Here’s the first one that I see: input(“\n\n\nPress enter to exit menu) And the second one is that you forgot to escape the newline in code at the end of this line: + \n by the patient himself/herself. 3 solved EOL while scanning string literal, file won’t run [closed]
You need to sort the other lists based on the order of ‘Functions’, and sort that list last. One way to implement this is using zip to combine e.g. ‘Functions’ and ‘Action’ into a single list [(‘Transfer Amount’, ‘N’), …], sort that, then extract the second value from each pair (using e.g. map and operator.itemgetter): … Read more
What you are after is an awk script of the following form: $ awk ‘(NR==FNR){a[FNR]=$0;next} !(FNR in a) { print “file2 has more lines than file1”; exit 1 } { print (($0 == a[FNR]) ? “matching” : “not matching”) } END { if (NR-FNR > FNR) print “file1 has more lines than file2”; exit 1}’ … Read more
Here’s one utility function that worked for me: def printList(L): # print list of objects using string representation if (len(L) == 0): print “[]” return s = “[” for i in range (len(L)-1): s += str(L[i]) + “, ” s += str(L[i+1]) + “]” print s This works ok for list of any object, for … Read more
Getting data out from a pdf file is pretty messy. If the pdf table is ordered and has got a unique pattern embedded along with it, the best way to get the data is by converting the pdf to xml. For this you can use: pdftohtml. Installation: sudo apt-get install pdftohtml Usage: pdftohtml -xml *Your … Read more
If I was developing this, I would: Keep all the words in a single set Skip the first question (you can determine word length by the length of the 2nd question) Set up a while loop for each question you ask the user so that it repeats the same question on invalid input. To check … Read more
I don`t really understand the question, but if you want the highest number of consecutive elements in list, maybe something like this from itertools import groupby list = [1,1,1,0,0,1,1,1,1,1] count_cons_val = [sum(1 for _ in group) for _, group in groupby(list)] print(max(count_cons_val)) Output: 5 solved Highest number of consecutively repeating values in a list
This is an indentation error. The if statements that print the password at the end are inside the symbol_want if statement. You need to unindent the print statements by one level, and it will work fine. import random def password(): normal_adjectives = [‘Funny’, ‘Amazing’, ‘Infinity’, ‘Fabulous’, ‘Red’, ‘Rainbow’, ‘Adorable’, ‘Adventurous’, ‘Impressive’, ‘Determined’, ‘Delighted’, ‘Scary’, ‘Active’, … Read more
You can do a program such as this: import datetime time_1 = datetime.datetime(2018, 5, 5, 12, 5, 1) time_2 = datetime.datetime(2018, 5, 5, 14, 5, 1) new_time = time_2 – time_1 seconds = new_time.total_seconds() time = seconds / 60 print(int(time)) The time difference is two hours just, similar to what you’re asking for. A timedelta … Read more
I would use the BeautifulSoup library. Here is how I would grap this info knowning that you already have the HTML file : from bs4 import BeautifulSoup with open(html_path) as html_file: html_page = BeautifulSoup(html_file, ‘html.parser’) div = html_page.find(‘div’, class_=’playbackTimeline__duration’) span = div.find(‘span’, {‘aria-hidden’: ‘true’}) text = span.get_text() I’m not sure if it works, but it … Read more
The package you’re looking for has an empty description on PyPI and installs nothing when you install it. If you click on the Homepage link, it’s a GitHub repo containing nothing but a README file saying: I really don’t have time to maintain this library further. If you want to fork it or use it … Read more
I think you are misunderstanding the “:” notation for the lists. upper[:3] gives the first 3 characters from upper whereas upper[3:] gives you the whole list but the 3 first characters. In the end you end up with : upperNew = upper[:3] + upper[3:] = ‘ABC’ + ‘DEFGHIJKLMNOPQRSTUVWXYZ’ When you sum them into upperNew, you … Read more
If you already have a list of filenames use collections.Counter: from collections import Counter files = [“foo.txt”,”foo.txt”,”foobar.c”] c = Counter(files) print (c) Counter({‘foo.txt’: 2, ‘foobar.c’: 1}) The keys will be your files and the values will be how many times the file appears in your list: solved I have a list of file names and … Read more
One reasonably simple approach: n = 5 s = sum(range(n+1)) for i in range(n): s -= i print(s) 15 14 12 9 5 1 solved Sum from 1 to n, 2 to n, … n in python
First, you should avoid using reserved words (such as input) as variables (now input is redefined and no longer calls the function input()). Also, a dictionary cannot be sorted. If you don’t need the keys, you can transform the dictionary into a list, and then sort it. The code would be like this: input_dict = … Read more