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I am trying to speed up the below operation by doing some sort of matrix/vector-multiplication, can anyone see a nice quick solution?
It should also work for a special case where a tensor has shape 0 (torch.Size([])) but i am not able to initialize such a tensor.
See the image below for the type of tensor i am referring to:
tensor to add to test
def adstock_geometric(x: torch.Tensor, theta: float):
x_decayed = torch.zeros_like(x)
x_decayed[0] = x[0]
for xi in range(1, len(x_decayed)):
x_decayed[xi] = x[xi] + theta * x_decayed[xi - 1]
return x_decayed
def adstock_multiple_samples(x: torch.Tensor, theta: torch.Tensor):
listtheta = theta.tolist()
if isinstance(listtheta, float):
return adstock_geometric(x=x,
theta=theta)
x_decayed = torch.zeros((100, 112, 1))
for idx, theta_ in enumerate(listtheta):
x_decayed_one_entry = adstock_geometric(x=x,
theta=theta_)
x_decayed[idx] = x_decayed_one_entry
return x_decayed
if __name__ == '__main__':
ones = torch.tensor([1])
hundreds = torch.tensor([idx for idx in range(100)])
x = torch.tensor([[idx] for idx in range(112)])
ones = adstock_multiple_samples(x=x,
theta=ones)
hundreds = adstock_multiple_samples(x=x,
theta=hundreds)
print(ones)
print(hundreds)
I came up with the following, which is 40 times faster on your example:
import torch
def adstock_multiple_samples(x: torch.Tensor, theta: torch.Tensor):
arange = torch.arange(len(x))
powers = (arange[:, None] - arange).clip(0)
return ((theta[:, None, None] ** powers[None, :, :]).tril() * x).sum(-1)
It behaves as expected:
>>> x = torch.arange(112)
>>> theta = torch.arange(100)
>>> adstock_multiple_samples(x, theta)
... # the same output
Note that I considered that x was a 1D-tensor, as for your example the second dimension was not needed.
It also works with theta = torch.empty((0,)), and it returns an empty tensor.
I have a function that takes a [32, 32, 3] tensor, and outputs a [256,256,3] tensor.
Specifically, the function interprets the smaller array as if it was a .svg file, and 'renders' it to a 256x256 array as a canvas using this algorithm
For an explanation of WHY I would want to do this, see This question
The function behaves exactly as intended, until I try to include it in the training loop of a GAN. The current error I'm seeing is:
NotImplementedError: Cannot convert a symbolic Tensor (mul:0) to a numpy array.
A lot of other answers to similar errors seem to boil down to "You need to re-write the function using tensorflow, not numpy"
Here's the working code using numpy - is it possible to re-write it to exclusively use tensorflow functions?
def convert_to_bitmap(input_tensor, target, j):
#implied conversion to nparray - the tensorflow docs seem to indicate this is okay, but the error is thrown here when training
array = input_tensor
outputArray = target
output = target
for i in range(32):
col = float(array[i,0,j])
if ((float(array[i,0,0]))+(float(array[i,0,1]))+(float(array[i,0,2]))/3)< 0:
continue
#slice only the red channel from the i line, multiply by 255
red_array = array[i,:,0]*255
#slice only the green channel, multiply by 255
green_array = array[i,:,1]*255
#combine and flatten them
combined_array = np.dstack((red_array, green_array)).flatten()
#remove the first two and last two indices of the combined array
index = [0,1,62,63]
clipped_array = np.delete(combined_array,index)
#filter array to remove values less than 0
filtered = clipped_array > 0
filtered_array = clipped_array[filtered]
#check array has an even number of values, delete the last index if it doesn't
if len(filtered_array) % 2 == 0:
pass
else:
filtered_array = np.delete(filtered_array,-1)
#convert into a set of tuples
l = filtered_array.tolist()
t = list(zip(l, l[1:] + l[:1]))
if not t:
continue
output = fill_polygon(t, outputArray, col)
return(output)
The 'fill polygon' function is copied from the 'mahotas' library:
def fill_polygon(polygon, canvas, color):
if not len(polygon):
return
min_y = min(int(y) for y,x in polygon)
max_y = max(int(y) for y,x in polygon)
polygon = [(float(y),float(x)) for y,x in polygon]
if max_y < canvas.shape[0]:
max_y += 1
for y in range(min_y, max_y):
nodes = []
j = -1
for i,p in enumerate(polygon):
pj = polygon[j]
if p[0] < y and pj[0] >= y or pj[0] < y and p[0] >= y:
dy = pj[0] - p[0]
if dy:
nodes.append( (p[1] + (y-p[0])/(pj[0]-p[0])*(pj[1]-p[1])) )
elif p[0] == y:
nodes.append(p[1])
j = i
nodes.sort()
for n,nn in zip(nodes[::2],nodes[1::2]):
nn += 1
canvas[y, int(n):int(nn)] = color
return(canvas)
NOTE: I'm not trying to get someone to convert the whole thing for me! There are some functions that are pretty obvious (tf.stack instead of np.dstack), but others that I don't even know how to start, like the last few lines of the fill_polygon function above.
Yes you can actually do this, you can use a python function in sth called tf.pyfunc. Its a python wrapper but its extremely slow in comparison to plain tensorflow. However, tensorflow and Cuda for example are so damn fast because they use stuff like vectorization, meaning you can rewrite a lot , really many of the loops in terms of mathematical tensor operations which are very fast.
In general:
If you want to use custom code as a custom layer, i would recommend you to rethink the algebra behind those loops and try to express them somehow different. If its just preprocessing before the training is going to start, you can use tensorflow but doing the same with numpy and other libraries is easier.
To your main question: Yes its possible, but better dont use loops. Tensorflow has a build-in loop optimizer but then you have to use tf.while() and thats anyoing (maybe just for me). I just blinked over your code, but it looks like you should be able to vectorize it quite good using the standard tensorflow vocabulary. If you want it fast, i mean really fast with GPU support write all in tensorflow, but nothing like 50/50 with tf.convert_to_tensor(), because than its going to be slow again. because than you switch between GPU and CPU and plain Python interpreter and the tensorflow low level API. Hope i could help you at least a bit
This code 'works', in that it only uses tensorflow functions, and does allow the model to train when used in a training loop:
def convert_image (x):
#split off the first column of the generator output, and store it for later (remove the 'colours' column)
colours_column = tf.slice(img_to_convert, tf.constant([0,0,0], dtype=tf.int32), tf.constant([32,1,3], dtype=tf.int32))
#split off the rest of the data, only keeping R + G, and discarding B
image_data_red = tf.slice(img_to_convert, tf.constant([0,1,0], dtype=tf.int32), tf.constant([32,31,1], dtype=tf.int32))
image_data_green = tf.slice(img_to_convert, tf.constant([0,1,1], dtype=tf.int32), tf.constant([32, 31,1], dtype=tf.int32))
#roll each row by 1 position, and make two more 2D tensors
rolled_red = tf.roll(image_data_red, shift=-1, axis=0)
rolled_green = tf.roll(image_data_green, shift=-1, axis=0)
#remove all values where either the red OR green channels are 0
zeroes = tf.constant(0, dtype=tf.float32)
#this is for the 'count_nonzero' command
boolean_red_data = tf.not_equal(image_data_red, zeroes)
boolean_green_data = tf.not_equal(image_data_green, zeroes)
initial_data_mask = tf.logical_and(boolean_red_data, boolean_green_data)
#count non-zero values per row and flatten it
count = tf.math.count_nonzero(initial_data_mask, 1)
count_flat = tf.reshape(count, [-1])
flat_red = tf.reshape(image_data_red, [-1])
flat_green = tf.reshape(image_data_green, [-1])
boolean_red = tf.math.logical_not(tf.equal(flat_red, tf.zeros_like(flat_red)))
boolean_green = tf.math.logical_not(tf.equal(flat_green, tf.zeros_like(flat_red)))
mask = tf.logical_and(boolean_red, boolean_green)
flat_red_without_zero = tf.boolean_mask(flat_red, mask)
flat_green_without_zero = tf.boolean_mask(flat_green, mask)
# create a ragged tensor
X0_ragged = tf.RaggedTensor.from_row_lengths(values=flat_red_without_zero, row_lengths=count_flat)
Y0_ragged = tf.RaggedTensor.from_row_lengths(values=flat_green_without_zero, row_lengths=count_flat)
#do the same for the rolled version
rolled_data_mask = tf.roll(initial_data_mask, shift=-1, axis=1)
flat_rolled_red = tf.reshape(rolled_red, [-1])
flat_rolled_green = tf.reshape(rolled_green, [-1])
#from SO "shift zeros to the end"
boolean_rolled_red = tf.math.logical_not(tf.equal(flat_rolled_red, tf.zeros_like(flat_rolled_red)))
boolean_rolled_green = tf.math.logical_not(tf.equal(flat_rolled_green, tf.zeros_like(flat_rolled_red)))
rolled_mask = tf.logical_and(boolean_rolled_red, boolean_rolled_green)
flat_rolled_red_without_zero = tf.boolean_mask(flat_rolled_red, rolled_mask)
flat_rolled_green_without_zero = tf.boolean_mask(flat_rolled_green, rolled_mask)
# create a ragged tensor
X1_ragged = tf.RaggedTensor.from_row_lengths(values=flat_rolled_red_without_zero, row_lengths=count_flat)
Y1_ragged = tf.RaggedTensor.from_row_lengths(values=flat_rolled_green_without_zero, row_lengths=count_flat)
#available outputs for future use are:
X0 = X0_ragged.to_tensor(default_value=0.)
Y0 = Y0_ragged.to_tensor(default_value=0.)
X1 = X1_ragged.to_tensor(default_value=0.)
Y1 = Y1_ragged.to_tensor(default_value=0.)
#Example tensor cel (replace with (x))
P = tf.cast(x, dtype=tf.float32)
#split out P.x and P.y, and fill a ragged tensor to the same shape as Rx
Px_value = tf.cast(x, dtype=tf.float32) - tf.cast((tf.math.floor(x/255)*255), dtype=tf.float32)
Py_value = tf.cast(tf.math.floor(x/255), dtype=tf.float32)
Px = tf.squeeze(tf.ones_like(X0)*Px_value)
Py = tf.squeeze(tf.ones_like(Y0)*Py_value)
#for each pair of values (Y0, Y1, make a vector, and check to see if it crosses the y-value (Py) either up or down
a = tf.math.less(Y0, Py)
b = tf.math.greater_equal(Y1, Py)
c = tf.logical_and(a, b)
d = tf.math.greater_equal(Y0, Py)
e = tf.math.less(Y1, Py)
f = tf.logical_and(d, e)
g = tf.logical_or(c, f)
#Makes boolean bitwise mask
#calculate the intersection of the line with the y-value, assuming it intersects
#P.x <= (G.x - R.x) * (P.y - R.y) / (G.y - R.y + R.x) - use tf.divide_no_nan for safe divide
h = tf.math.less(Px,(tf.math.divide_no_nan(((X1-X0)*(Py-Y0)),(Y1-Y0+X0))))
#combine using AND with the mask above
i = tf.logical_and(g,h)
#tf.count_nonzero
#reshape to make a column tensor with the same dimensions as the colours
#divide by 2 using tf.floor_mod (returns remainder of division - any remainder means the value is odd, and hence the point is IN the polygon)
final_count = tf.cast((tf.math.count_nonzero(i, 1)), dtype=tf.int32)
twos = tf.ones_like(final_count, dtype=tf.int32)*tf.constant([2], dtype=tf.int32)
divide = tf.cast(tf.math.floormod(final_count, twos), dtype=tf.int32)
index = tf.cast(tf.range(0,32, delta=1), dtype=tf.int32)
clipped_index = divide*index
sort = tf.sort(clipped_index)
reverse = tf.reverse(sort, [-1])
value = tf.slice(reverse, [0], [1])
pair = tf.constant([0], dtype=tf.int32)
slice_tensor = tf.reshape(tf.stack([value, pair, pair], axis=0),[-1])
output_colour = tf.slice(colours_column, slice_tensor, [1,1,3])
return output_colour
This is where the 'convert image' function is applied using tf.vectorize_map:
def convert_images(image_to_convert):
global img_to_convert
img_to_convert = image_to_convert
process_list = tf.reshape((tf.range(0,65536, delta=1, dtype=tf.int32)), [65536, 1])
output_line = tf.vectorized_map(convert_image, process_list)
output_line_squeezed = tf.squeeze(output_line)
output_reshape = (tf.reshape(output_line_squeezed, [256,256,3])/127.5)-1
output = tf.expand_dims(output_reshape, axis=0)
return output
It is PAINFULLY slow, though - It does not appear to be using the GPU, and looks to be single threaded as well.
I'm adding it as an answer to my own question because is clearly IS possible to do this numpy function entirely in tensorflow - it just probably shouldn't be done like this.
How do I implement this metric in Keras? My code below gives the wrong result!
Note that I'm undoing a previous log(x + 1) transformation via exp(x) - 1, also negative predictions are clipped to 0:
def rmsle_cust(y_true, y_pred):
first_log = K.clip(K.exp(y_pred) - 1.0, 0, None)
second_log = K.clip(K.exp(y_true) - 1.0, 0, None)
return K.sqrt(K.mean(K.square(K.log(first_log + 1.) - K.log(second_log + 1.)), axis=-1)
For comparison, here's the standard numpy implementation:
def rmsle_cust_py(y, y_pred, **kwargs):
# undo 1 + log
y = np.exp(y) - 1
y_pred = np.exp(y_pred) - 1
y_pred[y_pred < 0] = 0.0
to_sum = [(math.log(y_pred[i] + 1) - math.log(y[i] + 1)) ** 2.0 for i,pred in enumerate(y_pred)]
return (sum(to_sum) * (1.0/len(y))) ** 0.5
What I'm doing wrong? Thanks!
EDIT: Setting axis=0 seems to give a value very close to the correct one, but I'm not sure since all the code I've seem uses axis=-1.
I ran into the same problem and searched for it, here is what I found
https://www.kaggle.com/jpopham91/rmlse-vectorized
After modified a bit, this seems to work for me,rmsle_K method implemented with Keras and TensorFlow.
import numpy as np
import math
from keras import backend as K
import tensorflow as tf
def rmsle(y, y0):
assert len(y) == len(y0)
return np.sqrt(np.mean(np.power(np.log1p(y)-np.log1p(y0), 2)))
def rmsle_loop(y, y0):
assert len(y) == len(y0)
terms_to_sum = [(math.log(y0[i] + 1) - math.log(y[i] + 1)) ** 2.0 for i,pred in enumerate(y0)]
return (sum(terms_to_sum) * (1.0/len(y))) ** 0.5
def rmsle_K(y, y0):
return K.sqrt(K.mean(K.square(tf.log1p(y) - tf.log1p(y0))))
r = rmsle(y=[5, 20, 12], y0=[8, 16, 12])
r1 = rmsle_loop(y=[5, 20, 12], y0=[8, 16, 12])
r2 = rmsle_K(y=[5., 20., 12.], y0=[8., 16., 12.])
print(r)
print(r1)
sess = tf.Session()
print(sess.run(r2))
Result:
Using TensorFlow backend
0.263978210565
0.263978210565
0.263978
By the use of a list (to_sum) in the numpy implementation, I suspect your numpy array has shape (length,).
And on Keras, since you've got different results with axis=0 and axis=1, you probably got some shape like (length,1).
Also, when creating the to_sum list, you're using y[i] and y_pred[i], which means you're taking elements from the axis=0 in numpy implementation.
The numpy implementation also sums everything for calculating the mean in sum(to_sum). So, you really don't need to use any axis in the K.mean.
If you make sure your model's output shape is either (length,) or (length,1), you can use just K.mean(value) without passing the axis parameter.
Easiest thing might be for me to just post the numpy code that I'm trying to perform directly in Theano if it's possible:
tensor = shared(np.random.randn(7, 16, 16)).eval()
tensor2 = tensor[0,:,:].eval()
tensor2[tensor2 < 1] = 0.0
tensor2[tensor2 > 0] = 1.0
new_tensor = [tensor2]
for i in range(1, tensor.shape[0]):
new_tensor.append(np.multiply(tensor2, tensor[i,:,:].eval()))
output = np.array(new_tensor).reshape(7,16,16)
If it's not immediately obvious, what I'm trying to do is use the values from one matrix of a tensor made up of 7 different matrices and apply that to the other matrices in the tensor.
Really, the problem I'm solving is doing conditional statements in an objective function for a fully convoltional network in Keras. Basically the loss for some of the feature map values is going to be calculated (and subsequently weighted) differently from others depending on some of the values in one of the feature maps.
You can easily implement conditionals with switch statement.
Here would be the equivalent code:
import theano
from theano import tensor as T
import numpy as np
def _check_new(var):
shape = var.shape[0]
t_1, t_2 = T.split(var, [1, shape-1], 2, axis=0)
ones = T.ones_like(t_1)
cond = T.gt(t_1, ones)
mask = T.repeat(cond, t_2.shape[0], axis=0)
out = T.switch(mask, t_2, T.zeros_like(t_2))
output = T.join(0, cond, out)
return output
def _check_old(var):
tensor = var.eval()
tensor2 = tensor[0,:,:]
tensor2[tensor2 < 1] = 0.0
tensor2[tensor2 > 0] = 1.0
new_tensor = [tensor2]
for i in range(1, tensor.shape[0]):
new_tensor.append(np.multiply(tensor2, tensor[i,:,:]))
output = theano.shared(np.array(new_tensor).reshape(7,16,16))
return output
tensor = theano.shared(np.random.randn(7, 16, 16))
out1 = _check_new(tensor).eval()
out2 = _check_old(tensor).eval()
print out1
print '----------------'
print ((out1-out2) ** 2).mean()
Note: since your masking on the first filter, I needed to use split and join operations.
I've got zero experience with Python. I have looked around some tutorial materials, but it seems difficult to understand a advanced code. So I came here for a more specific answer.
For me the mission is to redo the code in my computer.
Here is the scenario:
I'm a graduate student studying tensor factorization in relation learning. A paper[1] providing a code to run this algorithm, as follows:
import logging, time
from numpy import dot, zeros, kron, array, eye, argmax
from numpy.linalg import qr, pinv, norm, inv
from scipy.linalg import eigh
from numpy.random import rand
__version__ = "0.1"
__all__ = ['rescal', 'rescal_with_random_restarts']
__DEF_MAXITER = 500
__DEF_INIT = 'nvecs'
__DEF_PROJ = True
__DEF_CONV = 1e-5
__DEF_LMBDA = 0
_log = logging.getLogger('RESCAL')
def rescal_with_random_restarts(X, rank, restarts=10, **kwargs):
"""
Restarts RESCAL multiple time from random starting point and
returns factorization with best fit.
"""
models = []
fits = []
for i in range(restarts):
res = rescal(X, rank, init='random', **kwargs)
models.append(res)
fits.append(res[2])
return models[argmax(fits)]
def rescal(X, rank, **kwargs):
"""
RESCAL
Factors a three-way tensor X such that each frontal slice
X_k = A * R_k * A.T. The frontal slices of a tensor are
N x N matrices that correspond to the adjecency matrices
of the relational graph for a particular relation.
For a full description of the algorithm see:
Maximilian Nickel, Volker Tresp, Hans-Peter-Kriegel,
"A Three-Way Model for Collective Learning on Multi-Relational Data",
ICML 2011, Bellevue, WA, USA
Parameters
----------
X : list
List of frontal slices X_k of the tensor X. The shape of each X_k is ('N', 'N')
rank : int
Rank of the factorization
lmbda : float, optional
Regularization parameter for A and R_k factor matrices. 0 by default
init : string, optional
Initialization method of the factor matrices. 'nvecs' (default)
initializes A based on the eigenvectors of X. 'random' initializes
the factor matrices randomly.
proj : boolean, optional
Whether or not to use the QR decomposition when computing R_k.
True by default
maxIter : int, optional
Maximium number of iterations of the ALS algorithm. 500 by default.
conv : float, optional
Stop when residual of factorization is less than conv. 1e-5 by default
Returns
-------
A : ndarray
array of shape ('N', 'rank') corresponding to the factor matrix A
R : list
list of 'M' arrays of shape ('rank', 'rank') corresponding to the factor matrices R_k
f : float
function value of the factorization
iter : int
number of iterations until convergence
exectimes : ndarray
execution times to compute the updates in each iteration
"""
# init options
ainit = kwargs.pop('init', __DEF_INIT)
proj = kwargs.pop('proj', __DEF_PROJ)
maxIter = kwargs.pop('maxIter', __DEF_MAXITER)
conv = kwargs.pop('conv', __DEF_CONV)
lmbda = kwargs.pop('lmbda', __DEF_LMBDA)
if not len(kwargs) == 0:
raise ValueError( 'Unknown keywords (%s)' % (kwargs.keys()) )
sz = X[0].shape
dtype = X[0].dtype
n = sz[0]
k = len(X)
_log.debug('[Config] rank: %d | maxIter: %d | conv: %7.1e | lmbda: %7.1e' % (rank,
maxIter, conv, lmbda))
_log.debug('[Config] dtype: %s' % dtype)
# precompute norms of X
normX = [norm(M)**2 for M in X]
Xflat = [M.flatten() for M in X]
sumNormX = sum(normX)
# initialize A
if ainit == 'random':
A = array(rand(n, rank), dtype=dtype)
elif ainit == 'nvecs':
S = zeros((n, n), dtype=dtype)
T = zeros((n, n), dtype=dtype)
for i in range(k):
T = X[i]
S = S + T + T.T
evals, A = eigh(S,eigvals=(n-rank,n-1))
else :
raise 'Unknown init option ("%s")' % ainit
# initialize R
if proj:
Q, A2 = qr(A)
X2 = __projectSlices(X, Q)
R = __updateR(X2, A2, lmbda)
else :
R = __updateR(X, A, lmbda)
# compute factorization
fit = fitchange = fitold = f = 0
exectimes = []
ARAt = zeros((n,n), dtype=dtype)
for iter in xrange(maxIter):
tic = time.clock()
fitold = fit
A = __updateA(X, A, R, lmbda)
if proj:
Q, A2 = qr(A)
X2 = __projectSlices(X, Q)
R = __updateR(X2, A2, lmbda)
else :
R = __updateR(X, A, lmbda)
# compute fit value
f = lmbda*(norm(A)**2)
for i in range(k):
ARAt = dot(A, dot(R[i], A.T))
f += normX[i] + norm(ARAt)**2 - 2*dot(Xflat[i], ARAt.flatten()) + lmbda*(R[i].flatten()**2).sum()
f *= 0.5
fit = 1 - f / sumNormX
fitchange = abs(fitold - fit)
toc = time.clock()
exectimes.append( toc - tic )
_log.debug('[%3d] fit: %.5f | delta: %7.1e | secs: %.5f' % (iter,
fit, fitchange, exectimes[-1]))
if iter > 1 and fitchange < conv:
break
return A, R, f, iter+1, array(exectimes)
def __updateA(X, A, R, lmbda):
n, rank = A.shape
F = zeros((n, rank), dtype=X[0].dtype)
E = zeros((rank, rank), dtype=X[0].dtype)
AtA = dot(A.T,A)
for i in range(len(X)):
F += dot(X[i], dot(A, R[i].T)) + dot(X[i].T, dot(A, R[i]))
E += dot(R[i], dot(AtA, R[i].T)) + dot(R[i].T, dot(AtA, R[i]))
A = dot(F, inv(lmbda * eye(rank) + E))
return A
def __updateR(X, A, lmbda):
r = A.shape[1]
R = []
At = A.T
if lmbda == 0:
ainv = dot(pinv(dot(At, A)), At)
for i in range(len(X)):
R.append( dot(ainv, dot(X[i], ainv.T)) )
else :
AtA = dot(At, A)
tmp = inv(kron(AtA, AtA) + lmbda * eye(r**2))
for i in range(len(X)):
AtXA = dot(At, dot(X[i], A))
R.append( dot(AtXA.flatten(), tmp).reshape(r, r) )
return R
def __projectSlices(X, Q):
q = Q.shape[1]
X2 = []
for i in range(len(X)):
X2.append( dot(Q.T, dot(X[i], Q)) )
return X2
It's boring to paste such a long code but there is no other way to figure out my problems. I'm sorry about this.
I import this module and pass them arguments according to the author's website:
import pickle, sys
from rescal import rescal
rank = sys.argv[1]
X = pickle.load('us-presidents.pickle')
A, R, f, iter, exectimes = rescal(X, rank, lmbda=1.0)
The dataset us-presidents.rdf can be found here.
My questions are:
According to the code note, the tensor X is a list. I don't quite understand this, how do I relate a list to a tensor in Python? Can I understand tensor = list in Python?
Should I convert RDF format to a triple(subject, predicate, object) format first? I'm not sure of the data structure of X. How do I assignment values to X by hand?
Then, how to run it?
I paste the author's code without his authorization, is it an act of infringement? if so, I am so sorry and I will delete it soon.
The problems may be a little bored, but these are important to me. Any help would be greatly appreciated.
[1] Maximilian Nickel, Volker Tresp, Hans-Peter Kriegel,
A Three-Way Model for Collective Learning on Multi-Relational Data,
in Proceedings of the 28th International Conference on Machine Learning, 2011 , Bellevue, WA, USA
To answer Q2: you need to transform the RDF and save it before you can load it from the file 'us-presidents.pickle'. The author of that code probably did that once because the Python native pickle format loads faster. As the pickle format includes the datatype of the data, it is possible that X is some numpy class instance and you would need either an example pickle file as used by this code, or some code doing the pickle.dump to figure out how to convert from RDF to this particular pickle file as rescal expects it.
So this might answer Q1: the tensor consists of a list of elements. From the code you can see that the X parameter to rescal has a length (k = len(X) ) and can be indexed (T = X[i]). So it elements are used as a list (even if it might be some other datatype, that just behaves as such.
As an aside: If you are not familiar with Python and are just interested in the result of the computation, you might get more help contacting the author of the software.
According to the code note, the tensor X is a list. I don't quite understand this, how do I relate a list to a tensor in Python? Can I
understand tensor = list in Python?
Not necessarily but the author of the code has decided to represent the tensor data as a list data structure. As the comments indicate, the list X contains:
List of frontal slices X_k of the tensor X. The shape of each X_k is ('N', 'N')
That means the tensor is repesented as a list of tuples: [(N, N), ..., (N, N)].
I'm not sure of the data structure of X. How do I assignment values to X by hand?
Now that we now the data structure of X, we can assign values to it using assignment. The following will assign the tuple (1, 3) to the first position in the list X (as the first position is at index 0, the second at position 1, et cetera):
X[0] = (1, 3)
Similarly, the following will assign the tuple (2, 4) to the second position:
X[1] = (2, 4)