我有一个一维numpy
数组-例如,
a = np.array([1, 4, 5, 7, 1, 2, 2, 4, 10])
我想获得第一个数字的索引,对于该数字,N个后续值都低于某个值x。
在这种情况下,对于N=3
和x=3
,我将搜索第一个数字,该数字后面的三个条目都小于3。这将是a[4]
。
可以简单地通过for
循环遍历所有值来轻松地实现这一点,但是我想知道是否有更干净,更有效的方法来实现此目的。
答案 0 :(得分:9)
方法1:
这是矢量化的NumPy方法-
def start_valid_island(a, thresh, window_size):
m = a<thresh
me = np.r_[False,m,False]
idx = np.flatnonzero(me[:-1]!=me[1:])
lens = idx[1::2]-idx[::2]
return idx[::2][(lens >= window_size).argmax()]
样品运行-
In [44]: a
Out[44]: array([ 1, 4, 5, 7, 1, 2, 2, 4, 10])
In [45]: start_valid_island(a, thresh=3, window_size=3)
Out[45]: 4
In [46]: a[:3] = 1
In [47]: start_valid_island(a, thresh=3, window_size=3)
Out[47]: 0
方法2:
from scipy.ndimage.morphology import binary_erosion
def start_valid_island_v2(a, thresh, window_size):
m = a<thresh
k = np.ones(window_size,dtype=bool)
return binary_erosion(m,k,origin=-(window_size//2)).argmax()
方法3:
要完成 set ,下面是一个基于短循环并使用numba
-
from numba import njit
@njit
def start_valid_island_v3(a, thresh, window_size):
n = len(a)
out = None
for i in range(n-window_size+1):
found = True
for j in range(window_size):
if a[i+j]>=thresh:
found = False
break
if found:
out = i
break
return out
时间-
In [142]: np.random.seed(0)
...: a = np.random.randint(0,10,(100000000))
In [145]: %timeit start_valid_island(a, thresh=3, window_size=3)
1 loop, best of 3: 810 ms per loop
In [146]: %timeit start_valid_island_v2(a, thresh=3, window_size=3)
1 loop, best of 3: 1.27 s per loop
In [147]: %timeit start_valid_island_v3(a, thresh=3, window_size=3)
1000000 loops, best of 3: 608 ns per loop
答案 1 :(得分:0)
像这样尝试,如果没有数字与条件匹配,将返回None
:
def func(a, n, x):
for i, e in enumerate(a):
nextN = a[i+1:i+n+1]
if len(nextN) < n:
return None
elif all([j < x for j in nextN]):
return e
答案 2 :(得分:0)
对于它的价值,这是在 vanilla-python 中进行的操作。
a = [1,4,5,7,1,2,2,4,10]
res = next(i for i in range(len(a)-3) if all(j<3 for j in a[i:i+3]))
print(res) # 4
不过,大多数Numpy
解决方案可能会更快。
还请注意,如果找不到解决方案,以上内容将抛出StopIteration
,因此请考虑将其包装在try
块中。