问题
I need to create a 2D array where each row may start and end with a different number. Assume that first and last element of each row is given and all other elements are just interpolated according to length of the rows In a simple case let's say I want to create a 3X3 array with same start at 0 but different end given by W below:
array([[ 0., 1., 2.],
[ 0., 2., 4.],
[ 0., 3., 6.]])
Is there a better way to do this than the following:
D=np.ones((3,3))*np.arange(0,3)
D=D/D[:,-1]
W=np.array([2,4,6]) # last element of each row assumed given
Res= (D.T*W).T
回答1:
Here's an approach using broadcasting -
def create_ranges(start, stop, N, endpoint=True):
if endpoint==1:
divisor = N-1
else:
divisor = N
steps = (1.0/divisor) * (stop - start)
return steps[:,None]*np.arange(N) + start[:,None]
Sample run -
In [22]: # Setup start, stop for each row and no. of elems in each row
...: start = np.array([1,4,2])
...: stop = np.array([6,7,6])
...: N = 5
...:
In [23]: create_ranges(start, stop, 5)
Out[23]:
array([[ 1. , 2.25, 3.5 , 4.75, 6. ],
[ 4. , 4.75, 5.5 , 6.25, 7. ],
[ 2. , 3. , 4. , 5. , 6. ]])
In [24]: create_ranges(start, stop, 5, endpoint=False)
Out[24]:
array([[ 1. , 2. , 3. , 4. , 5. ],
[ 4. , 4.6, 5.2, 5.8, 6.4],
[ 2. , 2.8, 3.6, 4.4, 5.2]])
Let's leverage multi-core!
We can leverage multi-core with numexpr module for large data and to gain memory efficiency and hence performance -
import numexpr as ne
def create_ranges_numexpr(start, stop, N, endpoint=True):
if endpoint==1:
divisor = N-1
else:
divisor = N
s0 = start[:,None]
s1 = stop[:,None]
r = np.arange(N)
return ne.evaluate('((1.0/divisor) * (s1 - s0))*r + s0')
回答2:
Like the OP's this use of linspace
assumes the start is 0 for all rows.
x=np.linspace(0,1,N)[:,None]*np.arange(0,2*N,2)
(edit - this is the transpose of what I should get; either transpose it or switch the use of [:,None]
)
For N=3000, it's noticeably faster than @Divaker's
solution. I'm not entirely sure why.
In [132]: timeit N=3000;x=np.linspace(0,1,N)[:,None]*np.arange(0,2*N,2)
10 loops, best of 3: 91.7 ms per loop
In [133]: timeit create_ranges(np.zeros(N),np.arange(0,2*N,2),N)
1 loop, best of 3: 197 ms per loop
In [134]: def foo(N):
...: D=np.ones((N,N))*np.arange(N)
...: D=D/D[:,-1]
...: W=np.arange(0,2*N,2)
...: return (D.T*W).T
...:
In [135]: timeit foo(3000)
1 loop, best of 3: 454 ms per loop
============
With starts and stops I could use:
In [201]: starts=np.array([1,4,2]); stops=np.array([6,7,8])
In [202]: x=(np.linspace(0,1,5)[:,None]*(stops-starts)+starts).T
In [203]: x
Out[203]:
array([[ 1. , 2.25, 3.5 , 4.75, 6. ],
[ 4. , 4.75, 5.5 , 6.25, 7. ],
[ 2. , 3.5 , 5. , 6.5 , 8. ]])
With the extra calculations that makes it a bit slower than create_ranges
.
In [208]: timeit N=3000;starts=np.zeros(N);stops=np.arange(0,2*N,2);x=(np.linspace(0,1,N)[:,None]*(stops-starts)+starts).T
1 loop, best of 3: 227 ms per loop
All these solutions are just variations the idea of doing a linear interpolation between the starts
and stops
.
回答3:
NumPy >= 1.16.0:
It is now possible to supply array-like values to start
and stop
parameters of the np.linspace.
For the example given in the question the syntax would be:
>>> np.linspace((0, 0, 0), (2, 4, 6), 3, axis=1)
array([[0., 1., 2.],
[0., 2., 4.],
[0., 3., 6.]])
New axis
parameter specifies in which direction data will be generated. By default it is 0
:
>>> np.linspace((0, 0, 0), (2, 4, 6), 3)
array([[0., 0., 0.],
[1., 2., 3.],
[2., 4., 6.]])
回答4:
I extended a bit of the functionality based on @Divakar's solutions. It sacrifices some speed but now is compatible for different lengths of N
instead of only scalar. Plus, this version it faster than @Saullo's sollution.
def create_ranges_divak(starts, stops, N, endpoint=True):
if endpoint==1:
divisor = N-1
else:
divisor = N
steps = (1.0/divisor) * (stops - starts)
uni_N = np.unique(N)
if len(uni_N) == 1:
return steps[:,None]*np.arange(uni_N) + starts[:,None]
else:
return [step * np.arange(n) + start for start, step, n in zip(starts, steps, N)]
来源:https://stackoverflow.com/questions/40624409/vectorized-numpy-linspace-for-multiple-start-and-stop-values