给定一个零的数组，称为：
In [38]: a
Out[38]: array([ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.])
我想用某些值填充某些指数。我有一个开始和结束索引列表，其中包含应在这些位置填充的相关值。它存储在列表中：fill_oneDim_array
[[1, 3, 500], [5, 7, 1000], [9, 15, 200]]
例如：[1,3,5]，填充数组a; a [1：3] = 500.重复[5,7,100]为[5：7] = 1000.
这是否有矢量化解决方案？我想尽可能避免循环。
到目前为止我的研究： - 据我所知，似乎没有一个明显的解决方案。
答案 0 :(得分：0)
这是一个灵感来自this post -
def fillval(a, fill):
info = np.asarray(fill)
start, stop, val = info.T
id_arr = np.zeros(len(a), dtype=int)
id_arr[start] = 1
id_arr[stop] = -1
a[id_arr.cumsum().astype(bool)] = np.repeat(val, stop - start)
return a
示例运行 -
In [676]: a = np.zeros(20, dtype=int)
...: fill = [[1, 3, 500], [5, 7, 1000], [9, 15, 200]]
In [677]: fillval(a, fill)
Out[677]:
array([ 0, 500, 500, 0, 0, 1000, 1000, 0, 0, 200, 200,
200, 200, 200, 200, 0, 0, 0, 0, 0])
修改/优化版
这可以进一步修改/优化，以最小的内存占用量完成输入的所有内容，如此 -
def fillval(a, fill):
fill = np.asarray(fill)
start, stop, val = fill[:,0], fill[:,1], fill[:,2]
a[start] = val
a[stop] = -val
return a.cumsum()
示例运行 -
In [830]: a = np.zeros(20, dtype=int)
...: fill = [[1, 3, 500], [5, 7, 1000], [9, 15, 200]]
In [831]: fillval(a, fill)
Out[831]:
array([ 0, 500, 500, 0, 0, 1000, 1000, 0, 0, 200, 200,
200, 200, 200, 200, 0, 0, 0, 0, 0])
其他方法 -
# Loopy one
def loopy(a, fill):
for start,stop,val in fill:
a[start:stop] = val
return a
# @Paul Panzer's soln
def multifill(target, spec):
spec = np.asarray(spec)
inds = np.zeros((2*len(spec) + 2,), dtype=int)
inds[-1] = len(target)
inds[1:-1] = spec[:, :2].astype(int).ravel()
lens = np.diff(inds)
mask = np.repeat((np.arange(len(lens), dtype=np.uint8)&1).view(bool), lens)
target[mask] = np.repeat(spec[:, 2], lens[1::2])
return target
计时 -
案例＃1：紧密间隔的短组
In [912]: # Setup inputs with group lengths at maximum extent of 10
...: L = 10000 # decides number of groups
...: np.random.seed(0)
...: s0 = np.random.randint(0,9,(L)) + 20*np.arange(L)
...: s1 = s0 + np.random.randint(2,10,(len(s0)))
...: fill = np.c_[s0,s1, np.random.randint(0,9,(len(s0)))].tolist()
...: len_a = fill[-1][1]+1
...: a0 = np.zeros(len_a, dtype=int)
...: a1 = a0.copy()
...: a2 = a0.copy()
In [913]: %timeit loopy(a0, fill)
...: %timeit multifill(a1, fill)
...: %timeit fillval(a2, fill)
100 loops, best of 3: 4.26 ms per loop
100 loops, best of 3: 4.49 ms per loop
100 loops, best of 3: 3.34 ms per loop
In [914]: # Setup inputs with group lengths at maximum extent of 10
...: L = 100000 # decides number of groups
In [915]: %timeit loopy(a0, fill)
...: %timeit multifill(a1, fill)
...: %timeit fillval(a2, fill)
10 loops, best of 3: 43.2 ms per loop
10 loops, best of 3: 49.4 ms per loop
10 loops, best of 3: 38.2 ms per loop
案例＃2：广泛分布的长组
In [916]: # Setup inputs with group lengths at maximum extent of 10
...: L = 10000 # decides number of groups
...: np.random.seed(0)
...: s0 = np.random.randint(0,9,(L)) + 100*np.arange(L)
...: s1 = s0 + np.random.randint(10,50,(len(s0)))
...: fill = np.c_[s0,s1, np.random.randint(0,9,(len(s0)))].tolist()
...: len_a = fill[-1][1]+1
...: a0 = np.zeros(len_a, dtype=int)
...: a1 = a0.copy()
...: a2 = a0.copy()
In [917]: %timeit loopy(a0, fill)
...: %timeit multifill(a1, fill)
...: %timeit fillval(a2, fill)
100 loops, best of 3: 4.51 ms per loop
100 loops, best of 3: 9.18 ms per loop
100 loops, best of 3: 5.16 ms per loop
In [921]: # Setup inputs with group lengths at maximum extent of 10
...: L = 100000 # decides number of groups
In [922]: %timeit loopy(a0, fill)
...: %timeit multifill(a1, fill)
...: %timeit fillval(a2, fill)
10 loops, best of 3: 44.9 ms per loop
10 loops, best of 3: 89 ms per loop
10 loops, best of 3: 58.3 ms per loop
因此，选择最快的一个取决于用例，特别是典型的组长度及其在输入数组中的扩展。
答案 1 :(得分：0)
您可以使用np.repeat构建蒙版并填充值：
import numpy as np
def multifill(target, spec):
inds = np.zeros((2*len(spec) + 2,), dtype=int)
inds[-1] = len(target)
inds[1:-1] = spec[:, :2].astype(int).ravel()
lens = np.diff(inds)
mask = np.repeat((np.arange(len(lens), dtype=np.uint8)&1).view(bool), lens)
target[mask] = np.repeat(spec[:, 2], lens[1::2])
target = np.zeros((16,))
spec = np.array([[1, 3, 500], [5, 7, 1000], [9, 15, 200]])
multifill(target, spec)
print(target)
# [ 0. 500. 500. 0. 0. 1000. 1000. 0. 0. 200.
# 200. 200. 200. 200. 200. 0.]
基准。 Divakar2速度最快，但它要求模板全为零。 PP和Divakar1更灵活。 更新：这些都是通过简单的循环来消除的，＆＃34;感谢＆＃34; @hpaulj。
# hpaulj 0.00256890 ms
# pp 0.01587310 ms
# D1 0.01193481 ms
# D2 0.00533720 ms
# n=100000
# hpaulj 0.03514440 ms
# pp 0.57968440 ms
# D1 0.87605349 ms
# D2 0.34365610 ms
# n=1000000
# hpaulj 0.50301510 ms
# pp 6.91325230 ms
# D1 8.96669030 ms
# D2 3.97435970 ms
代码：
import numpy as np
import types
from timeit import timeit
def f_hpaulj(target, spec):
for s, e, v in spec:
target[int(s):int(e)] = v
def f_pp(target, spec):
inds = np.zeros((2*len(spec) + 2,), dtype=int)
inds[-1] = len(target)
inds[1:-1:2] = spec[:, 0].astype(int)
inds[2:-1:2] = spec[:, 1].astype(int)
lens = np.diff(inds)
mask = np.repeat((np.arange(len(lens), dtype=np.uint8)&1).view(bool), lens)
target[mask] = np.repeat(spec[:, 2], lens[1::2])
def f_D1(a, info):
start, stop, val = info[:,0].astype(int), info[:,1].astype(int), info[:,2]
id_arr = np.zeros(len(a), dtype=int)
id_arr[start] = 1
id_arr[stop] = -1
a[id_arr.cumsum().astype(bool)] = np.repeat(val, stop - start)
def f_D2(a, info):
start, stop, val = info[:,0].astype(int), info[:,1].astype(int), info[:,2]
id_arr = np.zeros(len(a), dtype=val.dtype)
id_arr[start] = val
id_arr[stop] = -val
return id_arr.cumsum()
def setup_data(n, k):
inds = np.sort(np.random.randint(0, n-2*k, (2*k,)) + np.arange(2*k))
return np.c_[inds.reshape(-1, 2), np.random.randint(1, 10, (k,))].astype(float)
for n in (100, 100000, 1000000):
k = 3**(n.bit_length()>>3)
spec = setup_data(n, k)
ref = np.zeros((n,))
f_pp(ref, spec)
print(f"n={n}")
for name, func in list(globals().items()):
if not name.startswith('f_') or not isinstance(func, types.FunctionType):
continue
try:
res = np.zeros((n,))
ret = func(res, spec)
if not ret is None:
res = ret
assert np.allclose(ref, res)
print("{:16s}{:16.8f} ms".format(name[2:], timeit(
'f(a, spec)', 'a=np.zeros((n,))',
globals={'f':func, 'spec':spec, 'np':np, 'n':n}, number=10)*100))
except Exception:
print("{:16s} apparently failed".format(name[2:]))