[ad_1] You can use custom fonts in pygame. Just find one that uses the Greek alphabet and use that. Use pygame.font.SysFont() and load your new font and enjoy. 1 [ad_2] solved Is it possible to use greek letters in pygame?
[ad_1] This code extracts that particular site’s content a little better. def keyInfo(div): print(div.find(“h1”).get_text()) article = div.find(“article”) divText = article.find(“div”, id=”storytext”) [a.extract() for a in divText.findAll(“aside”)] [d.extract() for d in divText.findAll(“div”)] print(divText.get_text()) Approach After looking at the structure of the content using Chrome dev tools, I noticed the story content was in article > div[id=storytext], … Read more
[ad_1] Python’s comparison operators wisely refuse to work for variables of incompatible types. Decide on the criterion for sorting your list, encapsulate it in a function and pass it as the key option to sort(). For example, to sort by the repr of each element (a string): l.sort(key=repr) To sort by type first, then by … Read more
[ad_1] try this: print(“Between everyone the meal costs”, bill2 / int(people)) other solution is cast people input: people = int(input(“How many people are available to pay? “)) [ad_2] solved Python Restaurant Price + Tip [closed]
[ad_1] So the answer depends on your model setup. For example, if your models.py looks like this: class Post(models.Model) user = models.ForeignKey(User) content = models.CharField(max_length=500) … def __str__(self): return self.user Then if you wanted all of the posts of the user that is logged in, you would use this code, for example, in your request … Read more
[ad_1] There are 4 things I’d like to mention. Here’s some code and its output: def srk_func(words): current = [] lastc = [] for x in words: if len(current) == 0: current.append(int(x)) elif len(current) == 1: if current[0] < int(x): current.append(int(x)) else: if len(current) >= len(lastc): lastc = current current[:] = [] current.append(int(x)) elif len(current) … Read more
[ad_1] newlist = [] for sublist in yourlist: already_in_list = False for index, newsublist in enumerate(newlist): if sublist == newsublist[:-1]: newlist[index][2] += 1 already_in_list = True if not already_in_list: newlist.append(sublist+[1]) – >>>newlist [[‘personA’, ‘banana’, 1], [‘personB’, ‘banana’, 2], [‘personA’, ‘grape’, 1], [‘personA’, ‘lemon’, 2]] [ad_2] solved Python count list and types [closed]
[ad_1] Assuming you want to stick to arithmetic operations (and not strings), use the modulo operator with 10 to get the remainder of division by 10, i.e. the unit: 12345%10 output: 5 For an arbitrary number, you need to compute the position, you can use log10 and ceil: from math import log10, ceil N = … Read more
[ad_1] Just try to correct indentation like shown below: …. try: subdomain = row.find_all(‘td’)[4].text subdomain = subdomain.replace(“*.”,””) if subdomain not in self.foundURLsList: self.foundURLsList.append(subdomain) except Exception as e: pass … Current version of bs4 does not support python 2 Beautiful Soup’s support for Python 2 was discontinued on December 31, 2020: one year after the sunset … Read more
[ad_1] You can do this with map function as well, but if with loop, here’s your answer. We have to concatenate the first elements of all three list, second with second and third with third. So make a loop going from zero to len(<any_list>) [because all three list have same len] and concatenate all three. … Read more
[ad_1] To get individual blog details in the modal you need to pass the blog id to the modal as a prop. It can be done by adding blog.id in the button data-target prop and receive the blog.id in modal id prop. the solution in given below, {% for blog in blogs %} <div class=”col-xl-12 … Read more
[ad_1] You have to add this code after your code: desktop = pywinauto.Desktop(backend=’uia’, allow_magic_lookup=False) window = desktop.windows(title=”New Tab – Google Chrome”, control_type=”Pane”)[0] result = window.descendants(control_type=”StatusBar”)[0] print(result.texts()) 0 [ad_2] solved How to get the result of the Chrome search box? [closed]
[ad_1] Here is how you can use a nested for-loop with formmatted strings: list1 = [“ilovepython”,”ilovec#”] list2 = [“love”,”python”] for i in list1: # For every string in list11 print(f’in “{i}”:’) # Print the string for j in list2: # For every string in list2 num = i.count(j) # Store the number of occurrences of … Read more
[ad_1] I suspect that’s what you’re looking for: def hourglassSum(): arr = [] for i in range(0,6): arr1=map(int,input().split()) arr.append(list(arr1)) sum=[] for i in range(len(arr)-2): for j in range(min(len(arr[i]), len(arr[i+1]), len(arr[i+2]))-2): sum.append(arr[i][j]+arr[i][j+1]+arr[i][j+2]+arr[i+1][j]+arr[i+2][j]+arr[i+2][j+1]+arr[i+2][j+2]) maxi=max(sum) print(maxi) Two notes: (1) you want to persist each iteration of the first loop, what you were doing was just overwriting your variable … Read more
[ad_1] i = 1 while i <= 5: print(i) i += 1 This works in PyCharm, looking at the image it seems that you wrote the if statement and the print() on the same line. Which also gave me an error. 0 [ad_2] solved Why isn’t the while-loop working in Python? [closed]