Rock_Paper_Scissors | A great Rock Paper Scissor game against the computer

by SirWOH JavaScript Version: Current License: No License

X-Ray Key Features Code Snippets Community Discussions(10)Vulnerabilities Install Support

kandi X-RAY | Rock_Paper_Scissors Summary

Rock_Paper_Scissors is a JavaScript library. Rock_Paper_Scissors has no bugs, it has no vulnerabilities and it has low support. You can download it from GitHub.

A great Rock Paper Scissor game against the computer.

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Support

Rock_Paper_Scissors has a low active ecosystem.

It has 14 star(s) with 0 fork(s). There are no watchers for this library.

It had no major release in the last 6 months.

Rock_Paper_Scissors has no issues reported. There are no pull requests.

It has a neutral sentiment in the developer community.

The latest version of Rock_Paper_Scissors is current.

Quality

Rock_Paper_Scissors has no bugs reported.

Security

Rock_Paper_Scissors has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

License

Rock_Paper_Scissors does not have a standard license declared.

Check the repository for any license declaration and review the terms closely.

Without a license, all rights are reserved, and you cannot use the library in your applications.

Reuse

Rock_Paper_Scissors releases are not available. You will need to build from source code and install.

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Rock_Paper_Scissors Key Features

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Rock_Paper_Scissors Examples and Code Snippets

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Community Discussions

Trending Discussions on Rock_Paper_Scissors

Is there a way to limit which users response a discord bot reads after they have used a command?

Pass value from a button to a parameter in a function or to a variable

Python: Unable to use variable from another file even though the file is imported

What could be causing my program to go over a function twice even though it should be exiting after running over it once?

Why writing a correct input uppercase throws an error?

"Input 0 of layer sequential_1 is incompatible with the layer" with tf.data.Dataset

Simple Sequential model error : You must compile your model before using it

How to listen for an event that is happening to another element using JavaScript only?

When running this code, why does the loop not wait for my response?

Why am I getting an unexpected string when using .format to slice list?

QUESTION

Is there a way to limit which users response a discord bot reads after they have used a command?

Asked 2021-May-13 at 17:45

I am learning to write discord bots in python, so have settled on making a simple rock paper scissors bot. The way it currently works is the user types !RPS to start the game, and then has to type their choice, to which the bot replies with whether they have won or lost

...

ANSWER

Answered 2021-May-13 at 17:45

Inside your rock_paper_scissors() function, you can use the Bot.wait_for() method. You can create a check, to verify that the member who responds is the same who invoked the command.

Source https://stackoverflow.com/questions/67523593

QUESTION

Pass value from a button to a parameter in a function or to a variable

Asked 2021-Mar-08 at 12:07

im trying to make this "rock, paper, scissors" work using buttons.

I'm trying two different ways but i dont know how to achieve it:

1st way: Pass to the parameterplayerSelection in function playRound(playerSelection, computerSelection the correct parameter when we click on the buttons in the html

2nd way: Pass to variable var playerSelection inside function playRound(playerSelection, computerSelection the correct value when we click on the buttons in the html.

...

ANSWER

Answered 2021-Mar-08 at 12:07

You likely want something more like this

I use an array to count

Source https://stackoverflow.com/questions/66529209

QUESTION

Python: Unable to use variable from another file even though the file is imported

Asked 2021-Mar-08 at 09:24

I am trying to write a game of rock, paper, scissors, and I am getting an error about a file I imported not having an attribute of "quitGame", even though I should have that variable imported along with the others.

Here is the function that has the quitGame variable (uneeded parts filtered out):

...

ANSWER

Answered 2021-Mar-08 at 09:24

You have to use global before the variables is "requested" not "declared".

So if you want quitGame to be global you have to declare it were do you prefer (for example in main.py) and then were do you want to use it (as the example in a function) use "global quitGame".

You can tell to your function to access to a global var called quiteGame, instead of declaring quiteGame as global.

Source https://stackoverflow.com/questions/65996595

QUESTION

What could be causing my program to go over a function twice even though it should be exiting after running over it once?

Asked 2021-Feb-01 at 18:30

I am trying to make a game of rock, paper, scissors, and I am having issues figuring out why my program is reiterating over a function twice when it it supposed to go over it once and then start back at the beginning due to a while loop.

Here is the function that is causing the problem (unnecessary information filtered out):

...

ANSWER

Answered 2021-Feb-01 at 18:16

The function is being called the second time because you call the function again in your if statement:

Source https://stackoverflow.com/questions/65997953

QUESTION

Why writing a correct input uppercase throws an error?

Asked 2020-Dec-14 at 15:19

If user inputs a correct word uppercase I get a "You made a mistake" response. Why does user_choice.lower() not work like I intended? Where did I make a mistake?

...

ANSWER

Answered 2020-Dec-14 at 15:06

The condition if you set isn't correct. Try instead

Source https://stackoverflow.com/questions/65291303

QUESTION

"Input 0 of layer sequential_1 is incompatible with the layer" with tf.data.Dataset

Asked 2020-Oct-13 at 01:19

I am getting

ValueError: Input 0 of layer sequential_1 is incompatible with the layer: : expected min_ndim=4, found ndim=3. Full shape received: [300, 300, 3]. Model Summary shows 4 inputs but it says 3.

...

ANSWER

Answered 2020-Oct-11 at 12:19

You need to batch your dataset in order to get the right shape:

Source https://stackoverflow.com/questions/64290702

QUESTION

Simple Sequential model error : You must compile your model before using it

Asked 2020-May-02 at 15:28

This error seems to go if i remove the Flatten() layer.

I am trying to use my model but it is giving this runtime error : You must compile your model before using it.

I don't understand what is wrong, I've tried using lesser Dense layers but it is not working.

Can someone help me with this? PLZ explain with code.

...

ANSWER

Answered 2020-May-02 at 15:28

This error happens because your network is inconsistent, the input_shape parameter should be given to the first layer in your network, but you are giving it to the second layer.

The solution is simple, put the right input_shape to your Flatten layer.

Source https://stackoverflow.com/questions/61555935

QUESTION

How to listen for an event that is happening to another element using JavaScript only?

Asked 2020-Mar-25 at 22:42

So from my understanding of event listeners is that the action only applies to the element that you used the event listener on.

But say that I wanted to listen for a click on one element that is in a separate div, but the element I want to apply certain actions on is in a separate div. What would be the best way if possible in doing so?

To help visualize, im looking at this users example.

https://ryannathanwilson.github.io/Rock_Paper_Scissors/

And so from what it looks like it listens for a click on any of the buttons, and then the clicked button shows up in another area and then performs a specific action. Is this possible? Or am I understanding his code incorrectly and the elements that appear at the bottom when clicked is the original element?

...

ANSWER

Answered 2020-Mar-25 at 22:42

You can write a function like this:

Source https://stackoverflow.com/questions/60858316

QUESTION

When running this code, why does the loop not wait for my response?

Asked 2019-Dec-26 at 02:50

import random

def rock_paper_scissors(choice):
    computer_choice = random.choice(['rock', 'paper', 'scissors'])
    while True:
        if choice == computer_choice:
            print("Tie! Play again.")
            choice
        elif choice.lower() == 'quit':
            break

        elif choice.lower() == 'rock' and computer_choice.lower() == 'paper':
            print("TRY AGAIN!", computer_choice.lower(), "beats", choice.lower())
            choice
        elif choice.lower() == 'rock' and computer_choice.lower() == 'scissors':
            print("YOU WON!", choice.lower(), "beats", computer_choice.lower())
            choice
        elif choice.lower() == 'paper' and computer_choice.lower() == 'scissors':
            print("TRY AGAIN!", computer_choice.lower(), "beats", choice.lower())
            choice
        elif choice.lower() == 'paper' and computer_choice.lower() == 'rock':
            print("YOU WON!",  choice.lower(), "beats", computer_choice.lower())
            choice
        elif choice.lower() == 'scissors' and computer_choice.lower() == "rock":
            print("TRY AGAIN!", computer_choice.lower(), "beats", choice.lower())
            choice
        elif choice.lower() == 'scissors' and computer_choice.lower() == "paper":
            print("YOU WON!", choice.lower(), 'beats', computer_choice.lower())
            choice
        else:
            print("Sorry, invalid.")
            choice

print(rock_paper_scissors(input("Rock paper or scissors (enter 'q' to quit): ")))

...

ANSWER

Answered 2019-Dec-26 at 02:07

The input statement is not in the scope of the while loop, so it is only called once.

Once in the while loop, there is nothing changing the choice variable and hence the same print statement get triggered over and over.

Moving the input statement into the while loop along with the the computer choice initialisation (so the computer can choose a new option each go) solves your problem.

Source https://stackoverflow.com/questions/59483112

QUESTION

Why am I getting an unexpected string when using .format to slice list?

Asked 2019-Oct-23 at 14:35

I am trying to create a game of rock, paper, scissors. However, when I try to .format a string inside of an input function using a slice I am getting the following:Choose Rock, Sciss or Scissors:

I am using the following script:

...

ANSWER

Answered 2019-Oct-23 at 14:35

It's not the slicing that's going wrong, it's the string construction. You should indent the else: block a bit more. For reference, try print()-ing x before your input command.

What happened

Python for loops can have an else: clause. These are executed unless a loop is terminated by break or return. Therefore, your loop only did something on the first iteration while x was still an empty string, and then the else clause would modify your string with the option still bound from the last iteration of your loop.

Edit

Perhaps I'm a bit to used to Code Review, but I'd suggest doing the construction differently, more like this:

Source https://stackoverflow.com/questions/58524772

Community Discussions, Code Snippets contain sources that include Stack Exchange Network

Vulnerabilities

No vulnerabilities reported

Install Rock_Paper_Scissors

You can download it from GitHub.

Support

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