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kandi X-RAY | accu Summary
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QUESTION
I am trying to learn Common Lisp with the book Common Lisp: A gentle introduction to Symbolic Computation. In addition, I am using SBCL, Emacs, and Slime.
In the advanced section of chapter 8, the author presents the labels
special function. Actually, he draws a contrast between defining things on top-level (main and helper functions) versus using label
expression inside a function.
For instance, this would be a reverse
list function with tail call using the top-level approach:
ANSWER
Answered 2021-Jun-02 at 21:52Yes, there are good reasons. In Racket, we have define
In an internal-definition context, a
define
form introduces a local binding; see Internal Definitions. A the top level, the top-level binding forid
is created after evaluatingexpr
So, as you said, a define
in a local context (such as a function body) defines a local function, with access to the enclosing variables and which only exists during that function.
Now compare that to Common Lisp's defun
Defines a new function named function-name in the global environment.
So, regardless of where a defun
appears, it always defines a name at the global scope, with no access to local variables and with a name that becomes globally available. So your suggestion for a nested defun
is really equivalent to defining the defun
at the top-level (in the sense that the name is available at the top-level and in the sense that local variables are inaccessible), except that the name doesn't exist until you've called the original function at least once, which is, frankly, fairly unintuitive behavior.
Incidentally, the labels
approach is the thing you want. In Common Lisp, if we want local helper functions, we use flet
(for non-recursive functions) or labels
(for recursive functions).
As to why this is, Common Lisp tries to enforce a very clear variable scope at all times. In any function, local variables are introduced with (let ...)
and only exist inside of the block, and local functions are introduced with (flet ...)
and (labels ...)
. Racket has similar constructs but also allows the more Scheme-like paradigm (Racket is a Scheme dialect, after all) of using define
to define local variables for the rest of the current scope, similar to how you'd do it in more imperative languages.
QUESTION
I'm getting this error. Kindly note splits
is an array with values.
TypeError: Cannot read property 'push' of undefined
on
...ANSWER
Answered 2021-May-12 at 10:25null === undefined
is false
in Javascript.
QUESTION
I am using JGraphT for holding about 150,000 (or thereabouts) vertices in an in-memory graph. This a directed graph and each vertex is going to have { 0 | 1 } outgoing edges.
I want to find retrieve the set of vertices which have no outgoing edges. Here's what I have attempted:
...ANSWER
Answered 2021-Apr-24 at 18:14I'm not quite sure what's going on in the code, but the following would work fine:
QUESTION
I create the following algorithm in Rcpp and compile it in R.
...ANSWER
Answered 2021-Apr-14 at 13:41You are incrementing N
in place via +=
.
Your function fails to ensure it is initialized at zero. Rcpp
tends to do that by default (as I think it is prudent) -- but this can be suppressed for speed if you know you are doing.
A minimally repaired version of your code (with the correct header, and a call to .fill(0)
) follows.
QUESTION
I recently encountered this problem:
There is a (increasingly) sorted array formed by multiplying any two or more consecutive natural numbers.
...
ANSWER
Answered 2021-Apr-09 at 13:01The following implementation is based on a min-heap (std::priority_queue
in C++), that memorizes the "best" future candidates.
One important point is to treat the basic solutions k *(k+1)
differently. As it is likely that these numbers are in the majority, this allows to greatly reduce the size of the heap.
At each given time, we either decide to use a k(k+1)
number, or to use the current top value of the min-heap.
Each used value led to insertion of a new candidate in the min-heap.
Another aspect is to only insert in the heap the values less then the estimated max value, n(n+1)
.
The complexity is estimated to be O(n log M), where M is the mean size of the heap.
For n = 10^6
, the programme measures that the maximum size of the heap is equal to 9998, much less than n
.
On my PC, I get the result for n = 10^6
in 11 ms. Result: 977410038240
Here is the C++ code.
This code memorizes all the sequence, mainly for debugging. In practice, if we only need the nth value, such memorization can be avoided. The measurement of the maximum heap (useful for debugging) size can be removed too, if efficiency is still a concern.
QUESTION
I am using Dune for building OCaml projects and I'd like to build a standalone library for reusing it in other projects.
The root folder of the library is mylib
and the library was initialized with dune init lib mylib src
(after cd mylib
).
The directory tree of mylib
, the root of the project, is the following:
ANSWER
Answered 2021-Mar-07 at 16:19- In the root directory, these should be a file called
mylib.opam
with content similar to this one. Then, the(public_name mylib)
s-expression should be added tolibrary
in mylib/src/dune. Finally, making sure to be in the root of the project, the library shall be built withdune build
and installed withopam install .
.
For actually using the library in another project, even in a different dune workspace, all it's needed is to add(libraries mylib)
to the dune file of project that will use the library. - The
(private_modules priv)
s-expression should be added tolibrary
in mylib/src/dune. Then,mylib.ml
needs to be created with the following content:module Mymodule = Mymodule
: this will make sure that onlyMymodule
will be exposed under the packageMylib
.
QUESTION
I wrote the following code in SML with the NJ Compiler:
...ANSWER
Answered 2021-Mar-02 at 09:47As @kopecs pointed out, this is caused by case xs of
when you want case xs_left of
.
Here is a cleaned up (whitespace, naming) version of your function:
QUESTION
i have this CNN model with 3 block of VGG architecture
...ANSWER
Answered 2021-Jan-28 at 08:28- No - because valuation loss not increasing
- Your plots look fine. It is expected that the training process goes slower
- Yes, but it doesn't make sense. If you train any model for infinity - its performance will permanently improved - e.g. you can get 89.5% accuracy (which is better than 89.48%) if you train it for year.
- Try decaying learning rate with different schedules
QUESTION
I have correct shape of hidden layer for printing as below.
...ANSWER
Answered 2021-Jan-01 at 17:25The LSTM requires two hidden states, not one. So instead of
QUESTION
Let's say we have the following array of objects and we want to group them by property (here color)
Here is how I wrote it in JS
...ANSWER
Answered 2020-Dec-06 at 23:58Force the values of T
to be string | number | symbol
:
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