Grouping models without dictionary Announcing the arrival of Valued Associate #679: Cesar...
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Grouping models without dictionary
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Haskell Bencoded Dictionary ParserSwift complexities with DictionaryOrdered dictionary in swift 2Grouping an array by postId and userIdTime manipulation for notification remindersNotifying view controller of changes to any of five types of modelsSimple wrapper function grouping and summarising variableGrouping array elements into batches of at most threeExtension for grouping array into two dimensional array based on element ([] -> [[]])Grouping Date Objects by year and month
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
I want to group an array of structs only using arrays into an array of new structs. In this example I want to group the Person structs by Job and put them in new Department structs and put them in an array.
enum Job {
case developer
case programmer
case coder
}
struct Person {
let id: Int
let job: Job
}
struct Department {
let job: Job
var staff: [Person]
mutating func addStaff(_ person: Person) {
staff.append(person)
}
}
let initial = [
Person(id: 0, job: .developer),
Person(id: 1, job: .developer),
Person(id: 2, job: .programmer),
Person(id: 3, job: .programmer),
Person(id: 4, job: .coder)
]
func combinator(accumulator: [Department], current: Person) -> [Department] {
var accumulator = accumulator
if let index = accumulator.index(where: { $0.job == current.job }) {
var department = accumulator[index]
department.staff.append(current)
accumulator[index] = department
} else {
accumulator.append(Department(job: current.job, staff: [current]))
}
return accumulator
}
var departments: [Department] = initial.reduce([], combinator)
I feel like the if statement in the combinator(accumulator:current:) function is a bit clunky, is there a better way to write that part or any other part of this piece of code?
functional-programming swift swift3
asked Jun 22 '17 at 0:51
richyrichy
1134
$endgroup$
bumped to the homepage by Community♦ 1 min ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
$begingroup$
I want to group an array of structs only using arrays into an array of new structs. In this example I want to group the Person structs by Job and put them in new Department structs and put them in an array.
enum Job {
case developer
case programmer
case coder
}
struct Person {
let id: Int
let job: Job
}
struct Department {
let job: Job
var staff: [Person]
mutating func addStaff(_ person: Person) {
staff.append(person)
}
}
let initial = [
Person(id: 0, job: .developer),
Person(id: 1, job: .developer),
Person(id: 2, job: .programmer),
Person(id: 3, job: .programmer),
Person(id: 4, job: .coder)
]
func combinator(accumulator: [Department], current: Person) -> [Department] {
var accumulator = accumulator
if let index = accumulator.index(where: { $0.job == current.job }) {
var department = accumulator[index]
department.staff.append(current)
accumulator[index] = department
} else {
accumulator.append(Department(job: current.job, staff: [current]))
}
return accumulator
}
var departments: [Department] = initial.reduce([], combinator)
I feel like the if statement in the combinator(accumulator:current:) function is a bit clunky, is there a better way to write that part or any other part of this piece of code?
functional-programming swift swift3
asked Jun 22 '17 at 0:51
richyrichy
1134
$endgroup$
bumped to the homepage by Community♦ 1 min ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
$begingroup$
I want to group an array of structs only using arrays into an array of new structs. In this example I want to group the Person structs by Job and put them in new Department structs and put them in an array.
enum Job {
case developer
case programmer
case coder
}
struct Person {
let id: Int
let job: Job
}
struct Department {
let job: Job
var staff: [Person]
mutating func addStaff(_ person: Person) {
staff.append(person)
}
}
let initial = [
Person(id: 0, job: .developer),
Person(id: 1, job: .developer),
Person(id: 2, job: .programmer),
Person(id: 3, job: .programmer),
Person(id: 4, job: .coder)
]
func combinator(accumulator: [Department], current: Person) -> [Department] {
var accumulator = accumulator
if let index = accumulator.index(where: { $0.job == current.job }) {
var department = accumulator[index]
department.staff.append(current)
accumulator[index] = department
} else {
accumulator.append(Department(job: current.job, staff: [current]))
}
return accumulator
}
var departments: [Department] = initial.reduce([], combinator)
I feel like the if statement in the combinator(accumulator:current:) function is a bit clunky, is there a better way to write that part or any other part of this piece of code?
functional-programming swift swift3
asked Jun 22 '17 at 0:51
richyrichy
1134
$endgroup$
I want to group an array of structs only using arrays into an array of new structs. In this example I want to group the Person structs by Job and put them in new Department structs and put them in an array.
enum Job {
case developer
case programmer
case coder
}
struct Person {
let id: Int
let job: Job
}
struct Department {
let job: Job
var staff: [Person]
mutating func addStaff(_ person: Person) {
staff.append(person)
}
}
let initial = [
Person(id: 0, job: .developer),
Person(id: 1, job: .developer),
Person(id: 2, job: .programmer),
Person(id: 3, job: .programmer),
Person(id: 4, job: .coder)
]
func combinator(accumulator: [Department], current: Person) -> [Department] {
var accumulator = accumulator
if let index = accumulator.index(where: { $0.job == current.job }) {
var department = accumulator[index]
department.staff.append(current)
accumulator[index] = department
} else {
accumulator.append(Department(job: current.job, staff: [current]))
}
return accumulator
}
var departments: [Department] = initial.reduce([], combinator)
I feel like the if statement in the combinator(accumulator:current:) function is a bit clunky, is there a better way to write that part or any other part of this piece of code?
functional-programming swift swift3
functional-programming swift swift3
asked Jun 22 '17 at 0:51
richyrichy
1134
asked Jun 22 '17 at 0:51
richyrichy
1134
asked Jun 22 '17 at 0:51
richyrichy
1134
asked Jun 22 '17 at 0:51
richyrichy
1134
asked Jun 22 '17 at 0:51
richyrichy
1134
1134
bumped to the homepage by Community♦ 1 min ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 1 min ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Swift 4
i would use the new swift 4 init(grouping:by:)
since you create always new Departments then the following code is possible.
if you have an existing [Department] then the code not fit your request.
enum Job {
case developer
case programmer
case coder
}
struct Person {
let id: Int
let job: Job
}
struct Department {
let job: Job
var staff: [Person]
mutating func addStaff(_ person: Person) {
staff.append(person)
}
}
let initial = [
Person(id: 0, job: .developer),
Person(id: 1, job: .developer),
Person(id: 2, job: .programmer),
Person(id: 3, job: .programmer),
Person(id: 4, job: .coder)
]
let grouped = Dictionary(grouping: initial, by: { $0.job })
let departments = grouped.map{ key, value in Department( job: key, staff: value )}
( NOTE: i don't tested it in playground - just from head to the answer)
Swift 3
in Swift 3 you can use an extension to the array to resolve this grouping
public extension Sequence {
func group<U: Hashable>(by key: (Iterator.Element) -> U) -> [U:[Iterator.Element]] {
var categories: [U: [Iterator.Element]] = [:]
for element in self {
let key = key(element)
if case nil = categories[key]?.append(element) {
categories[key] = [element]
}
}
return categories
}
}
the extension is from here https://stackoverflow.com/a/31220067/1930509 and when you need a more performat grouping you can find it at the link
answered Jun 22 '17 at 21:49
mueschamuescha
1765
$endgroup$
1
$begingroup$
This is specifically tagged swift3...
$endgroup$
– Mr. Xcoder
Jun 25 '17 at 10:47
$begingroup$
There existing some collection extension for grouping by for swift3 if he like my 2 lines solution.
$endgroup$
– muescha
Jun 25 '17 at 15:48
$begingroup$
Yeah sorry, I was looking for a Swift 3 solution
$endgroup$
– richy
Jun 25 '17 at 19:08
$begingroup$
@richy then you can use this extension for swift3 to have a groupby in Swift 3: stackoverflow.com/a/31220067/1930509
$endgroup$
– muescha
Jun 26 '17 at 15:53
$begingroup$
with this a later upgrade to swift 4 is easy :)
$endgroup$
– muescha
Jun 26 '17 at 15:53
|
show 1 more comment
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Swift 4
i would use the new swift 4 init(grouping:by:)
since you create always new Departments then the following code is possible.
if you have an existing [Department] then the code not fit your request.
enum Job {
case developer
case programmer
case coder
}
struct Person {
let id: Int
let job: Job
}
struct Department {
let job: Job
var staff: [Person]
mutating func addStaff(_ person: Person) {
staff.append(person)
}
}
let initial = [
Person(id: 0, job: .developer),
Person(id: 1, job: .developer),
Person(id: 2, job: .programmer),
Person(id: 3, job: .programmer),
Person(id: 4, job: .coder)
]
let grouped = Dictionary(grouping: initial, by: { $0.job })
let departments = grouped.map{ key, value in Department( job: key, staff: value )}
( NOTE: i don't tested it in playground - just from head to the answer)
Swift 3
in Swift 3 you can use an extension to the array to resolve this grouping
public extension Sequence {
func group<U: Hashable>(by key: (Iterator.Element) -> U) -> [U:[Iterator.Element]] {
var categories: [U: [Iterator.Element]] = [:]
for element in self {
let key = key(element)
if case nil = categories[key]?.append(element) {
categories[key] = [element]
}
}
return categories
}
}
the extension is from here https://stackoverflow.com/a/31220067/1930509 and when you need a more performat grouping you can find it at the link
answered Jun 22 '17 at 21:49
mueschamuescha
1765
$endgroup$
1
$begingroup$
This is specifically tagged swift3...
$endgroup$
– Mr. Xcoder
Jun 25 '17 at 10:47
$begingroup$
There existing some collection extension for grouping by for swift3 if he like my 2 lines solution.
$endgroup$
– muescha
Jun 25 '17 at 15:48
$begingroup$
Yeah sorry, I was looking for a Swift 3 solution
$endgroup$
– richy
Jun 25 '17 at 19:08
$begingroup$
@richy then you can use this extension for swift3 to have a groupby in Swift 3: stackoverflow.com/a/31220067/1930509
$endgroup$
– muescha
Jun 26 '17 at 15:53
$begingroup$
with this a later upgrade to swift 4 is easy :)
$endgroup$
– muescha
Jun 26 '17 at 15:53
|
show 1 more comment
$begingroup$
Swift 4
i would use the new swift 4 init(grouping:by:)
since you create always new Departments then the following code is possible.
if you have an existing [Department] then the code not fit your request.
enum Job {
case developer
case programmer
case coder
}
struct Person {
let id: Int
let job: Job
}
struct Department {
let job: Job
var staff: [Person]
mutating func addStaff(_ person: Person) {
staff.append(person)
}
}
let initial = [
Person(id: 0, job: .developer),
Person(id: 1, job: .developer),
Person(id: 2, job: .programmer),
Person(id: 3, job: .programmer),
Person(id: 4, job: .coder)
]
let grouped = Dictionary(grouping: initial, by: { $0.job })
let departments = grouped.map{ key, value in Department( job: key, staff: value )}
( NOTE: i don't tested it in playground - just from head to the answer)
Swift 3
in Swift 3 you can use an extension to the array to resolve this grouping
public extension Sequence {
func group<U: Hashable>(by key: (Iterator.Element) -> U) -> [U:[Iterator.Element]] {
var categories: [U: [Iterator.Element]] = [:]
for element in self {
let key = key(element)
if case nil = categories[key]?.append(element) {
categories[key] = [element]
}
}
return categories
}
}
the extension is from here https://stackoverflow.com/a/31220067/1930509 and when you need a more performat grouping you can find it at the link
answered Jun 22 '17 at 21:49
mueschamuescha
1765
$endgroup$
1
$begingroup$
This is specifically tagged swift3...
$endgroup$
– Mr. Xcoder
Jun 25 '17 at 10:47
$begingroup$
There existing some collection extension for grouping by for swift3 if he like my 2 lines solution.
$endgroup$
– muescha
Jun 25 '17 at 15:48
$begingroup$
Yeah sorry, I was looking for a Swift 3 solution
$endgroup$
– richy
Jun 25 '17 at 19:08
$begingroup$
@richy then you can use this extension for swift3 to have a groupby in Swift 3: stackoverflow.com/a/31220067/1930509
$endgroup$
– muescha
Jun 26 '17 at 15:53
$begingroup$
with this a later upgrade to swift 4 is easy :)
$endgroup$
– muescha
Jun 26 '17 at 15:53
|
show 1 more comment
$begingroup$
Swift 4
i would use the new swift 4 init(grouping:by:)
since you create always new Departments then the following code is possible.
if you have an existing [Department] then the code not fit your request.
enum Job {
case developer
case programmer
case coder
}
struct Person {
let id: Int
let job: Job
}
struct Department {
let job: Job
var staff: [Person]
mutating func addStaff(_ person: Person) {
staff.append(person)
}
}
let initial = [
Person(id: 0, job: .developer),
Person(id: 1, job: .developer),
Person(id: 2, job: .programmer),
Person(id: 3, job: .programmer),
Person(id: 4, job: .coder)
]
let grouped = Dictionary(grouping: initial, by: { $0.job })
let departments = grouped.map{ key, value in Department( job: key, staff: value )}
( NOTE: i don't tested it in playground - just from head to the answer)
Swift 3
in Swift 3 you can use an extension to the array to resolve this grouping
public extension Sequence {
func group<U: Hashable>(by key: (Iterator.Element) -> U) -> [U:[Iterator.Element]] {
var categories: [U: [Iterator.Element]] = [:]
for element in self {
let key = key(element)
if case nil = categories[key]?.append(element) {
categories[key] = [element]
}
}
return categories
}
}
the extension is from here https://stackoverflow.com/a/31220067/1930509 and when you need a more performat grouping you can find it at the link
answered Jun 22 '17 at 21:49
mueschamuescha
1765
$endgroup$
Swift 4
i would use the new swift 4 init(grouping:by:)
since you create always new Departments then the following code is possible.
if you have an existing [Department] then the code not fit your request.
enum Job {
case developer
case programmer
case coder
}
struct Person {
let id: Int
let job: Job
}
struct Department {
let job: Job
var staff: [Person]
mutating func addStaff(_ person: Person) {
staff.append(person)
}
}
let initial = [
Person(id: 0, job: .developer),
Person(id: 1, job: .developer),
Person(id: 2, job: .programmer),
Person(id: 3, job: .programmer),
Person(id: 4, job: .coder)
]
let grouped = Dictionary(grouping: initial, by: { $0.job })
let departments = grouped.map{ key, value in Department( job: key, staff: value )}
( NOTE: i don't tested it in playground - just from head to the answer)
Swift 3
in Swift 3 you can use an extension to the array to resolve this grouping
public extension Sequence {
func group<U: Hashable>(by key: (Iterator.Element) -> U) -> [U:[Iterator.Element]] {
var categories: [U: [Iterator.Element]] = [:]
for element in self {
let key = key(element)
if case nil = categories[key]?.append(element) {
categories[key] = [element]
}
}
return categories
}
}
the extension is from here https://stackoverflow.com/a/31220067/1930509 and when you need a more performat grouping you can find it at the link
answered Jun 22 '17 at 21:49
mueschamuescha
1765
edited Jun 26 '17 at 16:01
answered Jun 22 '17 at 21:49
mueschamuescha
1765
answered Jun 22 '17 at 21:49
mueschamuescha
1765
answered Jun 22 '17 at 21:49
mueschamuescha
1765
1765
1
$begingroup$
This is specifically tagged swift3...
$endgroup$
– Mr. Xcoder
Jun 25 '17 at 10:47
$begingroup$
There existing some collection extension for grouping by for swift3 if he like my 2 lines solution.
$endgroup$
– muescha
Jun 25 '17 at 15:48
$begingroup$
Yeah sorry, I was looking for a Swift 3 solution
$endgroup$
– richy
Jun 25 '17 at 19:08
$begingroup$
@richy then you can use this extension for swift3 to have a groupby in Swift 3: stackoverflow.com/a/31220067/1930509
$endgroup$
– muescha
Jun 26 '17 at 15:53
$begingroup$
with this a later upgrade to swift 4 is easy :)
$endgroup$
– muescha
Jun 26 '17 at 15:53
|
show 1 more comment
1
$begingroup$
This is specifically tagged swift3...
$endgroup$
– Mr. Xcoder
Jun 25 '17 at 10:47
$begingroup$
There existing some collection extension for grouping by for swift3 if he like my 2 lines solution.
$endgroup$
– muescha
Jun 25 '17 at 15:48
$begingroup$
Yeah sorry, I was looking for a Swift 3 solution
$endgroup$
– richy
Jun 25 '17 at 19:08
$begingroup$
@richy then you can use this extension for swift3 to have a groupby in Swift 3: stackoverflow.com/a/31220067/1930509
$endgroup$
– muescha
Jun 26 '17 at 15:53
$begingroup$
with this a later upgrade to swift 4 is easy :)
$endgroup$
– muescha
Jun 26 '17 at 15:53
1
1
$begingroup$
This is specifically tagged swift3...
$endgroup$
– Mr. Xcoder
Jun 25 '17 at 10:47
$begingroup$
This is specifically tagged swift3...
$endgroup$
– Mr. Xcoder
Jun 25 '17 at 10:47
$begingroup$
There existing some collection extension for grouping by for swift3 if he like my 2 lines solution.
$endgroup$
– muescha
Jun 25 '17 at 15:48
$begingroup$
There existing some collection extension for grouping by for swift3 if he like my 2 lines solution.
$endgroup$
– muescha
Jun 25 '17 at 15:48
$begingroup$
Yeah sorry, I was looking for a Swift 3 solution
$endgroup$
– richy
Jun 25 '17 at 19:08
$begingroup$
Yeah sorry, I was looking for a Swift 3 solution
$endgroup$
– richy
Jun 25 '17 at 19:08
$begingroup$
@richy then you can use this extension for swift3 to have a groupby in Swift 3: stackoverflow.com/a/31220067/1930509
$endgroup$
– muescha
Jun 26 '17 at 15:53
$begingroup$
@richy then you can use this extension for swift3 to have a groupby in Swift 3: stackoverflow.com/a/31220067/1930509
$endgroup$
– muescha
Jun 26 '17 at 15:53
$begingroup$
with this a later upgrade to swift 4 is easy :)
$endgroup$
– muescha
Jun 26 '17 at 15:53
$begingroup$
with this a later upgrade to swift 4 is easy :)
$endgroup$
– muescha
Jun 26 '17 at 15:53
|
show 1 more comment
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