I fear being shot by my neighbors living this close in America. Will I feel different in Haarlem? Cause I love the look

When the teacher explained List comprehension and showed how mnuch you can shorten code I just had to use it for this. c = dict() fhand = open('mbox-short.txt') for line in fhand: line = line.rstrip() words = line.split() if len(words) < 3 or words[0] != 'From': continue name = words[1] c[name] = c.get(name,0) + 1 print(sorted([(v,k) for k,v in c.items()][:1]))

It's probably been explained to death already, but literally typing out the spaces on every line is not how that works. So a lot of programmers sort of gave a condescending laugh at the joke here. It's definitely pulled from real nitpicky debates, just demonstrated wrong... But good lord did I crack up at "OH GOD HELP US" at vim over emacs 🤣🤣 I forget they did get into more details than just "four keys vs one", nice. Also if we wanna be stereotypical here, Richard would totally be on the vim side of the equation.

2 spaces

Whats the point on You tube of all these vlogs on bike going from A to B?

Thank you so much for this video bro. You are a life saver! I was stuck with a difficult exercise problem for hours and was going nowhere. That's when I got desperate and started searching for the solution and came across your video.

please can you provide the questions in a pdf format?

the answer required in the assignment is 498.75 how it can be done?

My man, what happened to lofi code & chill videos?

The other side of the canal is actually nicer

i was struggling until you showed me that literally the entire verification was done purely with regex. thanks!!

first solution doesn't work, I can enter 1.5 and get no error message

great super thanksss!!

Hi Please help, I copy pasted all the exercises in the freecodecamp course to my sheet and every one them had an error output when I run them in the terminal no idea what to do :(

Thanks

Love you teaching partner

So relaxing!

I'm getting a NameError: name 'List' is not defined. Did you mean: 'list'?

Hello, sorry to bother you but could you upload or send me a link for the playlist "lofi coding" you uploaded it during covid. Please.. Thanks.

Thank you! I didn't want to watch someone just throw out the answer. It helped a lot to watch you work through it.

Great series!

Beautiful. Just beautiful!!!

I find these so helpful. Thank you!

It's been 7 years - and still same problem and same solution :O

Thank you this video you really helped solve my problem. Just curious, Is it possible to put the ranges for the grades in a try/except statement? I put the floating function in one to guard against input other than numbers and it worked but writing the values like 0.0 <= grade <= 1.0 was ignored completely.

At 22:00 can we just use .isalpha() to determine if the character is a letter of the alphabet?

at 17:30 why is it necessary to initialise k as None outside the for loop: Aren't we initialising it as max_address in the for loop. Or is it just in case we have an empty dictionary? Also, for question 4, say if we had more than one key with a max value of 5 in the dictionary, how would we get all those keys. I think this code only returns the first key with the maximum value. I noticed this when trying to get the min values (where there are several keys in the dictionary with values of 1) and haven't yet found a solution for getting all those keys with minimum values. Any ideas? Convert the dictionary to a sorted tuple and go from there?

Thanks for these. At 8:00, could we just split the string on its spaces and get the second item in the resulting list which will always be the number part of the string. I did this and it worked fine.

Thank you for the videos. At 6:44 how were you able to select all instances of print and edit them at once?

Nice video, but the camera needs to swivel upwards about 1 degree , you'll get more head view...

Why by importing the processed csv file back into ERP software - all my data end up in one cell?

The regular expression: /^(1\s?)?(\d{3}|\(\d{3}\))[\-\s]?\d{3}[\-\s]?\d{4}$/

ok Tom

Awesome

great work brother

Thanks man! You are doing great 👍

i have solved the work on exercise 4 chapter 9, am stuck in printing how many messages the person has

Really good explanation ❤👍

const currencyUnit = { "PENNY": 1, "NICKEL": 5, "DIME": 10, "QUARTER": 25, "ONE": 100, "FIVE": 500, "TEN": 1000, "TWENTY": 2000, "ONE HUNDRED": 10000 } function checkCashRegister(price, cash, cid) { let changeSum = cash * 100 - price * 100 ; let changeSumCheck = changeSum; let change = []; let status = ''; let cidSum = 0; let filteredCid = cid.filter(elem => elem[1] !== 0).reverse(); filteredCid.forEach(elem =>{ let curr = elem[0] let currSum = elem[1] * 100; cidSum += currSum; let amount = 0; while(changeSum >= currencyUnit[curr] && currSum > 0) { amount += currencyUnit[curr]; changeSum -= currencyUnit[curr]; currSum -= currencyUnit[curr]; } if (amount !== 0) { change.push([curr, amount /100]); } }); if(changeSum > 0 ) { status = 'INSUFICIENT_FUNDS'; change = []; }else if (changeSum == 0 && changeSumCheck == cidSum){ status = 'CLOSED'; change = cid; } else { status = 'OPEN'; } return{ 'status': status, 'change': change }; } console.log(checkCashRegister(19.5, 20, [["PENNY", 0.5], ["NICKEL",0 ], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]])); aqui tá dando erro

thank you for this video

Plz answer me ...I am waiting for u r reply I am fresher

Can u plz tell me from which plate form u got this knowledge I am very obliged to u

Unbelievable... amazing

thanks, this is nice!

Thank you very much, I learned more about regExp. in this video than in the course.

clear and functionaly syntax its look right to the concept. Thanks brò

Do you have a solution for Exercise 1? Thanks.

For exercise 1: The code only works with http URLs, not https. This is because the socket code provided is making a plain HTTP request on port 80. HTTPS requests require an SSL/TLS layer on top of the TCP connection, typically operating on port 443. Handling HTTPS requests requires additional libraries like ssl in Python and a more complex setup. Websites that begin with "https" use the HTTPS protocol, and thus this code would fail for such URLs. This code does work for any website: import socket import ssl def fetch_url(url): parts = url.split('/') if len(parts) < 3: raise ValueError("Invalid URL format") host = parts[2] path = '/' + '/'.join(parts[3:]) if len(parts) > 3 else '/' # Create a raw socket mysock = socket.socket(socket.AF_INET, socket.SOCK_STREAM) # Wrap socket with SSL/TLS for HTTPS if url.startswith('https'): context = ssl.create_default_context() mysock = context.wrap_socket(mysock, server_hostname=host) port = 443 else: port = 80 mysock.connect((host, port)) cmd = f'GET {path} HTTP/1.1 Host: {host} User-Agent: CustomSocketClient Connection: close ' mysock.send(cmd.encode()) data_received = [] while True: data = mysock.recv(512) if len(data) < 1: break data_received.append(data.decode(errors='ignore')) response = ''.join(data_received) header, body = response.split(' ', 1) # If the response is a redirect (e.g., 301 or 302), follow it if "302 Found" in header or "301 Moved Permanently" in header: for line in header.split(" "): if line.startswith("Location:"): redirect_url = line.split(": ", 1)[1] return fetch_url(redirect_url) return body try: url = input("Enter in URL: ") content = fetch_url(url) print(content) except ValueError: print("Invalid URL format")

There are really nice bicycle streets and paths ! So beautiful