Question

我有两个数组。一个是n p，另一个是d p r。我希望d的输出为n r B，我可以在构建下面的张量import numpy X = numpy.array([[1,2,3],[3,4,5],[5,6,7],[7,8,9]]) # n x p betas = numpy.array([[[1,2],[1,2],[1,2]], [[5,6],[5,6],[5,6]]]) # d x p x r print X.shape print betas.shape B = numpy.zeros((betas.shape[0],X.shape[0],betas.shape[2])) print B.shape for i in range(B.shape[0]): B[i,:,:] = numpy.dot(X, betas[i]) print "B",B C = numpy.tensordot(X, betas, axes=([1],[0])) print C.shape时轻松实现。但是，我想在没有这个循环的情况下这样做。

我尝试过各种方式让B与reshape匹配，但到目前为止，我一直没有成功。有没有一种方法不涉及对import {Subject} from 'rxjs/Subject'; import {Observable} from "rxjs/Observable"; import {VenueAdminInceptionModel} from '../../models/venueadmininceptionmodel/venueadmin.inception.model'; export class VenueAdminInceptionService{ // pk and url service private pkurlsend = new Subject<VenueAdminInceptionModel>(); sendurlpk(payload: VenueAdminInceptionModel){ this.pkurlsend.next(payload); } receiveurlpk(): Observable<VenueAdminInceptionModel>{ return this.pkurlsend.asObservable(); } }的调用？

Answer 1

由于dot规则是A的最后一个，而第二个到最后一个B＆＃39;，你可以X.dot(betas)获得一个（n，d，r）数组（这是共享p维度的总和。然后你需要一个转置来得到（d，n，r）

In [200]: X.dot(betas).transpose(1,0,2)
Out[200]: 
array([[[  6,  12],
        [ 12,  24],
        [ 18,  36],
        [ 24,  48]],

       [[ 30,  36],
        [ 60,  72],
        [ 90, 108],
        [120, 144]]])

我们也可以直接从维度规范中编写einsum版本：

np.einsum('np,dpr->dnr', X,betas)

matmul也是如此（最后两个轴上的dot，而d出现的时候是X@betas。

var sheet = SpreadsheetApp.getActive().getSheets()[0];

  var range = sheet.getRange(1, 1, sheet.getLastRow(), 1);
  var values = range.getValues();

  var sortBy = {  
    "a": 2,
    "b": 0,
    "c": 4,
    "d": 1,
    "e": 3 
  };

 values.sort(function(a, b){   
   return sortBy[[a][0]] - sortBy[[b][0]];
  });


 range.setValues(values);

如果任一参数是N-D，则N> 2，它被视为一堆   驻留在最后两个索引中的矩阵并相应地进行广播。

Answer 2

我们可以使用np.tensordot，然后需要置换轴 -

B = np.tensordot(betas, X, axes=(1,1)).swapaxes(1,2)
# Or np.tensordot(X, betas, axes=(1,1)).swapaxes(0,1)

Related post to understand tensordot

Answer 3

以下是使用 numpy.dot() 的另一种方法，按照您的要求返回视图，最重要的是比{{快4倍以上3}} 方法，特别是对于小型数组。但是，对于相当大的数组，tensordot比普通np.tensordot快得多。见下面的时间安排。

In [108]: X.shape
Out[108]: (4, 3)

In [109]: betas.shape
Out[109]: (2, 3, 2)

# use `np.dot` and roll the second axis to first position
In [110]: dot_prod = np.rollaxis(np.dot(X, betas), 1)

In [111]: dot_prod.shape
Out[111]: (2, 4, 2)

# @Divakar's approach
In [113]: B = np.tensordot(betas, X, axes=(1,1)).swapaxes(1,2)

# sanity check :)
In [115]: np.all(np.equal(dot_prod, B))
Out[115]: True

现在，两种方法的表现都是：

对于小尺寸阵列np.dot()比np.dot()

# @Divakar's approach
In [117]: %timeit B = np.tensordot(betas, X, axes=(1,1)).swapaxes(1,2)
10.6 µs ± 2.1 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

# @hpaulj's approach
In [151]: %timeit esum_dot = np.einsum('np, dpr -> dnr', X, betas)
4.16 µs ± 235 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

# proposed approach: more than 4x faster!!
In [118]: %timeit dot_prod = np.rollaxis(np.dot(X, betas), 1)
2.47 µs ± 11.7 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

对于相当大的数组，np.tensordot()比np.tensordot()

In [129]: X = np.random.randint(1, 10, (600, 500))
In [130]: betas = np.random.randint(1, 7, (300, 500, 300))

In [131]: %timeit B = np.tensordot(betas, X, axes=(1,1)).swapaxes(1,2)
18.2 s ± 2.41 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [132]: %timeit dot_prod = np.rollaxis(np.dot(X, betas), 1)
52.8 s ± 14.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

如何在numpy中沿特定尺寸拍摄点积？

3 个答案: