Commented for explanation of approach. Have found two addresses where ID from sheet2 comes back onto sheet1 import io sheeta = pd.read_csv(io.StringIO(“”” house_number street suburb 0 43 Smith Street Frewville 1 45 Smith Street Frewville 2 47 Smith Street Frewville 3 49 Smith Street Frewville 4 51 Smith Street Frewville 5 53 Smith Street Frewville … Read more
Well one suggestion is to use NLP Linguistic Features For ease, i will be using spacy import spacy nlp = spacy.load(‘en_core_web_md’) doc = nlp(‘Get me all sales numbers for August in the Delhi’) for token in doc: print(token.text,token.dep_,token.pos_) Output | ‘Word’ | ‘DEPENDENY’ | ‘POS’ | ———————————- |’Get’ | ‘ROOT’ | ‘AUX’ | |’me’, | … Read more
patientsList = [[‘sagar’,’9856782311′],[‘mahsh’,’7865423158′]] search = input(“Enter mobile number of patient to search: “) for patient in patientsList: ”’ patient looks like this [‘sagar’,’9856782311′] patient[0] is the name patient[1] is the phone number ”’ if search == patient[1]: print(“Required patient is”) print(patient) print(“is Appointed”) break 9 solved I have given nested list accessing via for loop, … Read more
The error occurs when you type in a letter instead of a digit. Also, you should test for the validity of the user input before trying to delete the number on that index: from numpy import random dice = [1, 2, 3, 4, 5, 6] first = [random.choice(dice) for i in range(5)] print(first) while True: … Read more
It is only run after the nested loop. Looking at the output from the post you referenced, the program starts with 1, prints one 1 without a newline, then exits the nested loop, then prints the newline. Then it enters the nested loop with 2, loops twice (prints a 2 without a newline, then prints … Read more
Strings don’t have a font size. Strings store sequences of characters (unicode strings) or bytes to be interpreted as characters (byte strings). Font size is an element of presentation, and is a function of whatever presentation and rendering system you are using. As you mention an email, you could create a multipart email with an … Read more
How can I solve this? mod_wsgi (pid=3445): Target WSGI script ‘/www/folder/index.py’ cannot be loaded as Python module solved How can I solve this? mod_wsgi (pid=3445): Target WSGI script ‘/www/folder/index.py’ cannot be loaded as Python module
You can do it by: a =[ [0,1,0,0,0,1], [4,1,0,3,2,0], [0,0,1,0,0,0], [0,1,0,0,1,0], [0,0,0,0,1,0], [0,0,0,0,0,0], ] b = [0, 1, 3] c = [[a[i][j] for j in b] for i in b] print(c) # ===> # [[0, 1, 0], [4, 1, 3], [0, 1, 0]] (Anyway, Python2 is no longer supported: link. I recommend you use Python3+NumPy, … Read more
It is still not really clear what you want. If you want multidimensional array you can use lists: >>> matrix = [[None]*10 for x in range(3)]#replace 10 with 1000 or what ever >>> matrix [[None, None, None, None, None, None, None, None, None, None], [None, None, None, None, None, None, None, None, None, None], [None, … Read more
Your problem is that you never eliminate the last number in numbers. If range_number is 21, then len(numbers) is 20 and len(numbers)-1 is 19. So this line here: for n in range(i + numbers[i], len(numbers)-1, numbers[i]): Never actually removes the number 20 from the list. You could have seen this if you’d printed out the … Read more
input returns a string, and 0 != “0” Either parse your strings immediately parx = int(input(“Write your parX: “)) pary = int(input(“Write your parY: “)) Or check against strings while pary != “0” and parx != “0”: Note too that you should be using some error checking. If the user enters a non-number, your program … Read more
Press Ctrl + C to send a terminating signal, this will close the Bokeh server so you can reuse the terminal. (This isn’t just for Bokeh, it also works for other applications!) 1 solved How to reuse the terminal after I ran a Bokeh app in the server
First of all, use .strip() to remove the newline characters. The answers comparison might not always work because of them. To ask every even row you can use the modulo (%) operator to get the remainder of the current line index after dividing it by two. If the answer is correct we add 1 to … Read more
Your xpath expressions aren’t correct. When you are using relative xpath expressions they need to start with a “./” and using class specifiers is much easier than indexing in my opinion. def parse(self, response): for row in response.xpath(‘//table[@class=”list”]//tr’): name = row.xpath(‘./td[@class=”name”]/a/text()’).get() address = row.xpath(‘./td[@class=”location”]/text()’).get() yield { ‘Name’:name, ‘Address’:address, } next_page = response.xpath(“//a[@class=”next-page”]/@href”).get() if next_page: yield … Read more
I believe you want is: Multiply each key with 2. Not the square of it. d = {2: (1,2), 8: (2,4), 30: (10,3)} d = {k*2: v for (k, v) in d.items()} print(d) OUTPUT: {4: (1, 2), 16: (2, 4), 60: (10, 3)} solved How to multiply dictionary keys? [duplicate]