for j in xrange(10,40,5): What this loop will generate? 10,15,20,25,30,35 Notice, it will generate 10 but not 40 as the first param is inclusive, but second one is exclusive. Then which of them is not divisible by 3? 10,20,25,35 Simple as that. For your added question, see this: for char in “ABC 123”: print char, … Read more
Yes there are. Search for “bigint” on crates.io and there are plenty of results. They are not in the standard library but that’s nothing unusual for Rust. Rust’s standard library is kept small intentionally to minimise binary size ease maintenance (among other reasons). Because Rust is not Python, and there’s no reason for it to … Read more
You can split the string into a list of substrings: >>> s=”xxxxxxxx,yyyyyyyyyyyy,zzzzzzzzz” >>> parts = s.split(‘,’) >>> parts [‘xxxxxxxx’, ‘yyyyyyyyyyyy’, ‘zzzzzzzzz’] Then you can re-order: >>> reordered = parts[0] + parts[2] + parts[1] >>> reordered [‘xxxxxxxx’, ‘zzzzzzzzz’, ‘yyyyyyyyyyyy’] And rejoin: >>> ‘,’.join(reordered) ‘xxxxxxxx,zzzzzzzzz,yyyyyyyyyyyy’ 1 solved How can i change the position of substring within a … Read more
The code needs to go through 11 inner loop iterations and 3 outer loop iterations to meet the condition of the country being ‘Brazil’ and the capital city being ‘Brasilia’. The print statement line is only executed the once during the entire script when those conditions are met. At the point of this line being … Read more
This is a number guessing Python code ,but it stops after entering the first guessing number. Can you tell me where is the fault? solved This is a number guessing Python code ,but it stops after entering the first guessing number. Can you tell me where is the fault?
if ‘ddd’ or ‘eee’ in test is evaluated as: if (‘ddd’) or (‘eee’ in test): as a non-empty string is always True, so or operations short-circuits and returns True. >>> bool(‘ddd’) True To solve this you can use either: if ‘ddd’ in test or ‘eee’ in test: or any: if any(x in test for x … Read more
You need to indent statements that are intended to be in if-else blocks: i = 21 j = 23 print (i) if i > j: print(‘greater’) else: print(‘lesser’) solved What does the error ‘IndentationError: expected an indented block in python’ mean? [closed]
You can implement a simple linear SVM with numpy only like below. BTW, please google before you ask question next time. There are lots of resources and tutorial online. import numpy as np def my_svm(dataset, label): rate = 1 # rate for gradient descent epochs = 10000 # no of iterations weights = np.zeros(dataset.shape[1]) # … Read more
Issues with Code : Object creation is not correct for password class. also argument self is not passed to class method . Fixed Code class password: def pas1(self): pas = [] tedad = int(input(‘how many do you have information ? : ‘)) for i in range(1,tedad): b=input(‘enter : ‘) pas.append(b) print(‘this is your pas —> … Read more
Use a for loop to iterate over the months: def prepareadres(city,y,m,d): months = [31,28,31,30,31,30,31,31,30,31,30,31] d = d + 1 d0 = 0 m0 = 0 for month in months: if d > month: d = d – month m += 1 if d == sum(months): print(“year complete”) #print(df.iloc[0, :-2]) if (m < 10) and (d … Read more
I’ll assume that your string is always of the form “<emoticon><alphabets>”. I’ll then splice the string. # Count number of alphabets first num = [c.isalpha() for c in string].count(True) # Splice string based on the result s1 = string[:-num] s2 = string[-num:] 5 solved Python split unicode characters and words
Dude It’s super easy: def triangle(n): for i in range(1, n +1): print ‘ ‘ * (n – i) + ‘x’ * i Or even: def triangle(n): for i in range(1, n +1): print (‘x’ * i).rjust(n, ‘ ‘) output for triangle(5): x xx xxx xxxx xxxxx Dont just copy this code without comprehending it, … Read more
From Python HTMLParser Documentation: from HTMLParser import HTMLParser # create a subclass and override the handler methods class MyHTMLParser(HTMLParser): def handle_starttag(self, tag, attrs): print “Encountered a start tag:”, tag def handle_endtag(self, tag): print “Encountered an end tag :”, tag def handle_data(self, data): print “Encountered some data :”, data # instantiate the parser and fed it … Read more
So using datetime you can actually get the minutes like this: from datetime import datetime time = datetime.now() minutes = time.minute Or in one line: from datetime import datetime minutes = datetime.now().minute Now that you have the minutes, the if statement can be simplified down, because you don’t look at the hour. if minutes == … Read more
[EDIT]The question was updated and the below was answered based on the original question. if you fix your input var1 then you can do something like this: var1 = ‘{“text”:”tool”,”pos”:”\\xd1\\x81\\xd1\\x83\\xd1\\x89\\xd0\\xb5\\xd1\\x81\\xd1\\x82\\xd0\\xb2\\xd0\\xb8\\xd1\\x82\\xd0\\xb5\\xd0\\xbb\\xd1\\x8c\\xd0\\xbd\\xd0\\xbe\\xd0\\xb5″}’ md = {} for e in var1[1:-1].split(‘,’): md[e.split(‘:’)[0][1:-1]] = e.split(‘:’)[1][1:-1] md[‘pos’] = (bytes.fromhex(”.join([h for h in md[‘pos’].split(‘\\x’)]))).decode(‘utf-8’) print(md) output: {‘text’: ‘tool’, ‘pos’: ‘существительное’} 2 solved … Read more