numpy: calculate average in a certain area - numpy

is there a way for calculating the average within a certain bbox. The difficulty is that the bbox may also contain float values, so that the bounds of the box values must be weighted. The center of each cell has integer values (the edges are x.5).
Sample:
[[ 1., 1., 1.],
[ 1., 1., 1.],
[ 1., 1., 1.]]
bbox = minx: -0.5, miny: -0.5, maxx: 1, maxy: 1
values = 1*1 + 0.5*1 + 0.5*1 + 0.25*2
weights = 1 + 0.5 + 0.5 + 0.25
average = values / weights = 1.1111...
I couldn't figure out how to do this with numpy.average, any ideas / solutions for this problem?
Thank you very much in advance.

Your question is unclear to me but it looks like you want to be formatting an array of weights and pass it to the np.average() function along with the array of data you want to average such as:
import numpy as np
values = np.array([[1, 1, 1],
[2, 2, 2],
[3, 3, 3]])
valueweights = np.array([[1, 1, 1],
[0.5, 0.5, 0.5],
[0.25, 0.25, 0.25]])
average = np.average(values, weights=valueweights)

Related

How to write a custom kernel in GPflow for the covariance matrix RBF plus noise only on the main block diagonal?

The required covariance matrix is
where t is 1D time and k={0, 1}
A sample from the kernel should look like:
with the orange sequence corresponding to k=0, and the blue one to k=1.
It sounds like you're looking for a kernel for different discrete outputs. You can achieve this in GPflow for example with the Coregion kernel, for which there is a tutorial notebook.
To construct a coregionalization kernel that is block-diagonal (all off-diagonal entries are zero), you can set rank=0. Note that you need to explicitly specify which kernel should act on which dimensions:
import gpflow
k_time = gpflow.kernels.SquaredExponential(active_dims=[0])
k_coreg = gpflow.kernels.Coregion(output_dim=2, rank=0, active_dims=[1])
You can combine them with * as in the notebook, or with + as specified in the question:
k = k_time + k_coreg
You can see that the k_coreg term is block-diagonal as you specified: Evaluating
test_inputs = np.array([
[0.1, 0.0],
[0.5, 0.0],
[0.7, 1.0],
[0.1, 1.0],
])
k_coreg(test_inputs)
returns
<tf.Tensor: shape=(4, 4), dtype=float64, numpy=
array([[1., 1., 0., 0.],
[1., 1., 0., 0.],
[0., 0., 1., 1.],
[0., 0., 1., 1.]])>
And you can get samples as in the graph in the question by running
import numpy as np
num_inputs = 51
num_outputs = 2
X = np.linspace(0, 5, num_inputs)
Q = np.arange(num_outputs)
XX, QQ = np.meshgrid(X, Q, indexing='ij')
pts = np.c_[XX.flatten(), QQ.flatten()]
K = k(pts)
L = np.linalg.cholesky(K + 1e-8 * np.eye(len(K)))
num_samples = 3
v = np.random.randn(len(L), num_samples)
f = L # v
import matplotlib.pyplot as plt
for i in range(num_samples):
plt.plot(X, f[:, i].reshape(num_inputs, num_outputs))
In GPflow, you can construct this kernel using a sum kernel consisting of a Squared Exponential (RBF) and a White kernel.
import gpflow
kernel = gpflow.kernels.SquaredExponential() + gpflow.kernels.White()

How to shift a tensor like pandas.shift in tensorflow / keras? (Without shift the last row to first row, like tf.roll)

I want to shift a tensor in a given axis. It's easy to do this in pandas or numpy. Like this:
import numpy as np
import pandas as pd
data = np.arange(0, 6).reshape(-1, 2)
pd.DataFrame(data).shift(1).fillna(0).values
Output is:
array([[0., 0.],
[0., 1.],
[2., 3.]])
But in tensorflow, the closest solution I found is tf.roll. But it shift the last row to the first row. (I don't want that). So I have to use something like
tf.roll + tf.slice(remove the last row) + tf.concat(add tf.zeros to the first row).
It's really ugly.
Is there a better way to handle shift in tensorflow or keras?
Thanks.
I think I find a better way for this problem.
We could use tf.roll, then apply tf.math.multiply to set the first row to zeros.
Sample code is as follows:
Original tensor:
A = tf.cast(tf.reshape(tf.range(27), (-1, 3, 3)), dtype=tf.float32)
A
Output:
<tf.Tensor: id=117, shape=(3, 3, 3), dtype=float32, numpy=
array([[[ 0., 1., 2.],
[ 3., 4., 5.],
[ 6., 7., 8.]],
[[ 9., 10., 11.],
[12., 13., 14.],
[15., 16., 17.]],
[[18., 19., 20.],
[21., 22., 23.],
[24., 25., 26.]]], dtype=float32)>
Shift (like pd.shift):
B = tf.concat((tf.zeros((1, 3)), tf.ones((2, 3))), axis=0)
C = tf.expand_dims(B, axis=0)
tf.math.multiply(tf.roll(A, 1, axis=1), C)
Output:
<tf.Tensor: id=128, shape=(3, 3, 3), dtype=float32, numpy=
array([[[ 0., 0., 0.],
[ 0., 1., 2.],
[ 3., 4., 5.]],
[[ 0., 0., 0.],
[ 9., 10., 11.],
[12., 13., 14.]],
[[ 0., 0., 0.],
[18., 19., 20.],
[21., 22., 23.]]], dtype=float32)>
Try this:
import tensorflow as tf
input = tf.constant([[0, 1, 3], [4, 5, 6], [7, 8, 9]])
shifted_0dim = input[1:]
shifted_1dim = input[:, 1:]
shifted2 = input[2:]
Generalizing the accepted answer to arbitrary tensor shapes, desired shift, and axis to shift:
import tensorflow as tf
def tf_shift(tensor, shift=1, axis=0):
dim = len(tensor.shape)
if axis > dim:
raise ValueError(
f'Value of axis ({axis}) must be <= number of tensor axes ({dim})'
)
mask_dim = dim - axis
mask_shape = tensor.shape[-mask_dim:]
zero_dim = min(shift, mask_shape[0])
mask = tf.concat(
[tf.zeros(tf.TensorShape(zero_dim) + mask_shape[1:]),
tf.ones(tf.TensorShape(mask_shape[0] - zero_dim) + mask_shape[1:])],
axis=0
)
for i in range(dim - mask_dim):
mask = tf.expand_dims(mask, axis=0)
return tf.multiply(
tf.roll(tensor, shift, axis),
mask
)
EDIT:
This code above doesn't allow for negative shift values, and is pretty slow. Here is a more efficient version utilizing tf.roll and tf.concat without creating a mask and multiplying the tensor of interest by it.
import tensorflow as tf
def tf_shift(values: tf.Tensor, shift: int = 1, axis: int = 0):
pad = tf.zeros([val if i != axis else abs(shift) for i, val in enumerate(values.shape)],
dtype=values.dtype)
size = [-1 if i != axis else val - abs(shift) for i, val in enumerate(values.shape)]
if shift > 0:
shifted = tf.concat(
[pad, tf.slice(values, [0] * len(values.shape), size)],
axis=axis
)
elif shift < 0:
shifted = tf.concat(
[tf.slice(values, [0 if i != axis else abs(shift) for i, _ in enumerate(values.shape)], size), pad],
axis=axis
)
else:
shifted = values
return shifted
Assuming a 2d tensor, this function should mimic a Dataframe shift:
def shift_tensor(tensor, periods, fill_value):
num_row = len(tensor)
num_col = len(tensor[0])
pad = tf.fill([periods, num_col], fill_value)
if periods > 0:
shifted_tensor = tf.concat((pad, tensor[:(num_row - periods), :]), axis=0)
else:
shifted_tensor = tf.concat((tensor[:(num_row - periods), :], pad), axis=0)
return shifted_tensor

How to train model with two kids functions for object detection?

I'm trying to implement the model described by Professor Andrew Ng for object detection (explanation starts at 10:00).
He describes the first element of the output vector as the probability that an object was detected, followed by the coordinates of the bounding box of the object matched (when one is matched). The last part of the output vector is a softmax of all the classes your model knows.
As he explains it, using a simple squared error for the case when there is a detection is fine, and just the squares difference of y^[0] - y[0]. I get that this is a naive approach. I'm just wanting to implement this for the learning experience.
My questions
How do I implement this conditional loss in tensorflow?
How do I handle this conditional about y^[0] when dealing with a batch.
How do I implement this conditional loss in tensorflow?
You can convert the loss function to:
Error = mask[0]*(y^[0]-y[0])**2 + mask[1]*(y^[1]-y[1])**2 ... mask[n]*(y^[n]-y[n])**2),
where mask = [1, 1,...1] for y[0] = 1 and [1, 0, ...0] for y[0] = 0
How do I handle this conditional about y^[0] when dealing with a
batch.
For a batch, you can construct the mask on the fly like:
mask = tf.concat([tf.ones((tf.shape(y)[0],1)),y[:,0][...,None]*y[:,1:]], axis=1)
Code:
y_hat_n = np.array([[3, 3, 3, 3], [3,3,3,3]])
y_1 = np.array([[1, 1, 1, 1], [1,1,1,1]])
y_0 = np.array([[0, 1, 1, 1], [0,1,1,1]])
y = tf.placeholder(tf.float32,[None, 4])
y_hat = tf.placeholder(tf.float32,[None, 4])
mask = tf.concat([tf.ones((tf.shape(y)[0],1)),y[:,0][...,None]*y[:,1:]], axis=1)
error = tf.losses.mean_squared_error(mask*y, mask*y_hat)
with tf.Session() as sess:
print(sess.run([mask,error], {y:y_0, y_hat:y_hat_n}))
print(sess.run([mask,error], {y:y_1, y_hat:y_hat_n}))
# Mask and error
#[array([[1., 0., 0., 0.],
# [1., 0., 0., 0.]], dtype=float32), 2.25]
#[array([[1., 1., 1., 1.],
# [1., 1., 1., 1.]], dtype=float32), 4.0]

Argmax on a tensor and ceiling in Tensorflow

Suppose I have a tensor in Tensorflow that its values are like:
A = [[0.7, 0.2, 0.1],[0.1, 0.4, 0.5]]
How can I change this tensor into the following:
B = [[1, 0, 0],[0, 0, 1]]
In other words I want to just keep the maximum and replace it with 1.
Any help would be appreciated.
I think that you can solve it with a one-liner:
import tensorflow as tf
import numpy as np
x_data = [[0.7, 0.2, 0.1],[0.1, 0.4, 0.5]]
# I am using hard-coded dimensions for simplicity
x = tf.placeholder(dtype=tf.float32, name="x", shape=(2,3))
session = tf.InteractiveSession()
session.run(tf.one_hot(tf.argmax(x, 1), 3), {x: x_data})
The result is the one that you expect:
Out[6]:
array([[ 1., 0., 0.],
[ 0., 0., 1.]], dtype=float32)

numpy - multiply each element in array by a scaling factor

I have a numpy array of values, and a list of scaling factors which I want to scale each value in the array by, down each column
values = [[ 0, 1, 2, 3 ],
[ 1, 1, 4, 3 ],
[ 2, 1, 6, 3 ],
[ 3, 1, 8, 3 ]]
ls_alloc = [ 0.1, 0.4, 0.3, 0.2]
# convert values into numpy array
import numpy as np
na_values = np.array(values, dtype=float)
Edit: To clarify:
na_values can is a 2-dimensional array of stock cumulative returns (ie: normalised to day 1), where each row represents a date, and each column a stock. The data is returned as an array for each date.
I want to now scale each stock's cumulative return by its allocation in the portfolio. So for each date (ie: each row of ndarray values, apply the respective element from ls_alloc to the array element-wise)
# scale each value by its allocation
na_components = [ ls_alloc[i] * na_values[:,i] for i in range(len(ls_alloc)) ]
This does what I want, but I can't help but feel there must be a way to have numpy do this for me automatically?
That is, I feel:
na_components = [ ls_alloc[i] * na_values[:,i] for i in range(len(ls_alloc)) ]
# display na_components
na_components
[array([ 0. , 0.1, 0.2, 0.3]), \
array([ 0.4, 0.4, 0.4, 0.4]), \
array([ 0.6, 1.2, 1.8, 2.4]), \
array([ 0.6, 0.6, 0.6, 0.6])]
should be able to be expressed as something like:
tmp = np.multiply(na_values, ls_alloc)
# display tmp
tmp
array([[ 0. , 0.4, 0.6, 0.6],
[ 0.1, 0.4, 1.2, 0.6],
[ 0.2, 0.4, 1.8, 0.6],
[ 0.3, 0.4, 2.4, 0.6]])
Is there a numpy function which will achieve what I want elegantly and succinctly?
Edit:
I see that my first solution has transposed my data, such that I am returned a list of ndarrays. na_components[0] now gives an ndarray of the stock values for the first stock, 1 element per date.
The next step that I perform with na_components is to calculate the total cumulative return for the portfolio by summing each individual component
na_pfo_cum_ret = np.sum(na_components, axis=0)
This works with the list of individual stock return ndarrays.
That order seems a little odd to me, but IIUC, all you need to do is to transpose the result of multiplying na_values by array(ls_alloc):
>>> v
array([[ 0., 1., 2., 3.],
[ 1., 1., 4., 3.],
[ 2., 1., 6., 3.],
[ 3., 1., 8., 3.]])
>>> a
array([ 0.1, 0.4, 0.3, 0.2])
>>> (v*a).T
array([[ 0. , 0.1, 0.2, 0.3],
[ 0.4, 0.4, 0.4, 0.4],
[ 0.6, 1.2, 1.8, 2.4],
[ 0.6, 0.6, 0.6, 0.6]])
It's not completely clear to me what you want to do, but the answer is probably in Broadcasting rules. I think you want:
values = np.array( [[ 0, 1, 2, 3 ],
[ 1, 1, 4, 3 ],
[ 2, 1, 6, 3 ],
[ 3, 1, 8, 3 ]] )
ls_alloc = np.array([ 0.1, 0.4, 0.3, 0.2])
and either:
na_components = values * ls_alloc
or:
na_components = values * ls_alloc[:,np.newaxis]