python - 如何使用生成器中的值填充 2D Python numpy 数组？

Question

基于答案here似乎没有一种简单的方法可以用生成器中的数据填充 2D numpy 数组。

但是，如果有人能想出一种方法来矢量化或以其他方式加速以下功能，我将不胜感激。

这里的区别是我想批量处理生成器中的值，而不是在内存中创建整个数组。我能想到的唯一方法是使用 for 循环。

import numpy as np
from itertools import permutations

permutations_of_values = permutations(range(1,20), 7)

def array_from_generator(generator, arr):
    """Fills the numpy array provided with values from
    the generator provided. Number of columns in arr
    must match the number of values yielded by the 
    generator."""
    count = 0
    for row in arr:
        try:
            item = next(generator)
        except StopIteration:
            break
        row[:] = item
        count += 1
    return arr[:count,:]

batch_size = 100000

empty_array = np.empty((batch_size, 7), dtype=int)
batch_of_values = array_from_generator(permutations_of_values, empty_array)

print(batch_of_values[0:5])

输出:

[[ 1  2  3  4  5  6  7]
 [ 1  2  3  4  5  6  8]
 [ 1  2  3  4  5  6  9]
 [ 1  2  3  4  5  6 10]
 [ 1  2  3  4  5  6 11]]

速度测试:

%timeit array_from_generator(permutations_of_values, empty_array)
10 loops, best of 3: 137 ms per loop

添加:

正如@COLDSPEED(谢谢)所建议的，这里是一个使用列表从生成器收集数据的版本。它的速度大约是上面代码的两倍。谁能对此进行改进:

permutations_of_values = permutations(range(1,20), 7)

def array_from_generator2(generator, rows=batch_size):
    """Creates a numpy array from a specified number 
    of values from the generator provided."""
    data = []
    for row in range(rows):
        try:
            data.append(next(generator))
        except StopIteration:
            break
    return np.array(data)

batch_size = 100000

batch_of_values = array_from_generator2(permutations_of_values, rows=100000)

print(batch_of_values[0:5])

输出:

[[ 1  2  3  4  5  6  7]
 [ 1  2  3  4  5  6  8]
 [ 1  2  3  4  5  6  9]
 [ 1  2  3  4  5  6 10]
 [ 1  2  3  4  5  6 11]]

速度测试:

%timeit array_from_generator2(permutations_of_values, rows=100000)
10 loops, best of 3: 85.6 ms per loop

Answer 1

您可以在基本恒定的时间内计算出 future 的大小。就这样做，并使用 numpy.fromiter:

In [1]: import math, from itertools import permutations, chain

In [2]: def n_chose_k(n, k, fac=math.factorial):
    ...:     return fac(n)/fac(n-k)
    ...:

In [3]: def permutations_to_array(r, k):
    ...:     n = len(r)
    ...:     size = int(n_chose_k(n, k))
    ...:     it = permutations(r, k)
    ...:     arr = np.fromiter(chain.from_iterable(it),
    ...:                       count=size,  dtype=int)
    ...:     arr.size = size//k, k
    ...:     return arr
    ...:

In [4]: arr = permutations_to_array(range(1,20), 7)

In [5]: arr.shape
Out[5]: (36279360, 7)

In [6]: arr[0:5]
Out[6]:
array([[ 1,  2,  3,  4,  5,  6,  7],
       [ 1,  2,  3,  4,  5,  6,  8],
       [ 1,  2,  3,  4,  5,  6,  9],
       [ 1,  2,  3,  4,  5,  6, 10],
       [ 1,  2,  3,  4,  5,  6, 11]])

只要 r 仅限于具有 len 的序列，此方法就有效。

编辑以添加我为 batchsize*k block 生成器编写的实现，并带有修剪选项!

import math
from itertools import repeat, chain

import numpy as np

def n_chose_k(n, k, fac=math.factorial):
    return fac(n)/fac(n-k)

def permutations_in_batches(r, k, batchsize=None, fill=0, dtype=int, trim=False):
    n = len(r)
    size = int(n_chose_k(n, k))
    if batchsize is None or batchsize > size:
        batchsize = size
    perms = chain.from_iterable(permutations(r, k))
    count = batchsize*k
    remaining = size - count
    while remaining > 0:
        current = np.fromiter(perms, count=count, dtype=dtype)
        current.shape = batchsize, k
        yield current
        remaining -= count
    if remaining: # remaining is negative
        remaining = -remaining
        if not trim:
            padding = repeat(fill, remaining)
            finalcount = count
            finalshape = batchsize, k
        else:
            q = remaining//k # always divisible q%k==0
            finalcount = q*k
            padding = repeat(fill, remaining)
            finalshape = q, k
        current =  np.fromiter(chain(perms, padding), count=finalcount, dtype=dtype)
        current.shape = finalshape
    else: # remaining is 0
        current = np.fromiter(perms, count=batchsize, dtype=dtype)
        current.shape = batchsize, k
    yield current