e learning – The best practice for displaying sections of web pages that are not yet available?

I am building a 2-part e-learning course. The user must finish part 1 before the second part "unlocks".

I currently have this section grayed out, with a photocopy that says "complete the first part to unlock this section",

Is there another way I can not think?

[GET] Bustle early learning book

SP: http://www.marketersnest.com/early-learning-book-bustle/

Here: https://www84.zippyshare.com/v/vqoeW5Ho/file.html

machine learning – He can not open the camera window and not take a picture

https://gist.github.com/AlokMishra051298/3e9db3e671e734fd61c2e1221c4789c6
import cv2
import numpy as np
face_classifier = cv2.CascadeClassifier (& # 39; C: /Users/HP/Anaconda3/Lib/site-packages/cv2/data/haarcascade_frontalface_default.xml')

def face_extractor (img):
#We convert to grayscale the image taken in rbg
#cvtColor (img, scaling in the obj converter)
img = np.full ((100,80,3), 12, np.uint8)
gray = cv2.cvtColor (img, cv2.COLOR_BGR2GRAY)
# now we take the image and detect the face with the help of detectMultiScale (img_from_which_we_detect_the_face, scale_factor, minNeothers)
The first parameter is a grayscale image
The #second parameter is Scale_factor: some faces may be closer to the camera, which appears larger than the faces at the back. So scale_factor compensates for that
The parameter #Third is minNeothers: parameter specifying the number of neighbors that each candidate repeater should have to keep it.
Higher values ​​result in less detection but higher quality
faces = face_classifier.detectMultiScale (gray, 1.3.5)
# we have the picture, now we have to check whether there is a picture or not
if faces is (): # if faces is not there, then
return None
#If the condition above is not satisfied, it means that a face is detected. We will now trim the faces.
for (x, y, w, h) in the faces:
cropped_face = img[y:y+h,x:x+w]Cropping is done according to the size of the image
return cropped_face

cap = cv2.VideoCapture (0) #camera starts to capture
count = 0 # variables used to count the number of photos taken
in truth:
ret, frame = cap.read () # this line returns an image that is captured
if face_extractor (frame) is not None: #means if face is present
count = count + 1 # increment the counter after capturing the image
# after detecting the face in a captured image, we need to resize this image
faces = cv2.resize (face_extractor (frame), (200,200)) # here resize the image first, we need to detect the face so that we can pass the frame as a parameter in the face_extractor, and resize the 39; image means convert to another measured value so here we too 300 * 300 pixels
faces = cv2.cvtColor (faces, cv2.COLOR_BGR2GRAY) #converts the resized image to grayscale
# we must give the name of the resized image to save it in any specified location
filename_path = "C: / Users / HP / Office / captures / alok" + str (count) + & # 39 ;. jpg & # 39;
#save the image (here called variable faces) in the location: path_file_name
cv2.imwrite (path_filename, faces)
#if we want to put a text that will inform the user of the number of images on which the user has clicked
# write you need to know
# 1.On which image,
# 2.Text what to write,
# 3.Where it will be written (i.e.origin),
# 4.In what font (optional),
# 5.Scale_factor,
# 6.Color of text,
Font size
cv2.putText (faces, str (count), (50,50), cv2.FONT_HERSHEY_COMPLEX, 1, (0,255,0), 4)
#we want to show the cropped image in order to use cv2.imshow (name_of_window, image)
cv2.imshow ("Face Cropper", Faces)
else: # it simply means that the face is not found
print ("face not found")
pass # pass this message
# —— we need a condition in which the camera stops capturing the shots, so you have to click ENTER (ASCII VALUE IN THE EAST 13) or a value maximum images.
if cv2.waitKey (1) == 13 or account == 100:
Pause
# ————— we need to release the camera without this creating a problem for our device
cap.release ()
# —- we have to close the windows and deallocate any associated memory usage
cv2.destroyAllWindows ()

print ("The collection of images is complete :)")

machine learning – Problem with this code

When I run the code, after inserting the photo, it does not tell me which race is the dog. Can someone help me?
That's the code:

                dobermann = {-> "Dobermann", -> "Dobermann", -> "Dobermann", ->
"Dobermann", -> "Dobermann"};
rottweiler = {-> "Rottweiler", -> "Rottweiler", -> "Rottweiler", ->
"Rottweiler", -> "Rottweiler"};
alano = {-> "Alano", -> "Alano", -> "Alano", -> "Alano", -> "Alano"};
labrador = {-> "Labrador", -> "Labrador", -> "Labrador", ->
"Labrador", -> "Labrador"};
husky = {-> "Husky", -> "Husky", -> "Husky", -> "Husky", -> "Husky"};
dogTypeSet = {dobermann, rottweiler, alano, labrador, husky};
dogTypes = Flatten[dogTypeSet];
class = classify[
      Flatten[{dobermann, rottweiler, alano, labrador, 
        husky}]]; Row[{FileNameSetter[Dynamic[file]], Dynamic[file]}]Button["Chargerlefichier"im=Import["LoadFile"im=Import["Chargerlefichier"im=Importer["Loadfile"im=Import[file]]Dynamic[im]

Which site is best for learning the selenium test?

Which site is best for learning the selenium test?

machine learning – what does this sentence mean: "forming a network of policies"

I know the basics (and maybe a substantial amount of basics) of imitation learning and reinforcement learning. In IL (Imitation), we are witnessing demonstrations by a confirmed expert, which, we suppose, has the most effective policy.

What does this statement mean? This will form a rule network using the expert dataset and record the results.

Question: Why do I need to form a network of rules? Why is it?
Are there any algorithms that come with this training a policy?

Thank you

machine learning – Creating interface on mathematica

I have developed the following code, however, in order to avoid problems on the cloud, I would like to create an interface on mathematica. How could I do?
The code is as follows:

dobermann = {-> "Dobermann", -> "Dobermann", -> "Dobermann", ->
"Dobermann", -> "Dobermann"};
rottweiler = {-> "Rottweiler", -> "Rottweiler", -> "Rottweiler", ->
"Rottweiler", -> "Rottweiler"};
alano = {-> "Alano", -> "Alano", -> "Alano", -> "Alano", -> "Alano"};
labrador = {-> "Labrador", -> "Labrador", -> "Labrador", ->
"Labrador", -> "Labrador"};
husky = {-> "Husky", -> "Husky", -> "Husky", -> "Husky", -> "Husky"};
dogTypeSet = {dobermann, rottweiler, alano, labrador, husky};
class = classify[
   Flatten[{dobermann, rottweiler, alano, labrador, husky}]];
CloudDeploy[
 FormPage["image" -> "Image", 
  Column[{#image, "This is a " <> ToString[class[#image]]}]And
Rules of appearance -> <|"Title" -> "What dog is this?"
"Description" ->
"Enter an image of a dog and I will tell you what it is,"
"SubmitLabel" -> "Classify" |>, PageTheme -> "Blue"],
Permissions -> "Public"]

machine learning – Problem identification image

I have developed this code that allows me to recognize which breed is the dog in the inserted image (for the moment developed only for five breeds of dogs), however, when I will insert the image into exit, he gives me a series of vectors. Could you tell me where I'm wrong?
I enclose the code.

dobermann = {-> "Dobermann", -> "Dobermann", -> "Dobermann", ->
"Dobermann", -> "Dobermann"};
rottweiler = {-> "Rottweiler", -> "Rottweiler", -> "Rottweiler", ->
"Rottweiler", -> "Rottweiler"};
alano = {-> "Alano", -> "Alano", -> "Alano", -> "Alano", -> "Alano"};
labrador = {-> "Labrador", -> "Labrador", -> "Labrador", ->
"Labrador", -> "Labrador"};
husky = {-> "Husky", -> "Husky", -> "Husky", -> "Husky", -> "Husky"};
dogTypeSet = {dobermann, rottweiler, alano, labrador, husky};
dogTypes = Flatten[dogTypeSet];
class = classify[
   Flatten[{dobermann, rottweiler, alano, labrador, husky}]];
CloudDeploy[
 FormPage["image" -> "Image", 
  Column[{image, "This is a " <> ToString[net[image]]<> "dog"}],
Rules of appearance -> <|"Title" -> "What dog is this?"
"Description" ->
"Enter a picture of a dog and I will tell you what it is" |>],
Permissions -> "Public"]

Before the arrows are inserted the images concerning each race of cana. I had to delete them otherwise the code would become incomprehensible.

Learning Javascript – Is this a bad implementation of a Javascript slider?

I had been learning JS for a few months now and I just finished creating a small image slider for a website. The sliders feature is as follows:

  • Browse images continuously and return to the beginning
  • Allow the user to click on a given section
  • When you click on it, the cycle timer must be reset and the loop must continue from that point.
    • Scroll a corresponding DOM element (in this case, it is a testimonial quote)

I am mainly looking for guidance on how to make my code more effective or on bad practices. Any help would be appreciated.

var container = document.getElementById (& # 39; slide-container & # 39;);
var button = document.getElementById (& # 39; cursor-buttons & # 39;);
var quote = document.getElementById ('quote-container');
var interval = 0;

sliderMove = function (position) {

// move the container to the next image
container.style.left = - (350 * position) + & # 39; px & # 39 ;;

// Remove active classes from all children
for (var i = 0; i < quote.children.length; i++) {
    quote.children[i].classList.remove("quote-active");
    button.children[i].classList.remove('active');
  };

  // Add active classes to specified children
  quote.children[position].classList.add("quote-active");
  button.children[position].classList.add('active');

  // Increase global interval variable by 1
  interval++;

  // Set interval back to 0 if it exceeds the amount of quotes
  if (interval >= quote.children.length) {
interval = 0;
};
};


// Add an event listener to the buttons
button.addEventListener ('click', function (event) {
clearInterval (timer); // Reset the counter when the user clicks on an image
automatic movement (); // Restart the timer
if (event.target.id> -1) {// Ensures the validity of the target
interval = event.target.id;
sliderMove (interval);
};
});

// Automatic rotation
autoMove function (index) {
timer = setInterval (function () {
index = interval;
sliderMove (index);
}, 5000);
}

autoMove (0);
sliderMove (0);

Auto Learning – How to Solve an Algorithm AO * Reducing Problems?

Algorithm AO * for reducing problems with AND OR GRAPH

visit here for more artificial intelligence info in India

Algorithm AO *
A * can be computer algorithms used in traversing drivers and graphs. It is used in the method of tracing an efficient guided path between several points called nodes.
In the AN * algorithmic rule, you traverse the tree in depth, keep moving forward and add the overall value to reach the target position from the current situation and add it to the value to reach the current position.
Algorithm AO *
In AO * algorithms, you follow the same method, although there are obstacles to cross specific approaches.
Once you have traversed these paths, the value of all paths from the previous node is added to this level where you search for the targeted position, despite the fact, Move.
REDUCTION OF THE PROBLEM:
Until now, we have been thinking about how to search the graph, through which we want to find a way to achieve a goal. This type of structure represents the real fact that we all know a way to move from the anode to the target position if we are told in a situation where the nodes with the stress of the branches will be explained.

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AND – OR Chart
AND – or graphic.
AND – OR graphic
AO * ALGORITHM:
Let G be the graph only with the introduction of the INIT node.
Repeat the follow up until you insert INIT in Solvi or H (INIT)> property.
A) Select the node suddenly from the path leading to INIT (it's called NODE)
B) Generate the successor of NODE. If none, set NODE = futility (that is, NODE fails); Otherwise, each SUCCESSOR that is not for the ancestors of NODE, does the following:
I add SUCCESSSOR to G
ii. If SUCCESSOR can be a terminal node, solve it and set HK (SUCCESSOR) = NULL.
iii. If SUCCESSPR is not a terminal node, calculate it.
C) Promote the newly discovered graph information by searching for the following: S should have a set of salt-like nodes or nodes whose H-values ​​have been changed and their parents should return the values ​​… S Repeat the trace until you reach the point. S is empty:
I remove the knot from the sand that lights it.
ii. Calculate the value of each bow that comes out. Assign the minimum value of your successor to HK
iii. Mark this route by marking the minimum arch value at step ii-iv. All nodes linked to the new labeled arc are fixed, then mark it as current.
v. If the label is solved or if its value has just been changed, broadcast your new value through the chart. So add all the ancestors from S. to S.

AND – OR graphic
AND – OR graphic
AND – OR graphic
In the form (A), the upper node A has been expanded to produce 2 zones that move to B and move to C-D. The numbers on each node represent the value of F on it (the value of reaching the target position from the current position). For the sake of simplicity, it is thought that each operation (ie the implementation of a rule) corresponds to the cost, that is to say that at each successor, the value of each component of each component. With the information available to date, it seems that C is the most promising node extended since F = 3, but going through B is preferable to using C, we use D and the value nine. Get also (3 + 4 + 1 + 1). He will be 6 (5 + 1) through B

To develop it, the selection of the next node does not depend only on a value, although the node is only the initial mode of the best current path. Figure (B) shows it well. The node of the form seems to be the most promising node with at least the F value. But to use the GG does not match the current beat path, the AO algorithm for problem reduction should always use GH with the value nine, it is then demanding that the arc be used (with the value 27). The route from A to B, E-F is the best with the full value (17 + one = 18). In this method, we will see that the following 3 things must be completed to see the AN & G graph.

Cross the graph starting from the initial node and follow the best current path, then submit all the nodes present on the path and not yet extended.

Select one of these amazing nodes and expand it. Add them to your successor graph and PCF (the cost of the remaining distance).

Modify the 'F estimate' of the new expanded node to reflect the new information provided by your successor. Boost this revision back across the chart. Check the current best path.

The estimate of the modified value is not required in the tree * A * algorithm. This can be {car} because extended nodes are tested again in the AO * algorithm so that the best current path can be selected. The work of AO * Formula is shown in the image as follows:

With reference to the figures, the initial node is expanded and D is initially marked due to the node. D Another E-F production order has been developed. AO Problem Reduction Algorithm: The value of the FD has been updated to ten. Going back, we will see that the Arch B-C finish is the best. It is currently marked because of the current best route. B and C should be reported. This method continues until all the methods have no answer or cause deadlocks, it shows that there is no answer. The AA algorithm is always the lowest cost from one node to the other and it is independent of the path across the opposite nodes.

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The algorithm to perform another horizontal search of the graph is given below. Unlike the A * algorithm that opens and closes 2 lists, the AO algorithm uses the same structure. The GG represents part of the search graph generated from this point. Each node reaches its immediate successor and its immediate predecessors, the problem reduction algorithm, in addition to the path cost value during all nodes of the response. The price to get the current node "live" from the starting node is not preserved in the formula A *.

Since it is out of the question to calculate this value, it may be possible for several methods to be used in the same state. The algorithm AO * predicts the quality of the node. Apart from this, the value called useless should be used. If the calculation of the answer exceeds the allowed cost, the inventions are too large to be applied