有效地为lil_matrix分配一行 [英] Efficiently assign a row to a lil_matrix

问题描述

如何有效地将行分配给lil_matrix?我当前正在使用:

How can I efficiently assign a row to a lil_matrix? I'm currently using:

Q[mid, :] = new_Q

其中new_Q是lil_matrix.getrow(x)

我使用Q.getrow(i)和Q[i, :]进行了测试，发现前者的速度快20倍.

I ran a test on using Q.getrow(i) vs. Q[i, :], and found the former to be 20x faster.

这是lil_matrix文档.

推荐答案

这些时间测试是在小型lil上进行的(密集，但我认为这并不重要)，表明x[i,:]不是问题设置.是的，由于某种原因，它用于获取行时速度很慢.

These time tests on small lil (dense, but I don't think that matters), suggest that x[i,:] is not a problem setting. Yes, for some reason, it is slow when used to fetch a row.

In [108]: x=sparse.lil_matrix(np.arange(120).reshape(24,5))

In [109]: timeit x[10,:]=10
1000 loops, best of 3: 235 us per loop

In [110]: timeit y=x.getrowview(10);y[:]=10
1000 loops, best of 3: 291 us per loop

In [111]: timeit x[10,:]
1000 loops, best of 3: 658 us per loop

In [112]: timeit x.getrowview(10)
10000 loops, best of 3: 51.4 us per loop

getrowview的来源很有启发性，显示了如何处理此矩阵的基础数据结构.

The source for getrowview is instructive, showing how the underlying data structures for this matrix are handled.

def getrowview(self, i):
    """Returns a view of the 'i'th row (without copying).
    """
    new = lil_matrix((1, self.shape[1]), dtype=self.dtype)
    new.rows[0] = self.rows[i]
    new.data[0] = self.data[i]
    return new

我认为x[10,:]使用x.__getitem__或x.__setitem__.这两个功能都比此getrowview更复杂.我猜x.__getitem__是很慢的，因为它也在列上建立索引(请参见x._get1). x[10,:]花费的时间与x[10,0:5]一样.

I think x[10,:] uses x.__getitem__ or x.__setitem__. Both of those functions are more complicated than this getrowview. I'm guessing x.__getitem__ is slow because it is also indexing on columns (see. x._get1). x[10,:] takes just as much time as x[10,0:5].

表明如果只需要设置一行并直接访问rows和data，可能会完成的事情:

Indicative of what might be accomplished if you only need to set one row, and access rows and data directly:

In [133]: timeit x.rows[10]=range(5);x.data[10]=[10]*5
1000000 loops, best of 3: 1.36 us per loop

这与一般情况相去甚远，但是可以让您了解在特殊情况下可以做什么.

This is far from general, but it gives an idea of what you can do in special cases.

更多时间:

In [156]: timeit x[10,:]=x.getrow(12)[:,:]
1000 loops, best of 3: 940 us per loop

In [157]: timeit x[10,:]=x.getrow(12)
1000 loops, best of 3: 259 us per loop

额外的[:,:]很慢. getrow已经返回了副本，因此可能不需要.

That extra [:,:] is slow. getrow already returns a copy, so it probably is not needed.

In [160]: timeit b=x.getrowview(10);b=x.getrow(12)
10000 loops, best of 3: 104 us per loop

In [169]: timeit x.rows[10],x.data[10] = x.rows[12][:],x.data[12][:]
1000000 loops, best of 3: 1.25 us per loop

rows和data的直接修改需要仔细测试.

Direct modification of rows and data needs to be tested carefully.

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