I have a tensor of shape [x, y] and I want to subtract the mean and divide by the standard deviation row-wise (i.e. I want to do it for each row). What is the most efficient way to do this in TensorFlow?
Of course I can loop through rows as follows:
new_tensor = [i - tf.reduce_mean(i) for i in old_tensor]
...to subtract the mean and then do something similar to find the standard deviation and divide by it, but is this the best way to do it in TensorFlow?
The TensorFlow tf.sub() and tf.div() operators support broadcasting, so you don't need to iterate through every row. Let's consider the mean, and leave standard deviation as an exercise:
old_tensor = ... # shape = (x, y)
mean = tf.reduce_mean(old_tensor, 1, keep_dims=True) # shape = (x, 1)
stdev = ... # shape = (x,)
stdev = tf.expand_dims(stdev, 1) # shape = (x, 1)
new_tensor = old_tensor - mean # shape = (x, y)
new_tensor = old_tensor / stdev # shape = (x, y)
The subtraction and division operators implicitly broadcast a tensor of shape (x, 1) along the column dimension to match the shape of the other argument, (x, y). For more details about how broadcasting works, see the NumPy documentation on the topic (TensorFlow implements NumPy broadcasting semantics).
calculate moments along axis 1 (y in your case) and keep dimensions, i.e. shape of mean and var is (len(x), 1)
subtract mean and divide by standard deviation (i.e. square root of variance)
mean, var = tf.nn.moments(old_tensor, [1], keep_dims=True)
new_tensor = tf.div(tf.subtract(old_tensor, mean), tf.sqrt(var))
Related
I have an idea for a tensor operation that would not be difficult to implement via iteration, with batch size one. However I would like to parallelize it as much as possible.
I have two tensors with shape (n, 5) called X and Y. X is actually supposed to represent 5 one-dimensional tensors with shape (n, 1): (x_1, ..., x_n). Ditto for Y.
I would like to compute a tensor with shape (n, 25) where each column represents the output of the tensor operation f(x_i, y_j), where f is fixed for all 1 <= i, j <= 5. The operation f has output shape (n, 1), just like x_i and y_i.
I feel it is important to clarify that f is essentially a fully-connected layer from the concatenated [...x_i, ...y_i] tensor with shape (1, 10), to an output layer with shape (1,5).
Again, it is easy to see how to do this manually with iteration and slicing. However this is probably very slow. Performing this operation in batches, where the tensors X, Y now have shape (n, 5, batch_size) is also desirable, particularly for mini-batch gradient descent.
It is difficult to really articulate here why I desire to create this network; I feel it is suited for my domain of 'itemized tabular data' and cuts down significantly on the number of weights per operation, compared to a fully connected network.
Is this possible using tensorflow? Certainly not using just keras.
Below is an example in numpy per AloneTogether's request
import numpy as np
features = 16
batch_size = 256
X_batch = np.random.random((features, 5, batch_size))
Y_batch = np.random.random((features, 5, batch_size))
# one tensor operation to reduce weights in this custom 'layer'
f = np.random.random((features, 2 * features))
for b in range(batch_size):
X = X_batch[:, :, b]
Y = Y_batch[:, :, b]
for i in range(5):
x_i = X[:, i:i+1]
for j in range(5):
y_j = Y[:, j:j+1]
x_i_y_j = np.concatenate([x_i, y_j], axis=0)
# f(x_i, y_j)
# implemented by a fully-connected layer
f_i_j = np.matmul(f, x_i_y_j)
All operations you need (concatenation and matrix multiplication) can be batched.
Difficult part here is, that you want to concatenate features of all items in X with features of all items in Y (all combinations).
My recommended solution is to expand the dimensions of X to [batch, features, 5, 1], expand dimensions of Y to [batch, features, 1, 5]
Than tf.repeat() both tensors so their shapes become [batch, features, 5, 5].
Now you can concatenate X and Y. You will have a tensor of shape [batch, 2*features, 5, 5]. Observe that this way all combinations are built.
Next step is matrix multiplication. tf.matmul() can also do batch matrix multiplication, but I use here tf.einsum() because I want more control over which dimensions are considered as batch.
Full code:
import tensorflow as tf
import numpy as np
batch_size=3
features=6
items=5
x = np.random.uniform(size=[batch_size,features,items])
y = np.random.uniform(size=[batch_size,features,items])
f = np.random.uniform(size=[2*features,features])
x_reps= tf.repeat(x[:,:,:,tf.newaxis], items, axis=3)
y_reps= tf.repeat(y[:,:,tf.newaxis,:], items, axis=2)
xy_conc = tf.concat([x_reps,y_reps], axis=1)
f_i_j = tf.einsum("bfij, fg->bgij", xy_conc,f)
f_i_j = tf.reshape(f_i_j , [batch_size,features,items*items])
I am trying to reconstruct the following matrix of shape (256 x 256 x 2) with SVD components as
U.shape = (256, 256, 256)
s.shape = (256, 2)
vh.shape = (256, 2, 2)
I have already tried methods from documentation of numpy and scipy to reconstruct the original matrix but failed multiple times, I think it maybe 3D matrix has a different way of reconstruction.
I am using numpy.linalg.svd for decompostion.
From np.linalg.svd's documentation:
"... If a has more than two dimensions, then broadcasting rules apply, as explained in :ref:routines.linalg-broadcasting. This means that SVD is
working in "stacked" mode: it iterates over all indices of the first
a.ndim - 2 dimensions and for each combination SVD is applied to the
last two indices."
This means that you only need to handle the s matrix (or tensor in general case) to obtain the right tensor. More precisely, what you need to do is pad s appropriately and then take only the first 2 columns (or generally, the number of rows of vh which should be equal to the number of columns of the returned s).
Here is a working code with example for your case:
import numpy as np
mat = np.random.randn(256, 256, 2) # Your matrix of dim 256 x 256 x2
u, s, vh = np.linalg.svd(mat) # Get the decomposition
# Pad the singular values' arrays, obtain diagonal matrix and take only first 2 columns:
s_rep = np.apply_along_axis(lambda _s: np.diag(np.pad(_s, (0, u.shape[1]-_s.shape[0])))[:, :_s.shape[0]], 1, s)
mat_reconstructed = u # s_rep # vh
mat_reconstructed equals to mat up to precision error.
I'm using the Kobe Bryant Dataset.
I wish to predict the shot_made_flag with KnnRegressor.
I've used game_date to extract year and month features:
# covert season to years
kobe_data_encoded['season'] = kobe_data_encoded['season'].apply(lambda x: int(re.compile('(\d+)-').findall(x)[0]))
# add year and month using game_date
kobe_data_encoded['year'] = kobe_data_encoded['game_date'].apply(lambda x: int(re.compile('(\d{4})').findall(x)[0]))
kobe_data_encoded['month'] = kobe_data_encoded['game_date'].apply(lambda x: int(re.compile('-(\d+)-').findall(x)[0]))
kobe_data_encoded = kobe_data_encoded.drop(columns=['game_date'])
and I wish to use season, year, month features to give them more weight in the distance function so events with closer date to the current event will be closer neighbors but still maintain reasonable distances to potential other datapoints, so for example I don't wish an event withing the same day would be the closest neighbor just because of the date features but it'll take into account the other features such as shot_range etc..
To give it more weight I've tried to use metric argument with custom distance function but the arguments of the function are just numpy array without column information of pandas so I'm not sure what I can do and how to implement what I'm trying to do.
EDIT:
Using larger weights for date features to find the optimal k with cv of 10 running on k from [1, 100]:
from IPython.display import display
from sklearn.neighbors import KNeighborsClassifier
from sklearn.model_selection import StratifiedKFold
from sklearn.model_selection import cross_val_score
# scaling
min_max_scaler = preprocessing.MinMaxScaler()
scaled_features_df = kobe_data_encoded.copy()
column_names = ['loc_x', 'loc_y', 'minutes_remaining', 'period',
'seconds_remaining', 'shot_distance', 'shot_type', 'shot_zone_range']
scaled_features = min_max_scaler.fit_transform(scaled_features_df[column_names])
scaled_features_df[column_names] = scaled_features
not_classified_df = scaled_features_df[scaled_features_df['shot_made_flag'].isnull()]
classified_df = scaled_features_df[scaled_features_df['shot_made_flag'].notnull()]
X = classified_df.drop(columns=['shot_made_flag'])
y = classified_df['shot_made_flag']
cv = StratifiedKFold(n_splits=10, shuffle=True)
neighbors = [x for x in range(1, 100)]
cv_scores = []
weight = np.ones((X.shape[1],))
weight[[X.columns.get_loc("season"),
X.columns.get_loc("year"),
X.columns.get_loc("month")
]] = 5
weight = weight/weight.sum() #Normalize weights
def my_distance(x, y):
dist = ((x-y)**2)
return np.dot(dist, weight)
for k in neighbors:
print('k: ', k)
knn = KNeighborsClassifier(n_neighbors=k, metric=my_distance)
cv_scores.append(np.mean(cross_val_score(knn, X, y, cv=cv, scoring='roc_auc')))
#optimal K
optimal_k_index = cv_scores.index(min(cv_scores))
optimal_k = neighbors[optimal_k_index]
print('best k: ', optimal_k)
plt.plot(neighbors, cv_scores)
plt.xlabel('Number of Neighbors K')
plt.ylabel('ROC AUC')
plt.show()
Runs really slow, any idea on how to make it faster?
The idea of the weighted features is to find neighbors more close to the data point date to avoid data leakage and cv for finding optimal k.
First, you have to prepare a numpy 1D weight array, specifying weight for each feature. You could do something like:
weight = np.ones((M,)) # M is no of features
weight[[1,7,10]] = 2 # Increase weight of 1st,7th and 10th features
weight = weight/weight.sum() #Normalize weights
You can use kobe_data_encoded.columns to find indexes of season, year, month features in your dataframe to replace 2nd line above.
Now define a distance function, which by guideline have to take two 1D numpy array.
def my_dist(x,y):
global weight #1D array, same shape as x or y
dist = ((x-y)**2) #1D array, same shape as x or y
return np.dot(dist,weight) # a scalar float
And initialize KNeighborsRegressor as:
knn = KNeighborsRegressor(metric=my_dist)
EDIT:
To make things efficient, you can precompute distance matrix, and reuse it in KNN. This should bring in significant speedup by reducing calls to my_dist, since this non-vectorized custom python distance function is quite slow. So now -
dist = np.zeros((len(X),len(X))) #Computing NXN distance matrix
for i in range(len(X)): # You can halve this by using the fact that dist[i,j] = dist[j,i]
for j in range(len(X)):
dist[i,j] = my_dist(X[i],X[j])
for k in neighbors:
print('k: ', k)
knn = KNeighborsClassifier(n_neighbors=k, metric='precomputed') #Note: metric='precomputed'
cv_scores.append(np.mean(cross_val_score(knn, dist, y, cv=cv, scoring='roc_auc'))) #Note: passing dist instead of X
I couldn't test it, so let me know if something isn't alright.
Just add on Shihab's answer regarding distance computation. Can use scipy pdist as suggested in this post, which is faster and more efficient.
from scipy.spatial.distance import pdist, minkowski, squareform
# create the custom weight array
weight = ...
# calculate pairwise distances, using Minkowski norm with custom weights
distances = pdist(X, minkowski, 2, weight)
# reformat the result as a square matrix
distances_as_2d_matrix = squareform(distances)
in addition to the MSE of y_true and y_predict i would like to use the second derivative of y_true in the cost function, because my model is currently very dynamic. Suppose I have y_predicted (256, 100, 1). The first dimension corresponds to the samples (delta_t between each sample is 0.1s). Now I would like to differentiate via the first dimension, i.e.
diff(diff(y_predicted[1, :, 1]))/delta_t**2
for each row (0-dim) in y_predictied.
Note, I only want to use y_predicted and delta_t to differentiate
Thank you very much,
Max
To calculate the second order derivative you could use tf.hessians as follow:
x = tf.Variable([7])
x2 = x * x
d2x2 = tf.hessians(x2, x)
Evaluating d2x2 yields:
[array([[2]], dtype=int32)]
In your case, you could do
loss += lam_l1 * tf.hessians(y_pred, xs)
where xs are the tensors with respect to which you would like to differentiate.
If you wish to use Keras directly, you can chain twice keras.backend.gradients(loss, variables), there is no Keras equivalent of tf.hessians.
I am looking for algorithm to solve the following problem :
I have two sets of vectors, and I want to find the matrix that best approximate the transformation from the input vectors to the output vectors.
vectors are 3x1, so matrix is 3x3.
This is the general problem. My particular problem is I have a set of RGB colors, and another set that contains the desired color. I am trying to find an RGB to RGB transformation that would give me colors closer to the desired ones.
There is correspondence between the input and output vectors, so computing an error function that should be minimized is the easy part. But how can I minimize this function ?
This is a classic linear algebra problem, the key phrase to search on is "multiple linear regression".
I've had to code some variation of this many times over the years. For example, code to calibrate a digitizer tablet or stylus touch-screen uses the same math.
Here's the math:
Let p be an input vector and q the corresponding output vector.
The transformation you want is a 3x3 matrix; call it A.
For a single input and output vector p and q, there is an error vector e
e = q - A x p
The square of the magnitude of the error is a scalar value:
eT x e = (q - A x p)T x (q - A x p)
(where the T operator is transpose).
What you really want to minimize is the sum of e values over the sets:
E = sum (e)
This minimum satisfies the matrix equation D = 0 where
D(i,j) = the partial derivative of E with respect to A(i,j)
Say you have N input and output vectors.
Your set of input 3-vectors is a 3xN matrix; call this matrix P.
The ith column of P is the ith input vector.
So is the set of output 3-vectors; call this matrix Q.
When you grind thru all of the algebra, the solution is
A = Q x PT x (P x PT) ^-1
(where ^-1 is the inverse operator -- sorry about no superscripts or subscripts)
Here's the algorithm:
Create the 3xN matrix P from the set of input vectors.
Create the 3xN matrix Q from the set of output vectors.
Matrix Multiply R = P x transpose (P)
Compute the inverseof R
Matrix Multiply A = Q x transpose(P) x inverse (R)
using the matrix multiplication and matrix inversion routines of your linear algebra library of choice.
However, a 3x3 affine transform matrix is capable of scaling and rotating the input vectors, but not doing any translation! This might not be general enough for your problem. It's usually a good idea to append a "1" on the end of each of the 3-vectors to make then a 4-vector, and look for the best 3x4 transform matrix that minimizes the error. This can't hurt; it can only lead to a better fit of the data.
You don't specify a language, but here's how I would approach the problem in Matlab.
v1 is a 3xn matrix, containing your input colors in vertical vectors
v2 is also a 3xn matrix containing your output colors
You want to solve the system
M*v1 = v2
M = v2*inv(v1)
However, v1 is not directly invertible, since it's not a square matrix. Matlab will solve this automatically with the mrdivide operation (M = v2/v1), where M is the best fit solution.
eg:
>> v1 = rand(3,10);
>> M = rand(3,3);
>> v2 = M * v1;
>> v2/v1 - M
ans =
1.0e-15 *
0.4510 0.4441 -0.5551
0.2220 0.1388 -0.3331
0.4441 0.2220 -0.4441
>> (v2 + randn(size(v2))*0.1)/v1 - M
ans =
0.0598 -0.1961 0.0931
-0.1684 0.0509 0.1465
-0.0931 -0.0009 0.0213
This gives a more language-agnostic solution on how to solve the problem.
Some linear algebra should be enough :
Write the average squared difference between inputs and outputs ( the sum of the squares of each difference between each input and output value ). I assume this as definition of "best approximate"
This is a quadratic function of your 9 unknown matrix coefficients.
To minimize it, derive it with respect to each of them.
You will get a linear system of 9 equations you have to solve to get the solution ( unique or a space variety depending on the input set )
When the difference function is not quadratic, you can do the same but you have to use an iterative method to solve the equation system.
This answer is better for beginners in my opinion:
Have the following scenario:
We don't know the matrix M, but we know the vector In and a corresponding output vector On. n can range from 3 and up.
If we had 3 input vectors and 3 output vectors (for 3x3 matrix), we could precisely compute the coefficients αr;c. This way we would have a fully specified system.
But we have more than 3 vectors and thus we have an overdetermined system of equations.
Let's write down these equations. Say that we have these vectors:
We know, that to get the vector On, we must perform matrix multiplication with vector In.In other words: M · I̅n = O̅n
If we expand this operation, we get (normal equations):
We do not know the alphas, but we know all the rest. In fact, there are 9 unknowns, but 12 equations. This is why the system is overdetermined. There are more equations than unknowns. We will approximate the unknowns using all the equations, and we will use the sum of squares to aggregate more equations into less unknowns.
So we will combine the above equations into a matrix form:
And with some least squares algebra magic (regression), we can solve for b̅:
This is what is happening behind that formula:
Transposing a matrix and multiplying it with its non-transposed part creates a square matrix, reduced to lower dimension ([12x9] · [9x12] = [9x9]).
Inverse of this result allows us to solve for b̅.
Multiplying vector y̅ with transposed x reduces the y̅ vector into lower [1x9] dimension. Then, by multiplying [9x9] inverse with [1x9] vector we solved the system for b̅.
Now, we take the [1x9] result vector and create a matrix from it. This is our approximated transformation matrix.
A python code:
import numpy as np
import numpy.linalg
INPUTS = [[5,6,2],[1,7,3],[2,6,5],[1,7,5]]
OUTPUTS = [[3,7,1],[3,7,1],[3,7,2],[3,7,2]]
def get_mat(inputs, outputs, entry_len):
n_of_vectors = inputs.__len__()
noe = n_of_vectors*entry_len# Number of equations
#We need to construct the input matrix.
#We need to linearize the matrix. SO we will flatten the matrix array such as [a11, a12, a21, a22]
#So for each row we combine the row's variables with each input vector.
X_mat = []
for in_n in range(0, n_of_vectors): #For each input vector
#populate all matrix flattened variables. for 2x2 matrix - 4 variables, for 3x3 - 9 variables and so on.
base = 0
for col_n in range(0, entry_len): #Each original unknown matrix's row must be matched to all entries in the input vector
row = [0 for i in range(0, entry_len ** 2)]
for entry in inputs[in_n]:
row[base] = entry
base+=1
X_mat.append(row)
Y_mat = [item for sublist in outputs for item in sublist]
X_np = np.array(X_mat)
Y_np = np.array([Y_mat]).T
solution = np.dot(np.dot(numpy.linalg.inv(np.dot(X_np.T,X_np)),X_np.T),Y_np)
var_mat = solution.reshape(entry_len, entry_len) #create square matrix
return var_mat
transf_mat = get_mat(INPUTS, OUTPUTS, 3) #3 means 3x3 matrix, and in/out vector size 3
print(transf_mat)
for i in range(0,INPUTS.__len__()):
o = np.dot(transf_mat, np.array([INPUTS[i]]).T)
print(f"{INPUTS[i]} x [M] = {o.T} ({OUTPUTS[i]})")
The output is as such:
[[ 0.13654096 0.35890767 0.09530002]
[ 0.31859558 0.83745124 0.22236671]
[ 0.08322497 -0.0526658 0.4417611 ]]
[5, 6, 2] x [M] = [[3.02675088 7.06241873 0.98365224]] ([3, 7, 1])
[1, 7, 3] x [M] = [[2.93479472 6.84785436 1.03984767]] ([3, 7, 1])
[2, 6, 5] x [M] = [[2.90302805 6.77373212 2.05926064]] ([3, 7, 2])
[1, 7, 5] x [M] = [[3.12539476 7.29258778 1.92336987]] ([3, 7, 2])
You can see, that it took all the specified inputs, got the transformed outputs and matched the outputs to the reference vectors. The results are not precise, since we have an approximation from the overspecified system. If we used INPUT and OUTPUT with only 3 vectors, the result would be exact.