Simple question here:
简单的问题:
I'm trying to get an array that alternates values (1, -1, 1, -1.....) for a given length. np.repeat just gives me (1, 1, 1, 1,-1, -1,-1, -1). Thoughts?
我正在尝试得到一个数组,它改变给定长度的值(1,1,1,1,1,1,1,1…)。np。重复,得到(1 1 1 1 1 -1 -1 -1 -1 -1 -1)想法吗?
7 个解决方案
#1
15
I like @Benjamin's solution. An alternative though is:
我喜欢@Benjamin的解决方案。另一种是:
import numpy as np
a = np.empty((15,))
a[::2] = 1
a[1::2] = -1
This also allows for odd-length lists.
这也允许奇数长度的列表。
EDIT: Also just to note speeds, for a array of 10000 elements
编辑:也只是为了注意速度,对于10000个元素的数组
import numpy as np
from timeit import Timer
if __name__ == '__main__':
setupstr="""
import numpy as np
N = 10000
"""
method1="""
a = np.empty((N,),int)
a[::2] = 1
a[1::2] = -1
"""
method2="""
a = np.tile([1,-1],N)
"""
method3="""
a = np.array([1,-1]*N)
"""
method4="""
a = np.array(list(itertools.islice(itertools.cycle((1,-1)), N)))
"""
nl = 1000
t1 = Timer(method1, setupstr).timeit(nl)
t2 = Timer(method2, setupstr).timeit(nl)
t3 = Timer(method3, setupstr).timeit(nl)
t4 = Timer(method4, setupstr).timeit(nl)
print 'method1', t1
print 'method2', t2
print 'method3', t3
print 'method4', t4
Results in timings of:
结果的时间:
method1 0.0130500793457
method2 0.114426136017
method3 4.30518102646
method4 2.84446692467
If N = 100
, things start to even out but starting with the empty numpy arrays is still significantly faster (nl
changed to 10000)
如果N = 100,那么事情就开始了,但从空的numpy数组开始,仍然要快得多(nl变为10000)
method1 0.05735206604
method2 0.323992013931
method3 0.556654930115
method4 0.46702003479
Numpy arrays are special awesome objects and should not be treated like python lists.
Numpy数组是特别棒的对象,不应该像python列表那样处理。
#2
7
use resize():
使用调整():
In [38]: np.resize([1,-1], 10) # 10 is the length of result array
Out[38]: array([ 1, -1, 1, -1, 1, -1, 1, -1, 1, -1])
it can produce odd-length array:
可以产生奇长阵列:
In [39]: np.resize([1,-1], 11)
Out[39]: array([ 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1])
#3
6
Use numpy.tile
!
使用numpy.tile !
import numpy
a = numpy.tile([1,-1], 15)
#4
4
use multiplication:
用乘法:
[1,-1] * n
#5
3
If you want a memory efficient solution, try this:
如果您想要一个内存有效的解决方案,可以尝试以下方法:
def alternator(n):
for i in xrange(n):
if i % 2 == 0:
yield 1
else:
yield -1
Then you can iterate over the answers like so:
然后你可以重复回答如下:
for i in alternator(n):
# do something with i
#6
2
Maybe you're looking for itertools.cycle?
也许您正在寻找itertools.cycle?
list_ = (1,-1,2,-2) # ,3,-3, ...
for n, item in enumerate(itertools.cycle(list_)):
if n==30:
break
print item
#7
0
I'll just throw these out there because they could be more useful in some circumstances.
我把这些扔到外面因为在某些情况下它们会更有用。
If you just want to alternate between positive and negative:
如果你只想在正负之间交替:
[(-1)**i for i in range(n)]
or for a more general solution
或者更一般的解
nums = [1, -1, 2]
[nums[i % len(nums)] for i in range(n)]
#1
15
I like @Benjamin's solution. An alternative though is:
我喜欢@Benjamin的解决方案。另一种是:
import numpy as np
a = np.empty((15,))
a[::2] = 1
a[1::2] = -1
This also allows for odd-length lists.
这也允许奇数长度的列表。
EDIT: Also just to note speeds, for a array of 10000 elements
编辑:也只是为了注意速度,对于10000个元素的数组
import numpy as np
from timeit import Timer
if __name__ == '__main__':
setupstr="""
import numpy as np
N = 10000
"""
method1="""
a = np.empty((N,),int)
a[::2] = 1
a[1::2] = -1
"""
method2="""
a = np.tile([1,-1],N)
"""
method3="""
a = np.array([1,-1]*N)
"""
method4="""
a = np.array(list(itertools.islice(itertools.cycle((1,-1)), N)))
"""
nl = 1000
t1 = Timer(method1, setupstr).timeit(nl)
t2 = Timer(method2, setupstr).timeit(nl)
t3 = Timer(method3, setupstr).timeit(nl)
t4 = Timer(method4, setupstr).timeit(nl)
print 'method1', t1
print 'method2', t2
print 'method3', t3
print 'method4', t4
Results in timings of:
结果的时间:
method1 0.0130500793457
method2 0.114426136017
method3 4.30518102646
method4 2.84446692467
If N = 100
, things start to even out but starting with the empty numpy arrays is still significantly faster (nl
changed to 10000)
如果N = 100,那么事情就开始了,但从空的numpy数组开始,仍然要快得多(nl变为10000)
method1 0.05735206604
method2 0.323992013931
method3 0.556654930115
method4 0.46702003479
Numpy arrays are special awesome objects and should not be treated like python lists.
Numpy数组是特别棒的对象,不应该像python列表那样处理。
#2
7
use resize():
使用调整():
In [38]: np.resize([1,-1], 10) # 10 is the length of result array
Out[38]: array([ 1, -1, 1, -1, 1, -1, 1, -1, 1, -1])
it can produce odd-length array:
可以产生奇长阵列:
In [39]: np.resize([1,-1], 11)
Out[39]: array([ 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1])
#3
6
Use numpy.tile
!
使用numpy.tile !
import numpy
a = numpy.tile([1,-1], 15)
#4
4
use multiplication:
用乘法:
[1,-1] * n
#5
3
If you want a memory efficient solution, try this:
如果您想要一个内存有效的解决方案,可以尝试以下方法:
def alternator(n):
for i in xrange(n):
if i % 2 == 0:
yield 1
else:
yield -1
Then you can iterate over the answers like so:
然后你可以重复回答如下:
for i in alternator(n):
# do something with i
#6
2
Maybe you're looking for itertools.cycle?
也许您正在寻找itertools.cycle?
list_ = (1,-1,2,-2) # ,3,-3, ...
for n, item in enumerate(itertools.cycle(list_)):
if n==30:
break
print item
#7
0
I'll just throw these out there because they could be more useful in some circumstances.
我把这些扔到外面因为在某些情况下它们会更有用。
If you just want to alternate between positive and negative:
如果你只想在正负之间交替:
[(-1)**i for i in range(n)]
or for a more general solution
或者更一般的解
nums = [1, -1, 2]
[nums[i % len(nums)] for i in range(n)]