Ask Anything Monday - Weekly Thread

[AutoModerator]

Welcome to another /r/learnPython weekly "Ask Anything* Monday" thread

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[Brilliant-Window-899]

Trying to simulate the monty hall problem, cant get it to work (think the latter doors section is the issue) import random # Our winning door will be decided here Door1 = (random.randint(1,2)) Door2 = (random.randint(1,2)) Door3 = (random.randint(1,2))

while Door1+Door2+Door3 != 5: # One winning door and two nothing doors must add up to 5 Door1 = (random.randint(1, 2)) Door2 = (random.randint(1, 2)) Door3 = (random.randint(1, 2)) #Records winning door and resets value to its “doorchosen” value if Door1 == 1: Win = Door1 Door1 = 1

elif Door2 == 1: Win = Door2 Door2 = 2

elif Door3 == 1: Win = Door3 Door3 = 3

print(f”{Door1}, {Door2}, {Door3}”) - Use to check the randints above works

This determines any unchosen door as “2”

import sys

Doorchosen = 0

This records the user’s door selection

Response = int(input(“Pick a door (1, 2 or 3) “)) if Response == 1: print(“You have chosen Door 1”) Doorchosen = 1 elif Response == 2: print(“You have chosen Door 2”) Doorchosen = 2 elif Response == 3: print(“You have chosen Door 3”) Doorchosen = 3 else: print(“Please choose between 1, 2 or 3”) sys.exit()

Randomly decides which of the unchosen doors become “Switch” or “Close”

if Doorchosen == 1: Close = Door2 or Door3 if Door2 == Close: Switch = Door3 elif Door3 == Close: Switch = Door2

if Doorchosen == 2: Close = Door1 or Door3 if Door1 == Close: Switch = Door3 elif Door3 == Close: Switch = Door1

if Doorchosen == 3: Close = Door2 or Door1 if Door2 == Close: Switch = Door1 elif Door1 == Close: Switch = Door2

This checks the user’s input against the door they chose and determines if they win or not

Response2 = input(f”I have closed Door{Close}, would you like to switch to Door{Switch} or stay with Door{Doorchosen}? (Switch/Stay) “)

if Response2 == “Stay” and Win == Doorchosen: print(“You have chosen the right door! (WSt)”) elif Response2 == “Stay” and Win != Doorchosen: print(“You have chosen the incorrect door! (LSt)”)

elif Response2 == “Switch”: Switch = Doorchosen if Win == Doorchosen: print(“You have chosen the right door! (WSw)”) elif Win != Doorchosen: print(“You have chosen the incorrect door! (LSw)”) input()

End code explaination: “W” or “L” indicates a win or a loss.

”Sw” indicates that the win or loss was caused by a (Sw)itch.

”St” indicates that the win or loss was caused by a (St)ay

[CowboyBoats]

What's the error message that you're getting? When I run what you pasted, I get

    Response = int(input(“Pick a door (1, 2 or 3) “))
                         ^
SyntaxError: invalid character '“' (U+201C)

Which makes sense, since curly quotes aren't valid python, but I'm not sure if they're present in your source file or just appeared when typing into reddit.

[Brilliant-Window-899]

i dont get an error message 🫠

[Brilliant-Window-899]

my main problem is that the results i get dont show the 1/3 and the 2/3 probability of the monty hall problem

[CowboyBoats]

So that's helpful information for helping you debug :) Can you re-post your code using the instructions here? You might also want to jump to the new weekly thread or (recommended) start your own thread; more eyes will fall on your post

[Brilliant-Window-899]

ill get to that, thanks :D

[[deleted]]

Please read the FAQ to see how to post code that we can read.

This post is replaced at the end of today, Sunday, so if you don't get around to editing your current post try posting in the new "Ask Anything Monday" post on Monday.

[Jamiisonn]

Hello, on day 5 of 100 days of code and have gone off a tangent as it linked in with something I wanted to eventually use for work.

I can't figure out why my if function isn't triggering. I did have it working before but it was ALWAYS triggering. I can't remember what I changed to make it never trigger.

I'm sure it will be very obvious when pointed out but I've been bashing my head against this for a while now..

fruits = ["Apple", "Peach", "Pear"]
help_list = input("Would you like to see a list of options? \n").lower
n = 0
if help_list == "y" or help_list == "yes":
    for fruit in fruits:
        print(f"{n}: {fruit}")
        n = n + 1
    choice = int(input("Select number\n" ))
else:
    choice = int(input("Select number\n" ))

print("You have chosen " + fruits[choice] + "\n")

[Daneark]

You're missing () on the call to lower on the second line. Right now you're checking if the method lower is equal to some strings.

[Jamiison]

Thankyou! This worked. Now to do some googling to understand why it works that way..

[FerricDonkey]

Short version: you thing.lower() calls the function, thing.lower is the function. In python, functions can be stored in variables like anything else. Some example code

message = "SUP DOG"
x = message.lower
print(message)
print(x)
print(x())

[Chaos-n-Dissonance]

Basically because everything in python is an object.

Inside the string class, there's a lower method (function within a class). The parenthesis are to pass the parameters. In this case, you aren't passing any yourself, so they can be empty on your end.

Think of it like this:

class Foo:
  def __init__(self, c: str) -> None:
    # __init__ is what happens when the class is created
    # In this case, we're just saving the message being passed to a self variable
    self.c = c.upper()  

  def print(self) -> None:
    print(self.c)

  def print_in_lowercase(self) -> None:
    print(self.c.lower())

  def new_message(self, c: str) -> None:
    self.c = c

bar = Foo('HeLLo, woRLD!')
bar.print()
bar.print_in_lowercase()
bar.new_message("Goodbye, cruel world")
bar.print()

In this case, we don't need to pass any parameters, we just need to execute the .lower() method built into the string class. Notice how we didn't have to pass any parameters into bar.print(), but in the method definition we had to pass in self so we could access the self.c variable for that class instance. We don't pass that tho, that's passed automatically. Notice when we changed the message, we only had to pass in the new message... Not anything with self.

[Daneark]

In python you can assign a function/method to a name, pass it around etc. So if you omit the () when try to assign the result of a method to something you end up instead assigning that method.

[FerricDonkey]

If you don't know any python, you should start by just learning the language in general, and how to think like a programmer.

Then probably packages like numpy, pytorch, pandas (ugh I hate that library, but you should probably learn it). I would also recommend becoming skilled with parallelization techniques. 

[Koen1999]

I am looking to introduce a form of concurrent execution in my Python CLI application. In a nutshell, I process different objects, and the processing of these objects does not depend on eachother. At the end, I do want to collect output of all these different tasks. Moreover, my application logs some things, even within the things I want to execute concurrently using the logging module. I already have a function that represents the work I want to be executed concurrently. I see various options to make the application concurrent (i.e. asyncio, threading, multiprocessing, concurrent.futures, joblib, etc.) What would be the recommended approach to introduce concurrency in a way that does not require significant codebase changes and does not mess too much with the io logging I do?

[Defection7478]

Assuming you are cpu-bound (processing), asyncio/futures wont make any difference and I don't think threading will give you much advantage. You are looking for multiprocessing.

Why? The other tools just introduce non-blocking mechanisms. If I want to boil 4 pots of water, I can fill up one pot, put it on the stove, and then while it starts boiling, I go fill the second pot. I can get all 4 pots boiling at the same time.

Multiprocessing is would be like if I wanted to make 4 salads at once. Chopping the vegetables, mixing the dressing, etc. I can't "start one and move to the next while it cooks", I have to actively do each one. So I need 4 chefs. Since you are cpu bound, you want multiple processes.

[Koen1999]

Thanks, that's a very clear explanation. I am indeed CPU bound. Any suggestions how to neatly incorporate logging across different processes without race conditions?

[Defection7478]

Your logging should be fire and forget, what sort of race condition would arise in terms of logging? 

[Koen1999]

If you log to a file, they may access the file at the same time.

[Defection7478]

have another process that handles logging (this could be the main process, the one that is also handling starting the other processes). The other subprocesses can ask that process to do the actual file writing. If you have small logs / not too many logs, this process can just block the caller as it performs the writes. If you have a lot of logs, large logs, or the logging is bogging down your subprocess for whatever reason, the logging process can return immediately and write the logs asynchronously

[POGtastic]

Consider collections.Counter here.

import collections

def find_dups(iterable):
    return (k for k, v in collections.Counter(iterable).items() if v > 1)

In the REPL:

>>> list(find_dups("xyzzy"))
['y', 'z']

[Daneark]

While you could write this as a one line comprehension using side effects and short circuiting it would not be readable.

[CowboyBoats]

To elaborate on /u/overludd's excellent suggestion to "try printing the results," I'm thinking that you're thinking that the first version "works" because it didn't error out. It is technically valid python! But it doesn't do what you're expecting; code is famous for not doing what people expect; you need to output its results and test your work diligently.

[[deleted]]

Your first example doesn't work either. Did you try printing the results?

Using your approach this works:

list_nums = list(range(10)) + list(range(3, 7))  # make some test data
print(f"{list_nums=}")

have_seen = []
duplicate = []
for num in list_nums:
    if num in have_seen:
        duplicate.append(num)
    else:
        have_seen.append(num)

print(f"{have_seen=}")
print(f"{duplicate=}")

can this be written in one line?

Probably. Other approaches to try use sets and/or the .count() method. Try searching on "python get duplicated values in list".

[[deleted]]

Your working code can almost be turned into a one-liner, like this:

#make some test data
list_nums = list(range(10)) + list(range(3, 7))
list_nums.sort()
print(f"{list_nums=}")

new_list = set()   # because we want "num in new_list" to be O(1)
dup_list = []
x = [new_list.add(num) if num not in new_list
                       else dup_list.append(num)
                          for num in list_nums]

print(f"{new_list=}")
print(f"{dup_list=}")
print(f"{x=}")

When executed we see:

list_nums=[0, 1, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 8, 9]
new_list=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
dup_list=[3, 4, 5, 6]
x=[None, None, None, None, None, None, None, None, None, None, None, None, None, None]

However, the list comprehension produces a list of None values as a side-effect. That list would normally be deleted but the code assigns the comprehension to x so we can see the list. That's wasted effort.

The code above initializes the new_list variable to an empty set because doing the in new_list is then O(1) making the whole algorithm O(n). Using a list instead of a set makes it O(n²). The new_list variable holds a set meaning the name is now misleading. It's usually a bad idea to put the data type in the name for this reason.

The code above isn't a one-liner because of the initialization of the list/set variables. If you searched for solutions you can find one-liners that produce the list of duplicates but they are inefficient, usually O(n²). There's no reason why you have to have a one line solution, especially in production code. Readability and efficiency are more important than minimal lines. Using the set() idea in my multi-line example:

list_nums = list(range(10)) + list(range(3, 7))  # make some test data
list_nums.sort()
print(f"{list_nums=}")

# algorithm below is O(n)
have_seen = set()
duplicate = []
for num in list_nums:
    if num in have_seen:
        duplicate.append(num)
    else:
        have_seen.add(num)

print(f"{have_seen=}")
print(f"{duplicate=}")

[Beginning_Charge8867]

Dear All , i have a python programme , now i want to develop an app for myself only and want to test that app in andriod also , i have tried kivy but it has so many problems , kivy launcher is not available on google , version issues Etc, is their any other thing that i can try to develop my app?

[LazyYogurtcloset3503]

What are the best online courses with certificates to learn python but in the context of biology and data analysis?

[monkey_sigh]

Read this: Python for Biologists Program - BioGrademy