[ad_1] 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 … Read more
[ad_1] 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
[ad_1] 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 [ad_2] solved I have given nested list accessing via … Read more
[ad_1] 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 … Read more
[ad_1] 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 … Read more
[ad_1] 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 … Read more
[ad_1] How can I solve this? mod_wsgi (pid=3445): Target WSGI script ‘/www/folder/index.py’ cannot be loaded as Python module [ad_2] solved How can I solve this? mod_wsgi (pid=3445): Target WSGI script ‘/www/folder/index.py’ cannot be loaded as Python module
[ad_1] 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 … Read more
[ad_1] 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], … Read more
[ad_1] 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 … Read more
[ad_1] 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 … Read more
[ad_1] 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 [ad_2] solved How to reuse the terminal after I ran a Bokeh app in the server
[ad_1] 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 … Read more
[ad_1] 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: … Read more
[ad_1] 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)} [ad_2] solved How to multiply dictionary keys? [duplicate]