以矢量化方式连接给定开始、停止数字的范围数组 - NumPy [英] Concatenate range arrays given start, stop numbers in a vectorized way - NumPy

问题描述

我有两个感兴趣的矩阵，第一个是词袋"矩阵，有两列:文档 ID 和术语 ID.例如:

I have two matrices of interest, the first is a "bag of words" matrix, with two columns: the document ID and the term ID. For example:

bow[0:10]

Out[1]:
    array([[ 0, 10],
           [ 0, 12],
           [ 0, 19],
           [ 0, 20],
           [ 1,  9],
           [ 1, 24],
           [ 2, 33],
           [ 2, 34],
           [ 2, 35],
           [ 3, 2]])

此外，我有一个索引"矩阵，其中矩阵中的每一行都包含词袋矩阵中给定文档 ID 的第一行和最后一行的索引.例如:第 0 行是 doc id 0 的第一个和最后一个索引.例如:

In addition, I have an "index" matrix, where every row in the matrix contains the index of the first and last row for a given document ID in the bag of words matrix. Ex: row 0 is the first and last index for doc id 0. For example:

index[0:4]

Out[2]:
    array([[ 0,  4],
           [ 4,  6],
           [ 6,  9],
           [ 9, 10]])

我想做的是随机抽取文档 ID 的样本，并获取这些文档 ID 的所有单词行包.词袋矩阵大约有 150M 行(~1.5Gb)，所以使用 numpy.in1d() 太慢了.我们需要快速返回这些以供下游任务使用.

What I'd like to do is take a random sample of document ID's and get all of the bag of word rows for those document ID's. The bag of words matrix is roughly 150M rows (~1.5Gb), so using numpy.in1d() is too slow. We need to return these rapidly for feeding into a downstream task.

我想出的幼稚解决方案如下:

The naive solution I have come up with is as follows:

def get_rows(ids):
    indices = np.concatenate([np.arange(x1, x2) for x1,x2 in index[ids]])
    return bow[indices]

get_rows([4,10,3,5])

通用示例

提出问题的通用示例是这样的 -

A generic sample to put forth the problem would be with something like this -

indices = np.array([[ 4, 7],
                    [10,16],
                    [11,18]]

预期的输出是 -

array([ 4,  5,  6, 10, 11, 12, 13, 14, 15, 11, 12, 13, 14, 15, 16, 17])

推荐答案

我想我终于用 cumsum 用于矢量化解决方案的技巧 -

Think I have cracked it finally with a cumsum trick for a vectorized solution -

def create_ranges(a):
    l = a[:,1] - a[:,0]
    clens = l.cumsum()
    ids = np.ones(clens[-1],dtype=int)
    ids[0] = a[0,0]
    ids[clens[:-1]] = a[1:,0] - a[:-1,1]+1
    out = ids.cumsum()
    return out

样品运行 -

In [416]: a = np.array([[4,7],[10,16],[11,18]])

In [417]: create_ranges(a)
Out[417]: array([ 4,  5,  6, 10, 11, 12, 13, 14, 15, 11, 12, 13, 14, 15, 16, 17])

In [425]: a = np.array([[-2,4],[-5,2],[11,12]])

In [426]: create_ranges(a)
Out[426]: array([-2, -1,  0,  1,  2,  3, -5, -4, -3, -2, -1,  0,  1, 11])

如果给定的开始和停止作为两个 1D 数组，我们只需要使用它们代替第一列和第二列.为了完整起见，这里是完整的代码 -

If we are given starts and stops as two 1D arrays, we just need to use those in place of the first and second columns. For completeness, here's the complete code -

def create_ranges(starts, ends):
    l = ends - starts
    clens = l.cumsum()
    ids = np.ones(clens[-1],dtype=int)
    ids[0] = starts[0]
    ids[clens[:-1]] = starts[1:] - ends[:-1]+1
    out = ids.cumsum()
    return out

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