Code

versioninfo()

Julia Version 1.10.0
Commit 3120989f39b (2023-12-25 18:01 UTC)
Build Info:
  Official https://julialang.org/ release
Platform Info:
  OS: macOS (arm64-apple-darwin22.4.0)
  CPU: 8 × Apple M1
  WORD_SIZE: 64
  LIBM: libopenlibm
  LLVM: libLLVM-15.0.7 (ORCJIT, apple-m1)
  Threads: 1 on 4 virtual cores
Environment:
  JULIA_NUM_THREADS =

よく間違えるjulia構文

配列のfor文

多次元配列と配列の配列を混同してfor文を間違えることがある. 適当な多次元配列\(x\)があるとする.

Code

x = rand(1:5, 10, 3)

10×3 Matrix{Int64}:
 4  2  4
 3  3  3
 1  5  1
 1  3  5
 2  4  3
 2  4  3
 4  3  4
 5  1  3
 2  4  1
 2  3  4

for文で多次元配列を回そうとすると全ての要素を縦になめる:

Code

for e in x
    println(e)
end

行ごとに回したければeachrowを使う:

Code

for row in eachrow(x)
    println(row)
end

[4, 2, 4]
[3, 3, 3]
[1, 5, 1]
[1, 3, 5]
[2, 4, 3]
[2, 4, 3]
[4, 3, 4]
[5, 1, 3]
[2, 4, 1]
[2, 3, 4]

Code

for col in eachcol(x)
    println(col)
end

[4, 3, 1, 1, 2, 2, 4, 5, 2, 2]
[2, 3, 5, 3, 4, 4, 3, 1, 4, 3]
[4, 3, 1, 5, 3, 3, 4, 3, 1, 4]

配列の配列の場合

配列の配列\(y\)があるとする

Code

y=[zeros(Int,3) for i=1:10]
for i=1:10
    for j=1:3
        y[i][j]=rand(1:5)
    end
end
y

10-element Vector{Vector{Int64}}:
 [3, 4, 4]
 [1, 4, 4]
 [3, 3, 2]
 [5, 5, 2]
 [5, 2, 2]
 [2, 1, 5]
 [3, 2, 1]
 [5, 5, 3]
 [5, 2, 2]
 [3, 2, 4]

この場合は次のように回せる:

Code

for row in y
    println(row)
end

[3, 4, 4]
[1, 4, 4]
[3, 3, 2]
[5, 5, 2]
[5, 2, 2]
[2, 1, 5]
[3, 2, 1]
[5, 5, 3]
[5, 2, 2]
[3, 2, 4]

部分配列

ある配列から特定の条件を満たす部分配列を求める場合filter関数を使いたくなる:

Code

?filter

Base.Meta.ParseError: ParseError:
# Error @ /Users/masaya/projects/notebooks2/posts/2021-07-12-julia_memo.ipynb:1:1
?filter
╙ ── not a unary operator
ParseError:

# Error @ /Users/masaya/projects/notebooks2/posts/2021-07-12-julia_memo.ipynb:1:1

?filter

╙ ── not a unary operator



Stacktrace:

 [1] top-level scope

   @ ~/projects/notebooks2/posts/2021-07-12-julia_memo.ipynb:1

多次元配列に対しfilter関数を使うと１次元配列が返る:

Code

 x =  reshape([rand(Int) for i=1:10*3], (:, 3))

10×3 Matrix{Int64}:
  1434155849139153490   8792201589749886045   -349363450898840745
  5845810908379303736  -8265185926902104011  -5065380767449374288
 -3447320615839289584   2389268345481233176  -6485329445295354633
 -5365742947389604229  -2326303111811756091   8698018437924141657
  3595047806792093185   1172421900134288996  -4750125639239034923
  1363143167553767130   7000736503973028484   5847702750677365622
 -9135160136868605419  -2968857547344022737  -3598011947351858499
 -7733185694899413747  -9217370134120495903  -7561702952325334262
 -7716510106645479528   5657573041425477116  -1159079173192823842
 -3856910791373303962   2502860500295826394   7127390806145534861

Code

filter(isodd, skipmissing(x))

15-element Vector{Int64}:
 -5365742947389604229
  3595047806792093185
 -9135160136868605419
 -7733185694899413747
  8792201589749886045
 -8265185926902104011
 -2326303111811756091
 -2968857547344022737
 -9217370134120495903
  -349363450898840745
 -6485329445295354633
  8698018437924141657
 -4750125639239034923
 -3598011947351858499
  7127390806145534861

従って2列目が偶数である部分配列を取り出そうとして次のコードを実行するとエラーになる:

Code

filter(x->iseven(x[2]), x)

BoundsError: BoundsError: attempt to access Int64 at index [2]
BoundsError: attempt to access Int64 at index [2]



Stacktrace:

 [1] getindex

   @ ./number.jl:98 [inlined]

 [2] #15

   @ ~/projects/notebooks2/posts/2021-07-12-julia_memo.ipynb:1 [inlined]

 [3] filter(f::var"#15#16", a::Matrix{Int64})

   @ Base ./array.jl:2675

 [4] top-level scope

   @ ~/projects/notebooks2/posts/2021-07-12-julia_memo.ipynb:1

filterを使えないので例えば次のようにする:

Code

x[x[:,2] .%2 .==0,:]

5×3 Matrix{Int64}:
 -3447320615839289584  2389268345481233176  -6485329445295354633
  3595047806792093185  1172421900134288996  -4750125639239034923
  1363143167553767130  7000736503973028484   5847702750677365622
 -7716510106645479528  5657573041425477116  -1159079173192823842
 -3856910791373303962  2502860500295826394   7127390806145534861

Code

x[iseven.(x[:,2]),:]

5×3 Matrix{Int64}:
 -3447320615839289584  2389268345481233176  -6485329445295354633
  3595047806792093185  1172421900134288996  -4750125639239034923
  1363143167553767130  7000736503973028484   5847702750677365622
 -7716510106645479528  5657573041425477116  -1159079173192823842
 -3856910791373303962  2502860500295826394   7127390806145534861

Code

x[findall(a -> iseven(x[a,2]), 1:size(x)[1]),:]

5×3 Matrix{Int64}:
 -3447320615839289584  2389268345481233176  -6485329445295354633
  3595047806792093185  1172421900134288996  -4750125639239034923
  1363143167553767130  7000736503973028484   5847702750677365622
 -7716510106645479528  5657573041425477116  -1159079173192823842
 -3856910791373303962  2502860500295826394   7127390806145534861

Code

x[findall(iseven,x[:,2]),:]

5×3 Matrix{Int64}:
 -3447320615839289584  2389268345481233176  -6485329445295354633
  3595047806792093185  1172421900134288996  -4750125639239034923
  1363143167553767130  7000736503973028484   5847702750677365622
 -7716510106645479528  5657573041425477116  -1159079173192823842
 -3856910791373303962  2502860500295826394   7127390806145534861

競プロ関連

listをつなげて文字列にして出力する場合はjoinの方が早い

と以前atcoderでハマったのでメモしようと思ったが試してみるとなぜかfor文ベタ書きが一番早い.

Code

l=rand(0:9,100);

Code

@time println(join(l))

0825670272040984510758647377189790987793993950097603858275133276948923626202573746492476422906277389
  0.021755 seconds (8.97 k allocations: 606.688 KiB, 97.98% compilation time: 13% of which was recompilation)

Code

s=""
for n in l
    s*=string(n)
end
@time println(s)

0825670272040984510758647377189790987793993950097603858275133276948923626202573746492476422906277389
  0.000111 seconds (18 allocations: 536 bytes)

Code

function j(l)
 println(join(l))
end
@time j(l)

0825670272040984510758647377189790987793993950097603858275133276948923626202573746492476422906277389
  0.083234 seconds (742 allocations: 46.375 KiB, 95.49% gc time, 4.17% compilation time)

Code

function jj(l)
s=""
for n in l
    s*=string(n)
end
println(s)
end
@time jj(l)

0825670272040984510758647377189790987793993950097603858275133276948923626202573746492476422906277389
  0.016438 seconds (2.65 k allocations: 170.320 KiB, 97.48% compilation time)

Code

@time begin
s=""
for n in l
    s*=string(n)
end
 println(s)
end

0825670272040984510758647377189790987793993950097603858275133276948923626202573746492476422906277389
  0.000128 seconds (418 allocations: 19.180 KiB)

n重ループの綺麗な回し方

複数の変数をforひとつで書ける

Code

for i=1:3, j=1:5
    println(i," ",j)
end

可変なn重ループはCartesianIndicesで上手にできる

Code


n=3
for c in CartesianIndices(ntuple(d->0:2, n))
    # vectorにする
    x=collect(c.I)
    println(x)
end

[0, 0, 0]
[1, 0, 0]
[2, 0, 0]
[0, 1, 0]
[1, 1, 0]
[2, 1, 0]
[0, 2, 0]
[1, 2, 0]
[2, 2, 0]
[0, 0, 1]
[1, 0, 1]
[2, 0, 1]
[0, 1, 1]
[1, 1, 1]
[2, 1, 1]
[0, 2, 1]
[1, 2, 1]
[2, 2, 1]
[0, 0, 2]
[1, 0, 2]
[2, 0, 2]
[0, 1, 2]
[1, 1, 2]
[2, 1, 2]
[0, 2, 2]
[1, 2, 2]
[2, 2, 2]

julia memo

Other Formats

よく間違えるjulia構文

配列のfor文

配列の配列の場合

部分配列

競プロ関連

listをつなげて文字列にして出力する場合はjoinの方が早い

n重ループの綺麗な回し方

複数の変数をforひとつで書ける

可変なn重ループはCartesianIndicesで上手にできる