Well this is simple. Ill only do it for the First File, but to show how many times it comes up, you can count, but this is a one word per line test. import os file1 = open(“file1.txt”, “r”) # Make sure you added the keywords in beforehand file2 = open(“file2.txt”, “r”) #These will open … Read more
Here is a cleaned-up version: from itertools import cycle BASE = ord(‘a’) – 1 def str_to_nums(s): “”” Convert a lowercase string to a list of integers in [1..26] “”” return [ord(ch) – BASE for ch in s] def nums_to_str(nums): “”” Convert a list of integers in [1..26] back to a lowercase string “”” return “”.join(chr(n … Read more
You’re getting the points out as a string, and need to convert them to an int before doing any computation with it, ideally as soon as possible: points = int(next_line(the_file)) in next_block should do the trick. Also, you are not adding the points to the score, you’re replacing it. total = sum(points + points) score … Read more
A simple while loop works, just make sure you divide the number by 2, otherwise you’ll get the NEXT power of 2. def function(number): x = 1 while x <= number: x *= 2 return x / 2 solved Defining function with an argument, employs while loop and returns largest power of 2 that is … Read more
You can make count global: count = 0 def intro(start): if start == “yes” or start == “y”: print(“Lets begin.”) else: print(“Thanks for checking it out! Bye Bye!”) def question1(): global count count += 1 return count def question2(): global count count += 1 return count def main(): print(“Hello There! Welcome to an all new … Read more
I’m not sure I completely understand what you’re trying to do, but I’ll give it a shot since you seem to have asked this question many times already (which usually isn’t a good idea). I believe that there’s a lack of clarity in your question – it would help a lot if you could clarify … Read more
You can use urllib2 and socket module for the task. import socket from urllib2 import urlopen, URLError, HTTPError socket.setdefaulttimeout( 23 ) # timeout in seconds url=”http://google.com/” try : response = urlopen( url ) except HTTPError, e: print ‘The server couldn\’t fulfill the request. Reason:’, str(e.code) except URLError, e: print ‘We failed to reach a server. … Read more
If the key argument is in switcher, the .get() method returns the value for the key. If the key is not in the dictionary, the method returns the optional “nothing”. def numbers_to_strings(argument): switcher = {0: “zero”, 1: “one”, 2: “two”} return switcher.get(argument, “nothing”) Calling the above function with a key that is in the dictionary: … Read more
Just create lists: somelist = [p1, p2] You probably want to append those to another data structure initialized outside of your loop: somelargerlist = [] for presumed_loop_variable in presumed_loop_not_shown: p1 = … p2 = … somelist = [p1, p2] somelargerlist.append(somelist) 17 solved putting float numbers in lists [closed]
Write a Python function histogram(l) that takes as input a list of integers with repetitions and returns a list of pairs [closed] solved Write a Python function histogram(l) that takes as input a list of integers with repetitions and returns a list of pairs [closed]
Try this, >>> import datetime >>> datetime.datetime.now().strftime(‘%d-%m-%Y %H:%M:%S’) ’03-08-2019 12:43:16′ If you max_startdate is string, >>> max_startdate=”2019-08-03 01:08:58.155000″ >>> max_startdate.split(‘.’)[0] ‘2019-08-03 01:08:58’ 3 solved How i can convert ISO time format to yyyy-mm-dd hh:mm:ss using python [closed]
So to fix your code as described: def collatz(number): number = input() # number is passed in – doesn’t make sense to get it again try: number = int(number) # Just interpret once except ValueError: # try must have an except clause return # exit on error if number%2 == 0: print(‘hi’) elif number%2 == … Read more
is this perhaps what you are trying to do?: if monthly_salary < 4000: deductions = 0.12 + 0.04 + 0.01 + 0.06 elif monthly_salary >= 4000 and monthly_salary < 8000: deductions = = 0.20 + 0.07 + 0.03 + 0.06 elif monthly_salary >= 8000 and monthly_salary < 16000: deductions = 0.30 + 0.09 + 0.05 … Read more
This is what evolution looks like. “We need string concatenation”. Sure. “hello” + ”’ ”’ + ‘world’ or, if you like, ‘ ‘.join([‘hello’, ‘world’]) “But “found ” + str(var) + ” matches” is kind of clumsy. I hear Lisp has princ…?” Oh okay, here: ‘found %i matches’ % var (… Time passes… ) “I hear … Read more
At the end I found a solution to my problem: text = “‘my_home1’ in houses.split(‘,’) and ‘2018’ in iphone.split(‘,’) and 14 < mask” dict = {‘house’ : ‘my_home1,my_home2’, ‘iphone’ : ‘2015,2018’, ‘mask’ : ’15’ } expression = ‘false’ for key in dict.keys(): if isinstance(dict.get(key), str): text = re.sub(key, ‘\'{}\”.format(dict.get(key)), text) else: text = re.sub(key, dict.get(key), … Read more