hash | C hash implementation based on khash | Hashing library
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kandi X-RAY | hash Summary
C hash implementation based on khash
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QUESTION
I created the default IntelliJ IDEA React project and got this:
...ANSWER
Answered 2021-Nov-15 at 00:32Failed to construct transformer: Error: error:0308010C:digital envelope routines::unsupported
The simplest and easiest solution to solve the above error is to downgrade Node.js to v14.18.1. And then just delete folder node_modules
and try to rebuild your project and your error must be solved.
QUESTION
Here are two measurements:
...ANSWER
Answered 2022-Mar-30 at 11:57Combining my comment and the comment by @khelwood:
TL;DR:
When analysing the bytecode for the two comparisons, it reveals the 'time'
and 'time'
strings are assigned to the same object. Therefore, an up-front identity check (at C-level) is the reason for the increased comparison speed.
The reason for the same object assignment is that, as an implementation detail, CPython interns strings which contain only 'name characters' (i.e. alpha and underscore characters). This enables the object's identity check.
Bytecode:
QUESTION
I have updated node
today and I'm getting this error:
ANSWER
Answered 2021-Oct-27 at 17:19Ran into the same issue with Node.js 17.0.0. To solve it, I downgraded to version 14.18.1, deleted node_modules
and reinstalled.
QUESTION
I am using Perceptual hashing technique to find near-duplicate and exact-duplicate images. The code is working perfectly for finding exact-duplicate images. However, finding near-duplicate and slightly modified images seems to be difficult. As the difference score between their hashing is generally similar to the hashing difference of completely different random images.
To tackle this, I tried to reduce the pixelation of the near-duplicate images to 50x50 pixel and make them black/white, but I still don't have what I need (small difference score).
This is a sample of a near duplicate image pair:
Image 1 (a1.jpg):
Image 2 (b1.jpg):
The difference between the hashing score of these images is : 24
When pixeld (50x50 pixels), they look like this:
rs_a1.jpg
rs_b1.jpg
The hashing difference score of the pixeled images is even bigger! : 26
Below two more examples of near duplicate image pairs as requested by @ann zen:
Pair 1
Pair 2
The code I use to reduce the image size is this :
...ANSWER
Answered 2022-Mar-22 at 12:48Rather than using pixelisation to process the images before finding the difference/similarity between them, simply give them some blur using the cv2.GaussianBlur()
method, and then use the cv2.matchTemplate()
method to find the similarity between them:
QUESTION
I got this error when learning Next.js, using npx create-next-app
command according to site documentation here https://nextjs.org/docs/api-reference/create-next-app. Everything works until I start the server,
Error stack:
...ANSWER
Answered 2021-Nov-24 at 21:38I found this solution https://github.com/webpack/webpack/issues/14532
if using bash just run
NODE_OPTIONS=--openssl-legacy-provider
before any commandadding
NODE_OPTIONS=--openssl-legacy-provider
to package.json
QUESTION
When I do this it works (these are the last 4 lines before the end of a method TWEAK. However, my first attempt had no line #3 and failed because %!columns was empty...
...ANSWER
Answered 2022-Feb-23 at 19:56Without the %!columns
on line three, the call to map
is lazy and thus never gets evaluated (the %!columns
call wants to check the current value of columns, which implies eagerness).
To more explicitly invoke eagerness, either use the eager
statement prefix (shown below) or switch to a for
loop, which is eager by default.
I think this code will behave the way you want it to:
QUESTION
[I ran into the issues that prompted this question and my previous question at the same time, but decided the two questions deserve to be separate.]
The docs describe using destructuring assignment with my
and our
variables, but don't mention whether it can be used with has
variables. But Raku is consistent enough that I decided to try, and it appears to work:
ANSWER
Answered 2022-Feb-10 at 18:47This is currently a known bug in Rakudo. The intended behavior is for has
to support list assignment, which would make syntax very much like that shown in the question work.
I am not sure if the supported syntax will be:
QUESTION
Whenever I am trying to run the docker images, it is exiting in immediately.
...ANSWER
Answered 2021-Aug-22 at 15:41Since you're already using Docker
, I'd suggest using a multi-stage build. Using a standard docker image like golang
one can build an executable asset which is guaranteed to work with other docker linux images:
QUESTION
I am trying to find a more efficient solution to a combinatorics problem than the solution I have already found.
Suppose I have a set of N objects (indexed 0..N-1) and wish to consider each subset of size K (0<=K<=N). There are S=C(N,K) (i.e., "N choose K") such subsets. I wish to map (or "encode") each such subset to a unique integer in the range 0..S-1.
Using N=7 (i.e., indexes are 0..6) and K=4 (S=35) as an example, the following mapping is the goal:
0 1 2 3 --> 0
0 1 2 4 --> 1
...
2 4 5 6 --> 33
3 4 5 6 --> 34
N and K were chosen small for the purposes of illustration. However, in my actual application, C(N,K) is far too large to obtain these mappings from a lookup table. They must be computed on-the-fly.
In the code that follows, combinations_table
is a pre-computed two-dimensional array for fast lookup of C(N,K) values.
All code given is compliant with the C++14 standard.
If the objects in a subset are ordered by increasing order of their indexes, the following code will compute that subset's encoding:
...ANSWER
Answered 2021-Oct-21 at 02:18Take a look at the recursive formula for combinations:
Suppose you have a combination space C(n,k)
. You can divide that space into two subspaces:
C(n-1,k-1)
all combinations, where the first element of the original set (of lengthn
) is presentC(n-1, k)
where first element is not preset
If you have an index X that corresponds to a combination from C(n,k)
, you can identify whether the first element of your original set belongs to the subset (which corresponds to X
), if you check whether X
belongs to either subspace:
X < C(n-1, k-1)
: belongsX >= C(n-1, k-1)
: doesn't belong
Then you can recursively apply the same approach for C(n-1, ...)
and so on, until you've found the answer for all n
elements of the original set.
Python code to illustrate this approach:
QUESTION
Every time I sign something, it display the next error:
...ANSWER
Answered 2021-Oct-24 at 23:50According to this bugreport, the error message seems to be a harmless bug introduced in version 2.3.3 on macOS. The report states it can safely be ignored.
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