RePr | My implementation of RePr training scheme in PyTorch | Machine Learning library
kandi X-RAY | RePr Summary
kandi X-RAY | RePr Summary
My implementation of RePr training scheme in PyTorch.
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Top functions reviewed by kandi - BETA
- Train the model
- Compute accuracy
- Update the sum
- Perform pruning
- Test the filter sparsity of the weights
- Validate the loss function
- Test if the filter weights are zero
RePr Key Features
RePr Examples and Code Snippets
Lines of Code : 6License : Non-SPDX (Apache License 2.0)
Copydef __repr__(self):
if self.loc.filename:
return '{}:{}:{}'.format(
os.path.split(self.loc.filename)[1], self.loc.lineno,
self.loc.col_offset)
return ':{}:{}'.format(self.loc.lineno, self.loc.col_offset) def __repr__(self):
return 'used: {}'.format(
self.return_used) def __repr__(self):
return '<_Continue(used: {}, var: {})>'.format(self.used,
self.control_var_name) Community Discussions
Trending Discussions on RePr
QUESTION
I would like to read a GRIB file downloaded from server using ecCodes library in Rust. However, my current solution results in segmentation fault. The extracted example, replicating the problem, is below.
I download the file using reqwest crate and get the response as Bytes1 using bytes(). To read the file with ecCodes I need to create a codes_handle using codes_grib_handle_new_from_file()2, which as argument requires *FILE usually get from fopen(). However, I would like to skip IO operations. So I figured I could use libc::fmemopen() to get *FILE from Bytes. But when I pass the *mut FILE from fmemopen() to codes_grib_handle_new_from_file() segmentation fault occurs.
I suspect the issue is when I get from Bytes a *mut c_void required by fmemopen(). I figured I can do this like that:
ANSWER
Answered 2021-Jun-12 at 13:291- Try changing
QUESTION
I tried to get the current word in a tkinter text widget. Here’s my code:
...ANSWER
Answered 2021-Jun-12 at 11:58I hope this answer is what you want since you didn't clear out what exactly you want. Remove the -1c in index_1 parameter to solve the extra space added to word retrieved and extra newlines.
QUESTION
I am trying to perform a sentiment analysis based on product reviews collected from various websites. I've been able to follow along with the below article until it gets to the model coefficient visualization step.
When I run my program, I get the following error:
...ANSWER
Answered 2021-Jun-09 at 19:25You've defined feature_names in terms of the features from a CountVectorizer with the default stop_words=None, but your model in the last bit of code is using a TfidfVectorizer with stop_words='english'. Use instead
QUESTION
I want to set a column default value for my custom column type.
Currently I have this:
...ANSWER
Answered 2021-Jun-09 at 10:20I have solved the problem by adjusting the init method:
QUESTION
I'm trying to code the Python equivalent of this Oracle SQL command:
...ANSWER
Answered 2021-Jun-08 at 06:18I think u could try this
QUESTION
I'm trying to implement a loop for this socket code, but I can't quite seem to wrap my head around it. Is anyone able to explain it for me?
Here's the code
...ANSWER
Answered 2021-Jun-08 at 04:43Do you possible mean this?
QUESTION
I have a wrapper around a C-API:
...ANSWER
Answered 2021-Jun-07 at 22:24Short answer: just cast it to *mut T and pass it to C.
Long answer:
It's best to first understand why casting *const T to *mut T is prone to undefined behaviour.
Rust's memory model ensures that a &mut T will not alias with anything else, so the compiler is free to, say, clobber T entirely and then restore its content, and the programmer could not observe that behaviour. If a &mut T and &T co-exists and point to the same location, undefined behaviour arises because what will happen if you read from &T while compiler clobbers &mut T? Similarly, if you have &T, the compiler assumes no one will modify it (excluding interior mutability through UnsafeCell), and undefined behaviour arise if the memory it points to is modified.
With the background, it's easy to see why *const T to *mut T is dangerous -- you cannot dereference the resulting pointer. If you ever dereference the *mut T, you've obtained a &mut T, and it'll be UB. However, the casting operation itself is safe, and you can safely cast the *mut T back to *const T and dereference it.
This is Rust semantics; on the C-side, the guarantee about T* is very weak. If you hold a T*, the compiler cannot assume there are no sharers. In fact, the compiler cannot even assert that it points to valid address (it could be null or past-the-end pointer). C compiler cannot generate store instructions to the memory location unless the code write to the pointer explicitly.
The weaker meaning of T* in C-side means that it won't violate Rust's assumption about semantics of &T. You can safely cast &T to *mut T and pass it to C, provided that C-side never modifies the memory pointed by the pointer.
Note that you can instruct the C compiler that the pointer won't alias with anything else with T * restrict, but as the C code you mentioned is not strict with const-correctness, it probably does not use restrict as well.
QUESTION
I have two pages:
ArticlePage and TimelinePage.
TimelinePage has an inline panel which is related to the model TimelinePageEntry.
TimelinePage displays all of it's child TimelinePageEntry(s) just fine. However when I try to get all the TimelinePageEntry(s) from ArticlePage, it fails with an exception.
ArticlePage fails with an exception when I try to do:
...ANSWER
Answered 2021-Jun-06 at 16:56This is due to the distinction between Page instances (which just contain the data common to all page types, such as title) and instances of your specific TimelinePage subclass (which provides access to all the methods, fields and relations defined on that class).
self.get_children() returns a queryset of type Page; this is necessary because the query doesn't know in advance what page types will be included in the results. .type(TimelinePage) filters this to just pages of the given type, but this only acts as a filter on the existing queryset - the results will still remain as Page objects, which are missing the timeline_entries relation.
If you rewrite the line
QUESTION
class LoggedError(Exception):
def __init__(self, value):
self.value = value
def __str__(self):
return repr(self.value)
class Member(object):
nextId = 0
def __init__(self):
self.id = Member.nextId
Member.nextId += 1
def getId(self):
return self.id
class Customer(Member):
def __init__(self, email, password):
self.email = email
self.password = password
self.logged = False
def signin(self, email, password):
if (self.email == email and self.password == password):
if (self.logged == True):
raise LoggedError("Already signed in.")
else:
self.logged = True
return True
else:
return False
def signout(self):
if (self.logged == False):
raise LoggedError("Already signed out.")
else:
self.logged = False
def __str__(self):
return str(self.email)
c1 = Customer("abc@abc.com", "12341234")
c2 = Customer("def@def.com", "56785678")
c3 = Customer("hello@world.com", "qwerty")
print("Customer 1 is {}".format(c1))
print("Customer 2 is {}".format(c2))
print("Customer 3 is {}".format(c3))
print("Customer 1's id is {}".format(c1.getId))
print("Customer 2's id is {}".format(c2.getId))
print("Customer 3's id is {}".format(c3.getId))
try:
print("Customer 1 sign-in {}".format(c1.signin("abc@abc.com", "12341234")))
except LoggedError as e:
print(e)
try:
print("Customer 2 sign-out {}".format(c2.signout()))
except LoggedError as e:
print(e)
try:
print("Customer 3 sign-in {}".format(c3.signin("abc@abc.com", "12341234")))
except LoggedError as e:
print(e)
ANSWER
Answered 2021-Jun-06 at 12:07You have to actually call the method:
QUESTION
I have a list of dicts in Python called "roles" inside the list the dicts look like this:
...ANSWER
Answered 2021-Jun-05 at 10:54My mistake was I was working with the original list so it caused unexpected behaviour - Moving Python Elements between Lists
This worked by using a copy of "roles"
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
Vulnerabilities
No vulnerabilities reported
Install RePr
You can use RePr like any standard Python library. You will need to make sure that you have a development environment consisting of a Python distribution including header files, a compiler, pip, and git installed. Make sure that your pip, setuptools, and wheel are up to date. When using pip it is generally recommended to install packages in a virtual environment to avoid changes to the system.
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