Juliaを使ってみる

00firstjulia.rst

Juliaをはじめてみる

計算をしてみる

1 + 2

3

log(10)

2.302585092994046

sqrt(36)

6.0

変数を使ってみる

a = 12

12

b = a/2

6.0

組み込みの変数

pi

π = 3.1415926535897...

e

e = 2.7182818284590...

変数の型

typeof(1)

Int64

typeof(1.0)

Float64

WORD_SIZE

64

typeof(0x01)

Uint8

typeof(0xffff)

Uint16

虚数

1 + 2im

1 + 2im

(1 + 2im)*(2 - 3im)

8 + 1im

文字列

typeof('x')

Char

typeof("aaa")

ASCIIString (constructor with 2 methods)

"ABCD"[2]

'B'

文字列の中に変数が埋め込める

"ABC$(a)D"

"ABC12D"

"TEST$(1+2)"

"TEST3"

正規表現

ismatch(r"\d{4}-\d\d-\d\d", "2014-08-31")

true

match(r"(\d{4})-(\d\d)-(\d\d)", "2014-08-31")

RegexMatch("2014-08-31", 1="2014", 2="08", 3="31")

関数

function f(x, y)
    x + y
end

f(1, 3)

4

g(x, y) = x + y
g(1,2)

3

+(1,2,3)

6

h(x ,y) = x+y, x*y
h(2,3)

(5,6)

無名関数

(x -> x^2 + 2x - 1)(2)

7

関数その他

function pp(x, y=10)
    return x * y
end

pp(2)

20

キーワード付き引数

function test1(x, y; a="test2", b="test2")
    print("$x, $y, $a, $b\n")
end
test1(1,2)
test1(3,2, b="kk")

1, 2, test2, test2
3, 2, test2, kk

Block syntax

map(x->begin
    if x > 10
        x^3
    else
        x^2
        end
    end
, [2,10,20])

3-element Array{Int64,1}:
    4
  100
 8000

map([2,10,20]) do x
    if (x > 10)
        x ^ 3
    else
        x^2
    end
end

3-element Array{Int64,1}:
    4
  100
 8000

01controlflow.rst

制御構文

http://docs.julialang.org/en/release-0.3/manual/control-flow/ から

複合表現

z = begin
    x = 1
    y = 2
    x + y
end

3

z = (x = 1; y = 2; x+y)

3

条件分岐

function test(x)
    if x < 10
        println("x is less than 10")
    elseif x > 10
        println("x is greater than 10")
    else
        println("x is 10")
    end
end

test(10)
test(11)
test(9)

x is 10
x is greater than 10
x is less than 10

is_greater_than_10(x) = x > 10 ? true : false
println(is_greater_than_10(10))
println(is_greater_than_10(11))

false
true

t(x) = (print("$x "); true)
f(x) = (print("$x "); false)

println(t(1) && t(2))
println(t(1) || t(2))
println(f(1) && t(2))
println(f(1) || t(2))

1 2 true
1 true
1 false
1 2 true

ループ

i = 1
while i < 5
    println(i)
    i += 1
end

1
2
3
4

for j = 1:5
    println(j)
end

1
2
3
4
5

j

j not defined
while loading In[18], in expression starting on line 1

i = 1
while true
    println(i)
    if i >= 5
        break
    end
    i += 1
end

1
2
3
4
5

エラー構文

lookup_or_0(x, i) = begin
    try
        x[i]
    catch e
        if isa(e, BoundsError)
            warn("Out of index")
            0
        end
    end
end

println(lookup_or_0([1,2,3], 5))
println(lookup_or_0([1,2,3], 1))
println(lookup_or_0([1,2,3], 3))

WARNING: Out of index

0
1
3

タスク

function test2()
    produce("start")
    for i = 1:3
        produce(i)
    end
    produce("stop")
end

p = Task(test2)
for i = 1:5
    println(consume(p))
end

start
1
2
3
stop