[ad_1] Use the ord and chr functions. Changing a letter to the next would be something like next_letter = chr(ord(letter) + 1) 2 [ad_2] solved How would I make an encrypt code in Python? [closed]
[ad_1] The memory is being managed by your numpy arrays. As soon as they go out of scope (most likely at the end of the PySparse constructor) the arrays cease to exist, and all your pointers are invalid. This applies to both large and small arrays, but presumably you just get lucky with small arrays. … Read more
[ad_1] To click on the element with text as Show 10 more deals on the page https://www.uswitch.com/broadband/compare/deals_and_offers/ you can use the following solution: Code Block: from selenium import webdriver from selenium.webdriver.support.wait import WebDriverWait from selenium.webdriver.support import expected_conditions as EC from selenium.webdriver.common.by import By url = “https://www.uswitch.com/broadband/compare/deals_and_offers/” options = webdriver.ChromeOptions() options.add_argument(“start-maximized”) options.add_argument(‘disable-infobars’) browser = webdriver.Chrome(chrome_options=options, executable_path=r’C:\Utility\BrowserDrivers\chromedriver.exe’) … Read more
[ad_1] This is a stack trace. It shows not just where the actual error occurred directly, but also what the program was doing while it occurred. The code df.loc[‘origin’ == ‘US’] (on line 6) is at the top of the trace, meaning it’s the root cause, but this by itself is not an error. This … Read more
[ad_1] Python has an eval() function that can do this: a = “https://stackoverflow.com/” b = “6” c = “3” print eval(b + a + c) However, please note that if you’re getting input from a remote source (like over a network), then passing such code to eval() is potentially very dangerous. It would allow network … Read more
[ad_1] This can be a start. This checks whether the number N is divisible by all numbers from 2 to int(sqrt(N)) + 1, where the int function truncates the square root of N. The all() function in python returns True if all members of a list satisfy some condition (here not zero). You should set … Read more
[ad_1] Replace last main() with main(game) and make this small changes, remember to read the python tutorials, good luck! import random import math game = int(input(“How many problems do you want?\n”)) num_1 = random.randint(1,10) num_2 = random.randint(1,10) def main(game): random.seed() count = 0 correct = 0 result = 0 #Here we initialized result to 0 … Read more
[ad_1] The build fails because you are using Windows instead of Linux. This is clearly stated in the installation instructions you claim to have followed. notes The build tasks rely on Linux shell commands such as pkill and rsync so are unlikely to run on other OS’s without some tweaks. If you want to run … Read more
[ad_1] To get the date-string in bash: echo “Current: ” $(date +”%Y”-“%m”-“%d”T”%H”-“%M”-“%S”) echo “+2 min : ” $(date –date=”@$(($(date +%s)+120))” +”%Y”-“%m”-“%d”T”%H”-“%M”-“%S”) prints Current: 2014-09-10T15-58-15 +2 min : 2014-09-10T16-00-15 Read the time from string and print +2min string str=”2014-09-10T15-58-15″ new=$(date –date=”@$(($( IFS=”-T” read y m d H M S <<< “$str”;date –date=”$y-$m-${d}T$H:$M:$S” +%s )+120))” +”%Y”-“%m”-“%d”T”%H”-“%M”-“%S”) echo … Read more
[ad_1] Create an intersection of the keys, then access the value in both dictionaries: {k: dict1[k] * dict2[k] for k in dict1.viewkeys() & dict2} This uses dictionary views which act as sets (and & creates a set intersection). In Python 3 you get dicitonary views via the default methods: {k: dict1[k] * dict2[k] for k … Read more
[ad_1] I agree that you should consider changing your single line comments to use the pound symbol. I ran your code and once I indented the multiline comments, the syntax error disappeared. As the other answers mentioned, following an if statement, the compiler expects either once-more indented code or an elif statement or new code … Read more
[ad_1] Assuming Benoît Latinier’s interpretation of your question is right (which, it looks like it is), then there will be some cases where a unique letter can’t be found, and in these cases you might throw an exception: def unique_chars(words): taken = set() uniques = [] for word in words: for char in word: if … Read more
[ad_1] You can pair adjacent numbers, calculate the differences between the pairs, and determine that the list forms an arithmetic progression if the number of unique differences is no greater than 1: from operator import sub def progression(l): return len(set(sub(*p) for p in zip(l, l[1:]))) <= 1 so that: print(progression([3])) print(progression([7,3,-1,-5])) print(progression([3,5,7,9,10])) outputs: True True … Read more
[ad_1] From the documentation: Indentation is rejected as inconsistent if a source file mixes tabs and spaces in a way that makes the meaning dependent on the worth of a tab in spaces; a TabError is raised in that case. You state the body of the with statement is indented with 3 tabs. If that … Read more
[ad_1] Let’s add some tracing so we can see order-of-operations: import sys def findIdx(ary, tgt): retval = ary.index(tgt) print(‘First instance of %r in %r is at position %r’ % (tgt, ary, retval), file=sys.stderr) return retval data1 = [1.48, -4.96] i = 0 mn = data1[1] k = findIdx(data1, mn) data1[i], data1[k] = data1[k], data1[i] print(“Prior … Read more