Today we learn List, an immutable, ordered collection of elements of the same type
The term "ordered" here means that the list maintains the sequence in which elements were added, didn't mean List is sorted
Create List
- Explicitly specifying elements
let numbers = [1; 2; 3; 4; 5]
let names = ["Alice"; "Bob"; "Charlie"]
the ; could be omit between elements if you write them in seperated lines
let list123 = [ 1 // first comment
2 // second comment
3 // third comment
]
all elements should be the same type
let list123 = ["1"; 2; 3]
the first elements is string, the rest is int, will get compile error:
All elements of a list must be implicitly convertible to the type of the first element, which here is 'string'. This element has type 'int'.(1,21)
- Using a range
let numbers = [1..5] // Equivalent to [1; 2; 3; 4; 5]
let evenNumbers = [2..2..10] // Equivalent to [2; 4; 6; 8; 10]
- Using list comprehensions
let doubles = [for x in 1..5 -> x * 2] // Equivalent to [2; 4; 6; 8; 10]
let doubles = [for x in 2..2..10 -> x * 2] // Equivalent to [4; 8; 12; 16; 20]
Get element of List
- Using index
let numbers = [1; 2; 3]
let firstNumber = numbers.[0] // first number 1
let thirdNumber = numbers.[2] // third number 3
printfn "firstNumber %i thirdNumber %i" firstNumber thirdNumber
if you use index not exist in List, will got index out of range exception
let numbers = [1; 2; 3]
let forthNumber = numbers.[4]
printfn "forthNumber %i" forthNumber
compile error:
Unhandled exception. System.ArgumentException: The index was outside the range of elements in the list
the browser environment have bug about this exception
- Using build-in method List.item
let numbers = [1; 2; 3]
let firstNumber = numbers |> List.item 0
printfn "firstNumber %i" firstNumber
- head and tail
List.head, get first element
let numbers = [1; 2; 3]
let head = numbers |> List.head
printfn "head %i" head
List.tail, return list after removing the first element, not the last element
let numbers = [1; 2; 3]
let tail = numbers |> List.tail
printfn "tail %A" tail // tail [2; 3]
printfn "numbers %A" numbers // numbers [1; 2; 3]
tail is a new List not including the first element, it won't change the value of original List numbers
Loop a List
- for...in
let numbers = [1; 2; 3]
for number in numbers do
printfn "Number: %d" number
- List.iter
let numbers = [1; 2; 3]
List.iter (fun number -> printfn "Number: %d" number) numbers
fun define a lambda function here, it's equivalent to
let numbers = [1; 2; 3]
let printElement number =
printfn "Number: %d" number
List.iter printElement numbers
Pattern Matching with List
let numbers0 = []
let numbers1 = [1]
let numbers3 = [1; 2; 3]
let describeList lst =
match lst with
| [] -> "The list is empty"
| [x] -> sprintf "The list has one element: %d" x
| [x; y] -> sprintf "The list has two elements: %d and %d" x y
| x :: xs -> sprintf "The list starts with %d and has more elements" x
printfn "%s" (describeList numbers0) // The list is empty
printfn "%s" (describeList numbers1) // The list has one element: 1
printfn "%s" (describeList numbers3) // The list starts with %d and has more elements
Operate List
lists are indeed immutable, which means that you cannot modify a list after it has been created. Any operation that appears to modify a list (such as appending or prepending an element) actually creates a new list
- Prepending an element
let originalList = [2; 3; 4]
let newList = 1 :: originalList
printfn "Original List: %A" originalList // Original List: [2; 3; 4]
printfn "New List: %A" newList // New List: [1; 2; 3; 4]
- Appending an element
let originalList = [1; 2; 3]
let newList = originalList @ [4]
printfn "Original List: %A" originalList // Original List: [1; 2; 3]
printfn "New List: %A" newList // New List: [1; 2; 3; 4]
- Concatenate Lists
let originalList = [1; 2; 3]
let newList = List.append originalList [4]
printfn "Original List: %A" originalList
printfn "New List: %A" newList
- Filter
you can not remove an element in List directly as List is immutable, but you can filter element based on some condition, put them into a new List
let numbers = [1; 2; 3; 4; 5]
let evenNumbers = List.filter (fun x -> x % 2 = 0) numbers
printfn "evenNumbers: %A" evenNumbers // evenNumbers: [2; 4]
- Map
apply an operation to every elements in List, you can use loop/List comprehensions to achieve that, or use Map
let numbers = [1; 2; 3; 4; 5]
let doubleNumbers = List.map (fun x -> 2 * x) numbers
printfn "doubleNumbers: %A" doubleNumbers // doubleNumbers: [2; 4; 6; 8; 10]
- Fold
"fold" or "compress" elements of List of into a single value by applying a function repeatedly
let numbers = [1; 2; 3; 4; 5]
let sum = List.fold (fun acc x -> acc + x) 0 numbers // 15
printfn "sum: %i" sum // sum: 15
step 1: acc = 0 and x = 1 New acc = 0 + 1 = 1
step 2: acc = 1 and x = 2 New acc = 1 + 2 = 3
step 3: acc = 3 and x = 3 New acc = 3 + 3 = 6
step 4: acc = 6 and x = 4 New acc = 6 + 4 = 10
step 5: acc = 10 and x = 5 New acc = 10 + 5 = 15
