– Luke
Nov 23 '18 at 16:56
@MatíasFidemraizer: Thanks, that's what I thought. I'm just not as up-to-date with the current state of EcmaScript, so I asked to be sure that I'm not reinventing the wheel.
– Ondra
Nov 23 '18 at 16:58
I tried myself the Symbol.iterator way, that is giving you what you wanted, unfortunatelly I was not able to transform it so it would work with the native reduce.
– Luke
Nov 23 '18 at 17:30
@MatíasFidemraizer I don't see any difference withxs.reduce((o, x) => (o + x), 10)or withArray.prototype.call(xs, (o, x) => (o + x), 10)if it is array like. That doesn't seem to answer the question.
– Dominique Fortin
Nov 23 '18 at 17:35
What's the problem with my answer? The answer is there's no current native way of reducing, mapping, filtering, ... iterables.
– Matías Fidemraizer
Nov 23 '18 at 16:50
What's the problem with my answer? The answer is there's no current native way of reducing, mapping, filtering, ... iterables.
– Matías Fidemraizer
Nov 23 '18 at 16:50
it´s a nice solution! +1
– Luke
Nov 23 '18 at 16:56
it´s a nice solution! +1
– Luke
Nov 23 '18 at 16:56
@MatíasFidemraizer: Thanks, that's what I thought. I'm just not as up-to-date with the current state of EcmaScript, so I asked to be sure that I'm not reinventing the wheel.
– Ondra
Nov 23 '18 at 16:58
@MatíasFidemraizer: Thanks, that's what I thought. I'm just not as up-to-date with the current state of EcmaScript, so I asked to be sure that I'm not reinventing the wheel.
– Ondra
Nov 23 '18 at 16:58
I tried myself the Symbol.iterator way, that is giving you what you wanted, unfortunatelly I was not able to transform it so it would work with the native reduce.
– Luke
Nov 23 '18 at 17:30
I tried myself the Symbol.iterator way, that is giving you what you wanted, unfortunatelly I was not able to transform it so it would work with the native reduce.
– Luke
Nov 23 '18 at 17:30
@MatíasFidemraizer I don't see any difference with
xs.reduce((o, x) => (o + x), 10) or with Array.prototype.call(xs, (o, x) => (o + x), 10) if it is array like. That doesn't seem to answer the question.– Dominique Fortin
Nov 23 '18 at 17:35
@MatíasFidemraizer I don't see any difference with
xs.reduce((o, x) => (o + x), 10) or with Array.prototype.call(xs, (o, x) => (o + x), 10) if it is array like. That doesn't seem to answer the question.– Dominique Fortin
Nov 23 '18 at 17:35
|
show 5 more comments
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You need to create a lazy evaluation array object/class with a lazy evaluation reduce.
– Dominique Fortin
Nov 23 '18 at 16:26