我知道如何删除numpy数组中的每个第四个元素：
frame = np.delete(frame,np.arange(4,frame.size,4))
方法 #1：以下是一种使用模数和布尔索引的方法 -
a[np.mod(np.arange(a.size),6)<3]
def select_in_groups(a, M, N): # Keep first M, delete next N and so on.
return a[np.mod(np.arange(a.size),M+N)<M]
样例逐步运行 -
# Input array
In [361]: a
Out[361]:
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,
18, 19, 20])
# Create a range array that spans along the length of array
In [362]: np.arange(a.size)
Out[362]:
array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16,
17, 18, 19])
# Use modulus to create "intervaled" version of it that shifts at
# the end of each group of 6 elements
In [363]: np.mod(np.arange(a.size),6)
Out[363]: array([0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1])
# We need to select the first three as valid ones, so compare against 3
# creating a boolean array or mask
In [364]: np.mod(np.arange(a.size),6) < 3
Out[364]:
array([ True, True, True, False, False, False, True, True, True,
False, False, False, True, True, True, False, False, False,
True, True], dtype=bool)
# Use the mask to select valid elements off array
In [365]: a[np.mod(np.arange(a.size),6)<3]
Out[365]: array([ 1, 2, 3, 7, 8, 9, 13, 14, 15, 19, 20])
方法二：为了提高性能，这里介绍另一种使用NumPy数组步幅的方法 -
def select_in_groups_strided(a, M, N): # Keep first M, delete next N and so on.
K = M+N
na = a.size
nrows = (1+((na-1)//K))
n = a.strides[0]
out = np.lib.index_tricks.as_strided(a, shape=(nrows,K), strides=(K*n,n))
N = M*(na//K) + (na - (K*(na//K)))
return out[:,:M].ravel()[:N]
样例运行 -
In [545]: a = np.arange(1,21)
In [546]: a
Out[546]:
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,
18, 19, 20])
In [547]: select_in_groups_strided(a,3,3)
Out[547]: array([ 1, 2, 3, 7, 8, 9, 13, 14, 15, 19, 20])
In [548]: a = np.arange(1,25)
In [549]: a
Out[549]:
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,
18, 19, 20, 21, 22, 23, 24])
In [550]: select_in_groups_strided(a,3,3)
Out[550]: array([ 1, 2, 3, 7, 8, 9, 13, 14, 15, 19, 20, 21])
运行时测试
使用与@Daniel Forsman的时间测试相同的设置-
In [637]: a = np.arange(1,21)
In [638]: %timeit block_delete(a,3,3)
10000 loops, best of 3: 21 µs per loop
In [639]: %timeit select_in_groups_strided(a,3,3)
100000 loops, best of 3: 6.44 µs per loop
In [640]: a = np.arange(1,2100)
In [641]: %timeit block_delete(a,3,3)
10000 loops, best of 3: 27 µs per loop
In [642]: %timeit select_in_groups_strided(a,3,3)
100000 loops, best of 3: 9.1 µs per loop
In [643]: a = np.arange(999999) + 1
In [644]: %timeit block_delete(a,3,3)
100 loops, best of 3: 2.24 ms per loop
In [645]: %timeit select_in_groups_strided(a,3,3)
1000 loops, best of 3: 1.12 ms per loop
Strided 在不同的尺寸上表现出色，如果你考虑性能方面。
def block_delete(a, n, m): #keep n, remove m
mask = np.tile(np.r_[np.ones(n), np.zeros(m)].astype(bool), a.size // (n + m) + 1)[:a.size]
return a[mask]
def mod_delete(a, n, m):
return a[np.mod(np.arange(a.size), n + m) < n]
a = np.arange(19) + 1
%timeit block_delete(a, 3, 4)
10000 loops, best of 3: 50.6 µs per loop
%timeit mod_delete(a, 3, 4)
The slowest run took 9.37 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 5.69 µs per loop
让我们试着使用更长的数组：
a = np.arange(999) + 1
%timeit block_delete(a, 3, 4)
The slowest run took 4.61 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 54.8 µs per loop
%timeit mod_delete(a, 3, 4)
The slowest run took 5.13 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 14.5 µs per loop
更长的：
a = np.arange(999999) + 1
%timeit block_delete(a, 3, 4)
100 loops, best of 3: 3.93 ms per loop
%timeit mod_delete(a, 3, 4)
100 loops, best of 3: 12.3 ms per loop
import numpy as np
a = np.array([10, 0, 0, 20, 0, 30, 40, 50, 0, 60, 70, 80, 90, 100,0])
print("Original array:")
print(a)
index=np.zeros(0)
for i in range(len(a)):
if a[i]==0:
index=np.append(index, i)
print("index=",index)
new_a=np.delete(a,index)
print("new_a=",new_a)
as_strided的答案，只是无法理解它。 - Daniel F