[ad_1] From what I can see of the html, there is no span with id=”sexo- button”, so BeautifulSoup(login_request.text, ‘lxml’).find(“span”,id=”sexo- button”) would have returned None, which is why you got the error from get_text. As for your second attempt, I don’t think bs4 Tags have a value property, which is why you’d be getting None that … Read more
[ad_1] Sure, just send a edit_message passing the new modified embed contents, and it should work. 4 [ad_2] solved Python Discordpy Is there a way to modify a message embed?
[ad_1] In R, using dplyr: data %>% group_by(Name, CoordinateX, CoordinateY) %>% arrange(desc(Date)) %>% distinct() %>% ungroup() Give the output: Name Date CoordinateX CoordinateY Aaa 2018-08-29 650000 134999 Bbb 2010-08-29 650000 134999 Bbb 2010-08-29 655600 134999 Ccc 2010-08-29 655600 134999 [ad_2] solved Remove duplicate on multiple columns keep newest [closed]
[ad_1] divided_list = [float(x)/2 for x in foo] This divides each value in a list by 2. def divide_list(foo, divisor): return [float(x)/divisor for x in foo] Where divisor is the number you want to divide by. 5 [ad_2] solved Create class in python [closed]
[ad_1] This question is a bit ambiguous as you did not provide example input or output. I will assume that you are going to ask the user multiple times to enter in an item and how many they need separated by a comma. If the input changes, then I will edit my answer. dictionary = … Read more
[ad_1] Here is a simple function to do what you want: def get_length_or_value(l, *i): item = l for index in i: item = item[index] return len(item) if isinstance(item, list) else item This function first finds the element at the given indices (if any), and then returns the appropriate value based off wether the element is … Read more
[ad_1] The code is almost perfect. It just can’t crop on the right side because of the scrollbar. And it does not consider some padding (which you liked according the comments). The thing I removed is the topology modification. import numpy as np import cv2 img = cv2.imread(‘cropwhitefromimage.png’) scrollbarright = 20 img = img[:, :-scrollbarright] … Read more
[ad_1] Use squaring = element * element against squaring = list[element] * list[element] because for the first element, for example, you’re consulting list[-9] and that index is out. 0 [ad_2] solved Given a sorted list of integers, square the elements and give the output in sorted order
[ad_1] The first output shows a comma because without it, 1 being the only element, (1) would be just a integer (parentheses are wrapping the expression 1), (1,) is shown to differentiate tuples and simple parentheses. in the second one, no trailing comma is needed to differentiate tuples, since there are more than one element. … Read more
[ad_1] You can use zip to process two lists in parallel. Zip will stop iteration at the shorter of the two lists. mashemup = [x.split()[0] + ‘ ‘ + y.split()[1] for x, y in zip(meeps, peeps)] print(mashempup) [‘Foghorn Cleese’, ‘Elmer Palin’, ‘Road Gilliam’, ‘Bugs Jones’, ‘Daffy Chapman’, ‘Tasmanian Idle’] Or if you need a list … Read more
[ad_1] Try: lst = [1, 2, 3, 4, 5, 6, 7] out = [sum(vv < v for vv in lst[:i]) for i, v in enumerate(lst)] print(out) Prints: [0, 1, 2, 3, 4, 5, 6] 4 [ad_2] solved How to find the count of numbers that are less than each element of a list [closed]
[ad_1] The mergesort function you call doesn’t modify its argument. Rather, it returns a new sorted list. A simple fix would be: def mergesort(data): if len(data) < 2: return data # Fix1 left = data[:len(data)//2] print(left) right = data[len(data)//2:] print(right) print(“left only now”) left = mergesort(left) # Fix2 print(“right now”) right = mergesort(right) # Fix3 … Read more
[ad_1] I am note sure if you can omit that loop in an efficient way, but when we do not use np.delete() but just use the indices to shrink payoff in every iteration I already got a big speed increase on my machine: # … arr = np.full(N+1, d/u) arr[0] = S*np.power(u, N) arr[1:] = … Read more
[ad_1] I figure this out finally. Assignment Expressions in Python 3.8 is what I am looking for. if name in name_dict: info = name_dict[‘name’] ## <– may be a more complicated function if info == ”: info = input(“Input your info: “) else: info = input(“Input your info: “) Can be turn into: if (name … Read more
[ad_1] I think you are referring to something like this: self.EPG_Channel=”Channel 5″ self.channel_str = [‘BBC One, BBC Two, ITV, Channel 4, Channel 5’] if self.EPG_Channel == self.channel_str[-1]: pass 1 [ad_2] solved Check if the element is last in the list