How to determine the maximum element of a matrix? For "matrix_integer_dense" there is height() [1]. For numpy matrix there is max function. How to do it in Sage matrices with one single "max" like instruction ?
[1] http://www.sagemath.org/doc/reference...
[2] Alternetive: m.numpy().max()
Not a single command, but you can try the following one-line possibilities:
sage: size = 1000
sage: m = random_matrix(RDF, size)
sage: max(max(i for i in rows) for rows in m)
sage: max(m[i,j] for i in range(size) for j in range(size))
sage: max(m.list())
sage: max(m._list())
From the slowest to the fastest (5.19 s, 3.1 s, 2.12 s, 2.08 s).
You forget "m.numpy().max()" which is around 2ms and is definitely the best solution in that case. The reason is that a matrix over RDF is a wrapper to a numpy matrix.
You are right, and the bad timings of the privious methods could come from there too, here are the timings for a matrix over `RR`: sage: size = 1000 sage: m = random_matrix(RR, size) m.numpy().max() # 2.14 s max(max(i for i in rows) for rows in m) # 862 ms max(m[i,j] for i in range(size) for j in range(size)) # 414 ms max(m.list()) # 232 ms max(m._list()) # 137 ms
And over `ZZ`, from the slowest to the fastest: sage: m = random_matrix(ZZ, size) m.numpy().max() # 2.53 s max(max(i for i in rows) for rows in m) # 1.03 s max(m.list()) # 430 ms max(m[i,j] for i in range(size) for j in range(size)) # 365 ms max(m._list()) # 359 ms
I tried to do this a while ago but couldn't do it in one step. I just went through all the columns and got their maximum. I also needed to find the location of the maximum.
Example for a matrix M with type 'sage.matrix.matrix_real_double_dense.Matrix_real_double_dense'
MAX_val=0
for i in range(len(M.columns())):
if (MAX_val< max(M.column(i))):
MAX_val=max(M.column(i))
MAX_val is then the maximum value in the matrix. I hope it was helpful. I ll follow to see if there is a better way.
Asked: 2013-07-11 05:44:26 +0200
Seen: 4,690 times
Last updated: Jul 11 '13
Copyright Sage, 2010. Some rights reserved under creative commons license. Content on this site is licensed under a Creative Commons Attribution Share Alike 3.0 license.
I do not think there is any... an other alternative would be `max(m.list())` which I do not like because it explicitely build the list of the entries of m.