Question

我想检索沿着N维数组A的最后一个维度的所有元素。也就是说，如果idx是一个（N-1）维元组，我想要A[idx...,:]。我已经弄明白了如何使用CartesianRange，它的工作方式如下所示

A = rand(2,3,4)
for idx in CartesianRange(size(A)[1:end-1])
    i = zeros(Int, length(idx))
    [i[bdx] = idx[bdx] for bdx in 1:length(idx)]
    @show(A[i...,:])
end

但是，必须有一种更简单的方法来创建上面显示的索引。 Splatting idx不起作用 - 我做错了什么？

Answer 1

您可以直接使用从CartesianRange生成的CartesianIndex进行索引！

function getNumber(array) {
    var counter = 0,
        landline = '',
        second = '';

    array.some(function (o) {
        if (o.type === 'landline') {
            landline = o.number;
            return true;
        }
        counter++;
        if (counter === 2) {
            second = o.number;
        }
    });
    return landline || second;
}

//should return the third number for this array
var phoneObject1 = [{ type: 'mobile',  number: '0123456789' }, { type: 'mobile', number:'78945610' }, { type: 'landline', number:'15975345' }];

// for this one it should return the first 
var phoneObject2 = [{ type: 'landline', number: '0123456789' }, { type: 'mobile', number:'78945610' }, { type:'landline', number:'15975345' }];

// for this one it should return the second number
var phoneObject3 = [{ type: 'mobile', number: '0123456789' }, { type: 'mobile', number:'78945610' }];

//and for this last one it should return a blank value
var phoneObject4 = [{ type: 'mobile', number: '0123456789' }];

console.log([phoneObject1, phoneObject2, phoneObject3, phoneObject4].map(getNumber));

我在此处提出的另一个建议是使用未导出的julia> for idx in CartesianRange(size(A)[1:end-1]) @show(A[idx,:]) end A[idx,:] = [0.0334735,0.216738,0.941401,0.973918] A[idx,:] = [0.842384,0.236736,0.103348,0.729471] A[idx,:] = [0.056548,0.283617,0.504253,0.718918] A[idx,:] = [0.551649,0.55043,0.126092,0.259216] A[idx,:] = [0.65623,0.738998,0.781989,0.160111] A[idx,:] = [0.177955,0.971617,0.942002,0.210386]函数从Base.front中提取前导维度，而不是将其编入索引。在这样的type-stable way中处理元组可能有点棘手，但是一旦掌握了它，它们就会非常快。

值得注意的是，Julia的数组是列专业的，因此像这样访问尾随维是going to be much slower而不是抓取列。

在julia中使用CartesianRange迭代多维数组

1 个答案: