Julia: Array元素过滤、元素替代、元素删除等酸爽操作
一、过滤：一个约束条件（略复杂的情况）
找出data中，满足第三列的元素在vet中的元素。
julia>data =[1 2 3;
4 5 6 ;
7 8 9 ;
10 11 12];
julia>vet =[6 ,9];
julia>condition =map(x->in(x,vet),data[:,3]);
julia>data[condition,:]
2×3 Array{Int64,2}:
4 5 6
7 8 9还有其它方法么？
二、过滤：两个约束条件的情况
第一列中值为4，第二列中值>=5的
julia>data[(data[:,2].>=5 & data[:,1].==4),:]
1×3 Array{Int64,2}:
4 5 6
```
几种error的情况：
```
julia> data[(data[:,2].>=5 & data[:,1]==4),:] # error! =>.
julia> data[data[:,2].>=5 & data[:,1].==4,:] # error!=>(),用括号理好&的优先次序
julia> data[(data[:,2].>=5 && data[:,1]==4),:] # error! =>&&
三、元素替代
1、把data中的元素为2的，全部改为0
julia> data =[1,2,3,4,5,2,3,1]
8-element Array{Int64,1}:
1
2
3
4
5
2
3
1
julia> data[data.==2] .=0
2-element view(::Array{Int64,1}, [2, 6]) with eltype Int64:
0
0
julia> data
8-element Array{Int64,1}:
1
0
3
4
5
0
3
12、如果要把以下data中“”，用0代替，有什么操作方式？
julia> data
9×12 Array{Any,2}:
8 "" "" "" "" "" "" "" "" "" "" ""
"" "" 3 6 "" "" "" "" "" "" "" ""
"" 7 "" "" 9 "" 2 "" "" "" "" ""
"" 5 "" "" "" 7 "" "" "" "" "" ""
"" "" "" "" 4 5 7 "" "" "" "" ""
"" "" "" 1 "" "" "" 3 "" "" "" ""
"" "" 1 "" "" "" "" 6 8 "" "" ""
"" "" 8 5 "" "" "" 1 "" "" "" ""
"" 9 "" "" "" "" 4 "" "" "" "" ""只需要：
julia> data[data .== ""] .= 0 # 其中两个点，均不能少！julia> data
9×12 Array{Any,2}:
8 0 0 0 0 0 0 0 0 0 0 0
0 0 3 6 0 0 0 0 0 0 0 0
0 7 0 0 9 0 2 0 0 0 0 0
0 5 0 0 0 7 0 0 0 0 0 0
0 0 0 0 4 5 7 0 0 0 0 0
0 0 0 1 0 0 0 3 0 0 0 0
0 0 1 0 0 0 0 6 8 0 0 0
0 0 8 5 0 0 0 1 0 0 0 0
0 9 0 0 0 0 4 0 0 0 0 0猜一猜如果：
julia> data[data .= ""] .= 0 # bug四、把Array 拆分成几个小Array
方法一： map
julia> data
9×9 Array{Any,2}:
8 0 0 0 0 0 0 0 0
0 0 3 6 0 0 0 0 0
0 7 0 0 9 0 2 0 0
0 5 0 0 0 7 0 0 0
0 0 0 0 4 5 7 0 0
0 0 0 1 0 0 0 3 0
0 0 1 0 0 0 0 6 8
0 0 8 5 0 0 0 1 0
0 9 0 0 0 0 4 0 0
尝试一：失败
A =map((p,q)->data[1:3,p:q], 1:3:7, 3:3:9) #
3-element Array{Array{Any,2},1}:
[8 0 0; 0 0 3; 0 7 0]
[0 0 0; 6 0 0; 0 9 0]
[0 0 0; 0 0 0; 2 0 0]
julia> A[1]
3×3 Array{Any,2}:
8 0 0
0 0 3
0 7 0
julia> A[2]
3×3 Array{Any,2}:
0 0 0
6 0 0
0 9 0
julia> A[3]
3×3 Array{Any,2}:
0 0 0
0 0 0
2 0 0有没有办法，一口气拆成N个 3x3的矩阵？
为什么才3个元素？ 因为是一一对应的关系，[1,4,7]<=>[1,4,7],不是循还。
尝试2：失败
julia> A =map((p,q)->data[p:p+2,q:q+2], [1,4,7],[1,4,7])
3-element Array{Array{Any,2},1}:
[8 0 0; 0 0 3; 0 7 0]
[0 0 7; 0 4 5; 1 0 0]
[0 6 8; 0 1 0; 4 0 0]
julia> A[1]
3×3 Array{Any,2}:
8 0 0
0 0 3
0 7 0
尝试3： 成功！双重map + flatten
julia> t =map(p->map(q->data[p:p+2,q:q+2],[1,4,7]),[1,4,7])
3-element Array{Array{Array{Any,2},1},1}:
[[8 0 0; 0 0 3; 0 7 0], [0 0 0; 6 0 0; 0 9 0], [0 0 0; 0 0 0; 2 0 0]]
[[0 5 0; 0 0 0; 0 0 0], [0 0 7; 0 4 5; 1 0 0], [0 0 0; 7 0 0; 0 3 0]]
[[0 0 1; 0 0 8; 0 9 0], [0 0 0; 5 0 0; 0 0 0], [0 6 8; 0 1 0; 4 0 0]]
julia> collect(Iterators.flatten(t))
9-element Array{Array{Any,2},1}:
[8 0 0; 0 0 3; 0 7 0]
[0 0 0; 6 0 0; 0 9 0]
[0 0 0; 0 0 0; 2 0 0]
[0 5 0; 0 0 0; 0 0 0]
[0 0 7; 0 4 5; 1 0 0]
[0 0 0; 7 0 0; 0 3 0]
[0 0 1; 0 0 8; 0 9 0]
[0 0 0; 5 0 0; 0 0 0]
[0 6 8; 0 1 0; 4 0 0]
尝试4：成功！大杀器，把repeat也搬出来
julia>using DelimitedFiles;
julia> map((i,j)->data[i:i+2,j:j+2],repeat([1,4,7],inner=3),repeat([1,4,7],outer=3))
9-element Array{Array{Any,2},1}:
[8 0 0; 0 0 3; 0 7 0]
[0 0 0; 6 0 0; 0 9 0]
[0 0 0; 0 0 0; 2 0 0]
[0 5 0; 0 0 0; 0 0 0]
[0 0 7; 0 4 5; 1 0 0]
[0 0 0; 7 0 0; 0 3 0]
[0 0 1; 0 0 8; 0 9 0]
[0 0 0; 5 0 0; 0 0 0]
[0 6 8; 0 1 0; 4 0 0]
2、列表推导式
简单，快捷。
尝试5： 成功！
julia> B=[data[p:p+2,q:q+2] for p in [1,4,7] for q in [1,4,7]]
9-element Array{Array{Any,2},1}:
[8 0 0; 0 0 3; 0 7 0]
[0 0 0; 6 0 0; 0 9 0]
[0 0 0; 0 0 0; 2 0 0]
[0 5 0; 0 0 0; 0 0 0]
[0 0 7; 0 4 5; 1 0 0]
[0 0 0; 7 0 0; 0 3 0]
[0 0 1; 0 0 8; 0 9 0]
[0 0 0; 5 0 0; 0 0 0]
[0 6 8; 0 1 0; 4 0 0]五、Array中元素的删除
1、filter!
把b中的为3的元素删除
julia> b =[1,2,3,4,5,6]
6-element Array{Int64,1}:
1
2
3
4
5
6
julia> filter!(x->x!=3,b)
5-element Array{Int64,1}:
1
2
4
5
62、deleteat!
也可以，大家可以试一下。删除特定位置的元素
julia> deleteat!([6, 5, 4, 3, 2, 1], 1:2:5)
3-element Array{Int64,1}:
5
3
1