我有一个整数值的数组。
a = [2,1,4,0,2]
我希望能够对a中的每个值应用一个arange函数,以便使其变成如下形式:
b = [0,1,0,0,1,2,3,1,2]
b "=" [arange(2),arange(1),arange(4),arange(0),arange(2)]
实际上,我使用np.repeat函数根据数组a重复数组行,并希望有一个标记i将每个重复值链接到原始值,并具有识别号以区分它们。
我尝试使用np.vectorize但没有成功。
有更多“numpythonic”的做法。一种可能的方法是这样的:
import numpy as np
from numpy.lib.stride_tricks import as_strided
def concatenated_ranges(ranges_list) :
ranges_list = np.array(ranges_list, copy=False)
base_range = np.arange(ranges_list.max())
base_range = as_strided(base_range,
shape=ranges_list.shape + base_range.shape,
strides=(0,) + base_range.strides)
return base_range[base_range < ranges_list[:, None]]
def junuxx(a) :
b = np.array([], dtype=np.uint8)
for x in a:
b = np.append(b, np.arange(x))
return b
def mr_e(a) :
return reduce(lambda x, y: x + range(y), a, [])
以下是一些时间统计:
In [2]: a = [2, 1, 4, 0 ,2] # the OP's original example
In [3]: concatenated_ranges(a) # show it works!
Out[3]: array([0, 1, 0, 0, 1, 2, 3, 0, 1])
In [4]: %timeit concatenated_ranges(a)
10000 loops, best of 3: 31.6 us per loop
In [5]: %timeit junuxx(a)
10000 loops, best of 3: 34 us per loop
In [6]: %timeit mr_e(a)
100000 loops, best of 3: 2.58 us per loop
In [7]: a = np.random.randint(1, 10, size=(10,))
In [8]: %timeit concatenated_ranges(a)
10000 loops, best of 3: 27.1 us per loop
In [9]: %timeit junuxx(a)
10000 loops, best of 3: 79.8 us per loop
In [10]: %timeit mr_e(a)
100000 loops, best of 3: 7.82 us per loop
In [11]: a = np.random.randint(1, 10, size=(100,))
In [12]: %timeit concatenated_ranges(a)
10000 loops, best of 3: 57.4 us per loop
In [13]: %timeit junuxx(a)
1000 loops, best of 3: 756 us per loop
In [14]: %timeit mr_e(a)
10000 loops, best of 3: 149 us per loop
In [15]: a = np.random.randint(1, 10, size=(1000,))
In [16]: %timeit concatenated_ranges(a)
1000 loops, best of 3: 358 us per loop
In [17]: %timeit junuxx(a)
100 loops, best of 3: 9.38 ms per loop
In [18]: %timeit mr_e(a)
100 loops, best of 3: 8.93 ms per loop
as_strided
函数,它让我感到困惑。 - Junuxxa = [2, 1, 4, 0 ,2]
reduce(lambda x, y: x+range(y), a, [])
[0, 1, 0, 0, 1, 2, 3, 0, 1]
np.repeat
,因为它与期望的行为相近,而不是特别针对numpy的问题。 - YXD这段代码实现了你所描述的功能,即将a
中的所有值连接起来形成一个aranges
数组。尽管这意味着b
中可能存在一些错误:
>>> a = [2, 1, 4, 0, 2]
>>> b = np.array([], dtype=np.uint8)
>>>for x in a:
>>> b = np.append(b, np.arange(x))
>>> print b
array([0,1,0,0,1,2,3,0,1,])
>>> a = [2, 1, 4, 0, 2]
>>> b = [np.arange(x) for x in a]
>>> print b
[array([0, 1]), array([0]), array([0, 1, 2, 3]), array([], dtype=int32),
array([0, 1])]
另一种更加内存高效的方法,通常稍微快一些:
import numpy as np
def concatenated_ranges2(ranges_list):
cumsum = np.append(0, np.cumsum(ranges_list[:-1]))
cumsum = np.repeat(cumsum, ranges_list)
return np.arange(cumsum.shape[0]) - cumsum
测试这个函数和之前的一个:
>>> a = np.random.randint(1, 10, size=(1000,))
>>> %timeit concatenated_ranges(a)
10000 loops, best of 3: 142 us per loop
>>> %timeit concatenated_ranges2(a)
10000 loops, best of 3: 72.6 us per loop
a
到b
的呢?最重要的是,你尝试过什么? - Inbar Rose