# The size of tensor a (2) must match the size of tensor b (3) while defining cross entropy function

S

#### Steamer

##### Guest
Steamer Asks: The size of tensor a (2) must match the size of tensor b (3) while defining cross entropy function
I am using the following snippet for defining cross entropy function

Code:
def my_cross_entropy(p, y):
"""
Args:
p: torch tensor with size of (N, C),
y (int): torch tensor with size of (N), the values range from 0 to C-1

Return:
loss: the cross_entropy of predicted values p and target y.
"""
loss = None
loss = -np.sum(y*np.log(p))
return loss/float(p.shape[0])

I am making the following method call

Code:
batch_size = 2
x = torch.randn(batch_size, 3, 32, 32)
tmp_tensor = torch.randint(3, (batch_size,))
torch_cross_entropy_out = F.cross_entropy(x[::, ::, 0, 0], tmp_tensor)

However, I am getting the following error

Code:
RuntimeError: The size of tensor a (2) must match the size of tensor b (3) at
non-singleton dimension 0

Can anyone help with this?

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#### How to perform feature engineering on columns with multiple categories?

Prateek Coder Asks: How to perform feature engineering on columns with multiple categories?
I want to perform feature engineering on a data that mostly contains textual data and lot of columns with multiple categories like Supervisor, location code, employee class, business unit, job title, etc. What should be the best approach to do this? I have binary encoded the column with 2-3 categories or performed one hot encoding, but these columns contains different categories from 8 to 150. Different location will play major role in prediction, so will the employee class or business unit etc and I believe they will be very important for my model.

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#### [Solved] Create Buffer but keep original Attribute table

J.Beer Asks: Create Buffer but keep original Attribute table
I want to create a buffer around a number of features. It is important for me to keep the attributes of the features in the new buffer.shp layer. When I try to write a new VectorLayer with a QgsVectorFileWriter, I do get the buffered polygons and also the fields of the original layer get copied over, however the attribute table stays empty.

How can I copy over the attributes of the original layer to the buffer layer?

As you can see I already tried to load the attributes using the . attributes() method on the QgsFeature object and setting the attributes. However I feel I am missing something here.

Code:
for f in feats:
attributes = f.attributes()
f.setAttributes(attributes)

See full code below:

Code:
path = '/Volumes/VAW_ETH-Z/M.Thesis/data/geospatial/'

sgi_filtered = QgsVectorLayer(path + 'sgi_filtered.shp')

## Create buffer around filtered sgi
outFn = path + 'gl_buffer3.shp'
bufferDist = 50 # in map units

fields = sgi_filtered.fields()
feats = sgi_filtered.getFeatures()

writer = QgsVectorFileWriter(
outFn,
'utf-8',
fields,
QgsWkbTypes.Polygon,
sgi_filtered.sourceCrs(), # loads the reference system of the source feature
'ESRI Shapefile')

for f in feats:
attributes = f.attributes()
f.setAttributes(attributes)
geom = f.geometry()
buff = geom.buffer(bufferDist, 5)
f.setGeometry(buff)

writer.addFeature(f)

Screenshot of the Attribute Table below (it shows the fields but no attributes):

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#### [Solved] Saving a Project Folder as a template (QGIS 3.28)

• Ben Sear
• Geography
• Replies: 0
Ben Sear Asks: Saving a Project Folder as a template (QGIS 3.28)
I am looking to create a template for easy layout integration for new QGIS project files, but when saving the project file to a template using save to > template, the file does not appear in the project templates as a template file, but just a new project file. I read that you need to restart GIS once you saved a project file as a template but restarting does not change the project file to being a template. Has anyone experienced saving template before and have any suggestions? Thank you.

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#### Method of cooling a gas without liquid evaporation for Sci-fi novel

• user2765977
• Chemistry
• Replies: 0
user2765977 Asks: Method of cooling a gas without liquid evaporation for Sci-fi novel
sorry for the amateur question here.

I know little about science, but regardless am writing a fictional amateur novel that deals with a lot of chemistry stuff, especially gasses.

My protagonist needs a clever method of cooling a gas. He has access to several gasses (CO2, O2, H2, nitrogen) in varying temperatures (70c and up). Unfortunately no water (at least not more than a few bottles). The atmosphere he is working under (planet atmosphere) is in the range of 150c and at a pressure of roughly 150kpa in the novel.

He also has access to sci-fi style manufacturing machines that can simply 'make' parts, like pipes, valves, pumps, that sort of thing.

I am looking for a clever way or really any way at all, that he can use to cool down some O2 to at least under 40c in such a situation.

My amateur reading online suggests the main way, is via liquid evaporation, but I doubt that would work in the above scenario easily.

Does anybody have any ideas? Is this even possible?

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#### Is there a "fundamental problem of thermodynamics"?

• agaminon
• Physics
• Replies: 0
agaminon Asks: Is there a "fundamental problem of thermodynamics"?
The "fundamental problem of mechanics" can be boiled down to finding and solving the equation of motion of a system. Similar statements can be said for quantum mechanics for the Schrödinger equation and for electrodynamics and Maxwell's equations, etc. But is there such a thing for thermodynamics? Is there a formulation that allows for this kind of perspective?

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#### What does this vertical line notation mean?

• Lambda
• Physics
• Replies: 0
Lambda Asks: What does this vertical line notation mean?
Here is the definition of the Noether momentum in my script.

$$I = \left.\frac{\partial L}{\partial \dot{x}} \frac{d x}{d \alpha} \right|_{\alpha=0} = \frac{\partial L}{\partial \dot{x}} = m \dot{x} = p_x.$$ But I don't understand exactly what this vertical line with $\alpha=0$ means, I would have interpreted it as $\frac{dx}{d0}$ now.

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#### Does diffusion of a spring depend on the spring’s equilibrium length?

• Robert S. Pierre
• Physics
• Replies: 0
Robert S. Pierre Asks: Does diffusion of a spring depend on the spring’s equilibrium length?
Say we perform a random walk on $\mathbb{Z}$ with two equal particles that are connected by a spring with spring constant $K$ and equilibrium length $a$, where at each time step each particle has probability $\frac{1}{2}$ of making a step $\pm1$. However, at each step a random number $q\in[0,1]$ is sampled in order to refuse the move if $$q\geq\exp\left(\Delta E\right),$$ where $\Delta E$ is the difference in elastic energy between the two time steps.

Does the diffusion coefficient depend on the equilibrium length $a$?

I know that the diffusion is independent of the spring constant $K$, so we find the same diffusion coefficient for all $K$, namely that of two independent particles. I assume the same holds for the equilibrium length, but is this true and how can I see this?
And furthermore, what role does the initial extension of the spring play? Say the initial length of the spring $x_0$ is not equal to $a$, then the whole system will need a while to equilibrize, as opposed to when we start with $x_0=a$.

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