如何在numpy中有效地连接多个arange调用?
发布时间:2020-12-16 22:58:22 所属栏目:Python 来源:网络整理
导读:我想在cnt值的向量上向numpy.arange(0,cnt_i)这样的调用进行向量化,并像这个片段一样连接结果: import numpycnts = [1,2,3]numpy.concatenate([numpy.arange(cnt) for cnt in cnts])array([0,1,2]) 不幸的是,由于临时数组和列表推导循环,上面的代码非常低效
我想在cnt值的向量上向numpy.arange(0,cnt_i)这样的调用进行向量化,并像这个片段一样连接结果:
import numpy cnts = [1,2,3] numpy.concatenate([numpy.arange(cnt) for cnt in cnts]) array([0,1,2]) 不幸的是,由于临时数组和列表推导循环,上面的代码非常低效. 有没有办法在numpy中更有效地做到这一点? 解决方法
这是一个完全矢量化的函数:
def multirange(counts): counts = np.asarray(counts) # Remove the following line if counts is always strictly positive. counts = counts[counts != 0] counts1 = counts[:-1] reset_index = np.cumsum(counts1) incr = np.ones(counts.sum(),dtype=int) incr[0] = 0 incr[reset_index] = 1 - counts1 # Reuse the incr array for the final result. incr.cumsum(out=incr) return incr 这是@ Developer的答案的变体,它只调用一次范围: def multirange_loop(counts): counts = np.asarray(counts) ranges = np.empty(counts.sum(),dtype=int) seq = np.arange(counts.max()) starts = np.zeros(len(counts),dtype=int) starts[1:] = np.cumsum(counts[:-1]) for start,count in zip(starts,counts): ranges[start:start + count] = seq[:count] return ranges 这是原始版本,作为函数编写: def multirange_original(counts): ranges = np.concatenate([np.arange(count) for count in counts]) return ranges 演示: In [296]: multirange_original([1,3]) Out[296]: array([0,2]) In [297]: multirange_loop([1,3]) Out[297]: array([0,2]) In [298]: multirange([1,3]) Out[298]: array([0,2]) 使用更多的计数比较时间: In [299]: counts = np.random.randint(1,50,size=50) In [300]: %timeit multirange_original(counts) 10000 loops,best of 3: 114 μs per loop In [301]: %timeit multirange_loop(counts) 10000 loops,best of 3: 76.2 μs per loop In [302]: %timeit multirange(counts) 10000 loops,best of 3: 26.4 μs per loop (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |