a=QQ(-1)
matrix(QQ,1000,1000,lambda i,j:a)
oddly enough the following is faster:
from itertools import repeat
matrix(QQ,1000,1000,repeat(a,1000000))
Of course, select the appropriate ring in your own applications.
Or you could try numpy which is roughly 50 times faster
import time
import numpy as np
a=QQ(-1)
elapsed_time=0
for i in range(100):
start=time.time()
test = a*np.ones([1000,1000])
end=time.time()
elapsed_time=elapsed_time+(end-start)
print(elapsed_time/100)
But keep in mind that the example
from itertools import repeat
matrix(QQ,1000,1000,repeat(a,1000000))
generates a different type than a numpy array does
type(matrix(QQ,1000,1000,repeat(a,1000000)))
<type 'sage.matrix.matrix_rational_dense.Matrix_rational_dense'>
type(a*np.ones([1000,1000]))
<type 'numpy.ndarray'>
As another alternative, you can use ones_matrix, which returns a matrix with all entries equal to 1. So, to build, say, an $1000\times1000$ matrix with all entries being $-1$, you can simply write -ones_matrix(1000,1000).
For the sake of comparison, once done all the imports and set m=1000, let us test the time required by each option:
%%timeit
matrix(m,m,lambda i,j:-1)
1 loop, best of 3: 431 ms per loop
%%timeit
matrix(m,m,repeat(-1,m*m))
10 loops, best of 3: 98.3 ms per loop
%%timeit
-np.ones([m,m])
1000 loops, best of 3: 1.79 ms per loop
%%timeit
-ones_matrix(m,m)
10 loops, best of 3: 72 ms per loop
In a different computer, results will be different, of course. Except in the case of the Numpy array, the elements of the resulting matrix belong to the Integer Ring.
It makes quite a difference to make the element a:=-1 outside of the function and to specify the base ring of the matrix so that sage doesn't have to figure that out itself. So
a=-1
matrix(ZZ,m,m,lambda i,j:a)
is measurably faster. This is mainly because in sage, -1 is not a literal but gets converted to Integer(-1) by the preparser.
If you want to remove a further bit of overhead, you can define the lambda via
def make_lambda():
a=-1
return lambda i,j: a
fn=make_lambda()
matrix(ZZ,m,m,fn)
@nbruin Yes, you are right. Now %%timeit yields 10 loops, best of 3: 146 ms per loop. To be fair, an analog time reduction happens in the second alternative:
a = -1
matrix(ZZ,m,m,repeat(a,m*m))
10 loops, best of 3: 74.2 ms per loop
which is as fast as -ones_matrix(ZZ,m,m).
Asked: 2019-03-06 03:44:27 +0100
Seen: 536 times
Last updated: Mar 06 '19
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