Refactor matrix permutations in numpy's style
I wrote the following code to do multiplication of matrix permutations and I was wondering if it can be written in a numpy style, such that I can get rid of the two for loops:
Z = np.empty([new_d, X.shape[1]])
Z = np.ndarray(shape=(new_d, X.shape[1]))
Z = np.concatenate((X, X**2))
res =
for i in range(0, d):
for j in range(i+1, d):
res.append(np.array(X.T[:,i]* X.T[:,j]))
Z = np.concatenate((Z, res))
while: X shape is (7, 1000), d = 7, new_d=35
any suggestion ?
python numpy machine-learning vectorization
edited Nov 24 '18 at 14:28
Divakar
157k1488179
asked Nov 24 '18 at 11:30
user181452user181452
11712
add a comment |
I wrote the following code to do multiplication of matrix permutations and I was wondering if it can be written in a numpy style, such that I can get rid of the two for loops:
Z = np.empty([new_d, X.shape[1]])
Z = np.ndarray(shape=(new_d, X.shape[1]))
Z = np.concatenate((X, X**2))
res =
for i in range(0, d):
for j in range(i+1, d):
res.append(np.array(X.T[:,i]* X.T[:,j]))
Z = np.concatenate((Z, res))
while: X shape is (7, 1000), d = 7, new_d=35
any suggestion ?
python numpy machine-learning vectorization
edited Nov 24 '18 at 14:28
Divakar
157k1488179
asked Nov 24 '18 at 11:30
user181452user181452
11712
add a comment |
I wrote the following code to do multiplication of matrix permutations and I was wondering if it can be written in a numpy style, such that I can get rid of the two for loops:
Z = np.empty([new_d, X.shape[1]])
Z = np.ndarray(shape=(new_d, X.shape[1]))
Z = np.concatenate((X, X**2))
res =
for i in range(0, d):
for j in range(i+1, d):
res.append(np.array(X.T[:,i]* X.T[:,j]))
Z = np.concatenate((Z, res))
while: X shape is (7, 1000), d = 7, new_d=35
any suggestion ?
python numpy machine-learning vectorization
edited Nov 24 '18 at 14:28
Divakar
157k1488179
asked Nov 24 '18 at 11:30
user181452user181452
11712
I wrote the following code to do multiplication of matrix permutations and I was wondering if it can be written in a numpy style, such that I can get rid of the two for loops:
Z = np.empty([new_d, X.shape[1]])
Z = np.ndarray(shape=(new_d, X.shape[1]))
Z = np.concatenate((X, X**2))
res =
for i in range(0, d):
for j in range(i+1, d):
res.append(np.array(X.T[:,i]* X.T[:,j]))
Z = np.concatenate((Z, res))
while: X shape is (7, 1000), d = 7, new_d=35
any suggestion ?
python numpy machine-learning vectorization
python numpy machine-learning vectorization
edited Nov 24 '18 at 14:28
Divakar
157k1488179
asked Nov 24 '18 at 11:30
user181452user181452
11712
edited Nov 24 '18 at 14:28
Divakar
157k1488179
asked Nov 24 '18 at 11:30
user181452user181452
11712
edited Nov 24 '18 at 14:28
Divakar
157k1488179
edited Nov 24 '18 at 14:28
Divakar
157k1488179
edited Nov 24 '18 at 14:28
Divakar
157k1488179
157k1488179
asked Nov 24 '18 at 11:30
user181452user181452
11712
asked Nov 24 '18 at 11:30
user181452user181452
11712
asked Nov 24 '18 at 11:30
user181452user181452
11712
11712
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Approach #1
We could use np.triu_indices to get those pair-wise permutation-indices and then simply perform elementwise multiplicatons of row-indexed arrays -
r,c = np.triu_indices(d,1)
res = X[r]*X[c]
Approach #2
For memory efficiency and hence performance especially on large arrays, we are better off slicing the input array and run a single loop with each iteration working on chunks of data, like so -
n = d-1
idx = np.concatenate(( [0], np.arange(n,0,-1).cumsum() ))
start, stop = idx[:-1], idx[1:]
L = n*(n+1)//2
res_out = np.empty((L,X.shape[1]), dtype=X.dtype)
for i,(s0,s1) in enumerate(zip(start,stop)):
res_out[s0:s1] = X[i] * X[i+1:]
To get Z directly and thus avoid all those concatenations, we could modify the earlier posted approach, like so -
n = d-1
N = len(X)
idx = 2*N + np.concatenate(( [0], np.arange(n,0,-1).cumsum() ))
start, stop = idx[:-1], idx[1:]
L = n*(n+1)//2
Z_out = np.empty((2*N + L,X.shape[1]), dtype=X.dtype)
Z_out[:N] = X
Z_out[N:2*N] = X**2
for i,(s0,s1) in enumerate(zip(start,stop)):
Z_out[s0:s1] = X[i] * X[i+1:]
answered Nov 24 '18 at 11:34
DivakarDivakar
157k1488179
Your first approach is so smart !! Whenever I write too much code in python (Matrices), I know it can be written in a shorter way but because I'm still beginner I'm not aware of all of those functions. Many thanks :))
– user181452
Nov 24 '18 at 14:19
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Approach #1
We could use np.triu_indices to get those pair-wise permutation-indices and then simply perform elementwise multiplicatons of row-indexed arrays -
r,c = np.triu_indices(d,1)
res = X[r]*X[c]
Approach #2
For memory efficiency and hence performance especially on large arrays, we are better off slicing the input array and run a single loop with each iteration working on chunks of data, like so -
n = d-1
idx = np.concatenate(( [0], np.arange(n,0,-1).cumsum() ))
start, stop = idx[:-1], idx[1:]
L = n*(n+1)//2
res_out = np.empty((L,X.shape[1]), dtype=X.dtype)
for i,(s0,s1) in enumerate(zip(start,stop)):
res_out[s0:s1] = X[i] * X[i+1:]
To get Z directly and thus avoid all those concatenations, we could modify the earlier posted approach, like so -
n = d-1
N = len(X)
idx = 2*N + np.concatenate(( [0], np.arange(n,0,-1).cumsum() ))
start, stop = idx[:-1], idx[1:]
L = n*(n+1)//2
Z_out = np.empty((2*N + L,X.shape[1]), dtype=X.dtype)
Z_out[:N] = X
Z_out[N:2*N] = X**2
for i,(s0,s1) in enumerate(zip(start,stop)):
Z_out[s0:s1] = X[i] * X[i+1:]
answered Nov 24 '18 at 11:34
DivakarDivakar
157k1488179
Your first approach is so smart !! Whenever I write too much code in python (Matrices), I know it can be written in a shorter way but because I'm still beginner I'm not aware of all of those functions. Many thanks :))
– user181452
Nov 24 '18 at 14:19
add a comment |
Approach #1
We could use np.triu_indices to get those pair-wise permutation-indices and then simply perform elementwise multiplicatons of row-indexed arrays -
r,c = np.triu_indices(d,1)
res = X[r]*X[c]
Approach #2
For memory efficiency and hence performance especially on large arrays, we are better off slicing the input array and run a single loop with each iteration working on chunks of data, like so -
n = d-1
idx = np.concatenate(( [0], np.arange(n,0,-1).cumsum() ))
start, stop = idx[:-1], idx[1:]
L = n*(n+1)//2
res_out = np.empty((L,X.shape[1]), dtype=X.dtype)
for i,(s0,s1) in enumerate(zip(start,stop)):
res_out[s0:s1] = X[i] * X[i+1:]
To get Z directly and thus avoid all those concatenations, we could modify the earlier posted approach, like so -
n = d-1
N = len(X)
idx = 2*N + np.concatenate(( [0], np.arange(n,0,-1).cumsum() ))
start, stop = idx[:-1], idx[1:]
L = n*(n+1)//2
Z_out = np.empty((2*N + L,X.shape[1]), dtype=X.dtype)
Z_out[:N] = X
Z_out[N:2*N] = X**2
for i,(s0,s1) in enumerate(zip(start,stop)):
Z_out[s0:s1] = X[i] * X[i+1:]
answered Nov 24 '18 at 11:34
DivakarDivakar
157k1488179
Your first approach is so smart !! Whenever I write too much code in python (Matrices), I know it can be written in a shorter way but because I'm still beginner I'm not aware of all of those functions. Many thanks :))
– user181452
Nov 24 '18 at 14:19
add a comment |
Approach #1
We could use np.triu_indices to get those pair-wise permutation-indices and then simply perform elementwise multiplicatons of row-indexed arrays -
r,c = np.triu_indices(d,1)
res = X[r]*X[c]
Approach #2
For memory efficiency and hence performance especially on large arrays, we are better off slicing the input array and run a single loop with each iteration working on chunks of data, like so -
n = d-1
idx = np.concatenate(( [0], np.arange(n,0,-1).cumsum() ))
start, stop = idx[:-1], idx[1:]
L = n*(n+1)//2
res_out = np.empty((L,X.shape[1]), dtype=X.dtype)
for i,(s0,s1) in enumerate(zip(start,stop)):
res_out[s0:s1] = X[i] * X[i+1:]
To get Z directly and thus avoid all those concatenations, we could modify the earlier posted approach, like so -
n = d-1
N = len(X)
idx = 2*N + np.concatenate(( [0], np.arange(n,0,-1).cumsum() ))
start, stop = idx[:-1], idx[1:]
L = n*(n+1)//2
Z_out = np.empty((2*N + L,X.shape[1]), dtype=X.dtype)
Z_out[:N] = X
Z_out[N:2*N] = X**2
for i,(s0,s1) in enumerate(zip(start,stop)):
Z_out[s0:s1] = X[i] * X[i+1:]
answered Nov 24 '18 at 11:34
DivakarDivakar
157k1488179
Approach #1
We could use np.triu_indices to get those pair-wise permutation-indices and then simply perform elementwise multiplicatons of row-indexed arrays -
r,c = np.triu_indices(d,1)
res = X[r]*X[c]
Approach #2
For memory efficiency and hence performance especially on large arrays, we are better off slicing the input array and run a single loop with each iteration working on chunks of data, like so -
n = d-1
idx = np.concatenate(( [0], np.arange(n,0,-1).cumsum() ))
start, stop = idx[:-1], idx[1:]
L = n*(n+1)//2
res_out = np.empty((L,X.shape[1]), dtype=X.dtype)
for i,(s0,s1) in enumerate(zip(start,stop)):
res_out[s0:s1] = X[i] * X[i+1:]
To get Z directly and thus avoid all those concatenations, we could modify the earlier posted approach, like so -
n = d-1
N = len(X)
idx = 2*N + np.concatenate(( [0], np.arange(n,0,-1).cumsum() ))
start, stop = idx[:-1], idx[1:]
L = n*(n+1)//2
Z_out = np.empty((2*N + L,X.shape[1]), dtype=X.dtype)
Z_out[:N] = X
Z_out[N:2*N] = X**2
for i,(s0,s1) in enumerate(zip(start,stop)):
Z_out[s0:s1] = X[i] * X[i+1:]
answered Nov 24 '18 at 11:34
DivakarDivakar
157k1488179
edited Nov 24 '18 at 12:08
answered Nov 24 '18 at 11:34
DivakarDivakar
157k1488179
answered Nov 24 '18 at 11:34
DivakarDivakar
157k1488179
answered Nov 24 '18 at 11:34
DivakarDivakar
157k1488179
157k1488179
Your first approach is so smart !! Whenever I write too much code in python (Matrices), I know it can be written in a shorter way but because I'm still beginner I'm not aware of all of those functions. Many thanks :))
– user181452
Nov 24 '18 at 14:19
add a comment |
Your first approach is so smart !! Whenever I write too much code in python (Matrices), I know it can be written in a shorter way but because I'm still beginner I'm not aware of all of those functions. Many thanks :))
– user181452
Nov 24 '18 at 14:19
Your first approach is so smart !! Whenever I write too much code in python (Matrices), I know it can be written in a shorter way but because I'm still beginner I'm not aware of all of those functions. Many thanks :))
– user181452
Nov 24 '18 at 14:19
Your first approach is so smart !! Whenever I write too much code in python (Matrices), I know it can be written in a shorter way but because I'm still beginner I'm not aware of all of those functions. Many thanks :))
– user181452
Nov 24 '18 at 14:19
add a comment |
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