L2 | style Tensor+Autograd library | Machine Learning library
kandi X-RAY | L2 Summary
kandi X-RAY | L2 Summary
l2 is named after the l2 or Euclidean distance, a popular distance function in deep learning. l2 is a Pytorch-style Tensor+Autograd library written in Rust. It contains a multidimensional array class, Tensor, with support for strided arrays, numpy-style array slicing, broadcasting, and most major math operations (including fast, BLAS-accelerated matrix multiplication!). On top of this, l2 has a built-in efficient graph-based autograd engine that keeps track of all operations performed on a tensor and topologically sorts and traverses the graph to compute the gradients.
Support
Quality
Security
License
Reuse
Top functions reviewed by kandi - BETA
Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of L2
L2 Key Features
L2 Examples and Code Snippets
def _contrib_layers_l2_regularizer_transformer(
parent, node, full_name, name, logs):
"""Replace slim l2 regularizer with Keras one, with l=0.5*scale.
Also drops the scope argument.
"""
def _replace_scale_node(parent, old_value):
"""
def l2_normalize(x, axis=None):
"""Normalizes a tensor wrt the L2 norm alongside the specified axis.
Args:
x: Tensor or variable.
axis: axis along which to perform normalization.
Returns:
A tensor.
"""
return nn.l2_norma
def _l2_loss_flops(graph, node):
"""Compute flops for L2Loss operation."""
in_shape = graph_util.tensor_shape_from_node_def_name(graph, node.input[0])
in_shape.assert_is_fully_defined()
# Tensorflow uses inefficient implementation, with (3*N-
Community Discussions
Trending Discussions on L2
QUESTION
I am having a hard time to solve the following issue: I have to merge N lists. Each list contains some string objects. For each list, although I do not know which is the ordering function, I know that it is ordered. Moreover, the final list should respect all the ordering of the child that generated it. For instance:
...ANSWER
Answered 2021-Jun-14 at 17:05The main difficulty with your problem is if we have a case like this:
QUESTION
In R, I can take a df and get a list of frequency tables for each sample/group by doing the following:
...ANSWER
Answered 2021-Jun-15 at 02:03In pandas
you can do groupby
with pd.crosstab
QUESTION
I have two lists of strings and i want to check if the words within one list contain the strings within the other list throughout
Here is an example of what i mean:
...ANSWER
Answered 2021-Jun-14 at 19:33This will work:
QUESTION
a <- structure(c(1017, 63, 9184, 9782, 42, 1467, 84, 11829, 11989,
49, 1459, 93, 11399, 11302, 42, 188, 14, 1530, 1463, 7, 189,
20, 1647, 1569, 6), .Dim = c(5L, 5L), .Dimnames = list(c("NC_013663.1_297_-",
"NC_013663.1_553_-", "NC_013663.1_553_+", "NC_013663.1_554_-",
"NC_013663.1_555_-"), c("CL1", "CL2", "CL3", "CL4", "CL5")))
...ANSWER
Answered 2021-Jun-14 at 19:08Get the index of columns with substring L1 to L3 using grep
. Extract the columns, multiply with the value 0.1343109 and update by assigning back to the columns. (Note - multiplying by 1 returns the same value, thus it is left as such)
QUESTION
I am writing code to answer the following question: "for two linked lists l1 and l2 that contain positive integers (and have at least one node), return their sum as a linked list. "
For example:
...ANSWER
Answered 2021-Jun-14 at 13:51The reason for the error is that you cannot assume that both prev
and prev2
are not None
. Your while
condition only guarantees that at least one of them is not None
, but not both. So that means you still need to have a None
check inside the body of the loop.
There are also some other issues:
l1==[0]:
is not correct. The left side of this comparison is a linked list instance, while the right side is a standard list. These comparisons will always beFalse
, and so you can just omit themYou are adding the
carry
to the wrong sum. You should add thecarry
from the previous iteration to the current sum, so things are currently happing in the wrong order.The resulting list should be reversed. You can do this on the fly by prefixing each node to
l3
, instead of appending it. This will also make it unnecessary to create a dummyListNode(0)
before the final loop. You can just start withNone
.Your code mutates the original input lists. This will be acceptable on code challenge sites, and be efficient, but it is bad practice. The caller should not be unpleasantly surprised that the lists they provide as input have been reversed by a call to this
addTwoNumbers
function. You can solve this in several ways, but it is probably easiest to reverse them a second time to undo that operation.To avoid repetition of code, you should create a
reverse
function
I'll assume that the ListNode
constructor can take a second argument for initialising its next
reference:
QUESTION
Hello I have a list of dataframes where I want to add new columns to each of those dataframe. My current for-loop approach gets the job done, however I was looking for an elegant approach, something from apply
family of functions.
Here is a reprex-
...ANSWER
Answered 2021-Jun-14 at 13:31The function week_no
is not vectorised so you would need some kind of loop to iterate over each value after strsplit
. In the for
loop you use sapply
, so we can use the same here.
QUESTION
How to remove first line with names level_0
and all and convert index 0
as columns.
ANSWER
Answered 2021-Jun-14 at 08:42I suggest first create MultiIndex in columns
by header=[0,1]
by convert first 2 headers rows:
QUESTION
Give numpy array A with is MxN and B which is DxN calculate the euclidean distance s.t. ijth element of AB=L2(A[i,:]-B[:,j]) without a for loop.
I thought you could do something liek
...ANSWER
Answered 2021-Jun-14 at 03:24import numpy as np
M, D, N = 100, 50, 5
A = np.random.random((M, N))
B = np.random.random((D, N))
# A.K.A AB
euclidean = np.sqrt(np.sum((A[:, None, :] - B[None, :, :]) ** 2, axis=2))
QUESTION
A month into google apps/googlesheets. I've got some of the basics down, however struggling to put a lot of basic concepts together.
Step 1) Create and check if Spreadsheet exists in folder. If it doesn't exist create one based on the name in Cell A1 and COPY ActiveSpreadsheet() data to that new FILE with sheet name TODAY() date.
Step 2) If a spreadsheet with name exists, copy from ActiveSpreadsheet() to the spreadsheet named in Cell A1 with a NEW SHEET named after today's date.
So far I have got pieces of stuff together but I am MISSING basic knowledge of trying to put it altogether. Sorry if its a COMPLETE mess I'm trying to piece it together as I go. ANY HELP WILL be appreciated or websites/resources to lead me in the right direction.
...ANSWER
Answered 2021-Jun-13 at 20:59Probably you need something like this:
QUESTION
These codes convert uppercase letters ("letters only") to lowercase letters and lowercase letters to uppercase. My question is that I want to print them as well and keep them unchanged, if any (non-verbal symbols and actors). With the cmp and ... commands that you see in the program
...ANSWER
Answered 2021-Jun-13 at 16:03You need to restrict the ranges for the uppercase and lowercase characters by specifying a lower limit and a higher limit, not just the one value (96) that your current code uses.
Uppercase characters [A,Z] are in [65,90]
Lowercase characters [a,z] are in [97,122]
The nice thing of course is that you don't actually need to write these numbers in your code. You can just write the relevant characters and the assembler will substitute them for you:
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
Vulnerabilities
No vulnerabilities reported
Install L2
Support
Reuse Trending Solutions
Find, review, and download reusable Libraries, Code Snippets, Cloud APIs from over 650 million Knowledge Items
Find more librariesStay Updated
Subscribe to our newsletter for trending solutions and developer bootcamps
Share this Page