[ad_1] go and download packages from pypi. After that transport packages to your offline pc. open cmd and use this command. “pip install [Your packages path]”. 0 [ad_2] solved Installing python packages offline [closed]
[ad_1] I would say it does not. The lack of agreement with the earlier values for Jan-Sep is a bad sign. The large negative value for Nov suggests that there’s something wrong. What do these numbers represent? Is a negative value physically possible? It’s a mistake to treat a function like a black box. I … Read more
[ad_1] Print works differently in Python 3 Try changing print “Warning: Url load error ” + str(e) + ” for url ” + url to print(‘Warning: Url load error {} for url {}’.format(e, url)) 0 [ad_2] solved urllib2.urlopen – error opening url [closed]
[ad_1] One way would be to read each line of the file into a list. Then find the index of the first list item to remove and the last list item to remove and use del[first:last+1] to remove those, then use list.insert() to put in the new rows starting at first. [ad_2] solved Deleting lines … Read more
[ad_1] I wouldn’t recommend this course of action. You will probably get all kind of replies telling how you can do this using setattr and friends, but in practice dynamically-created variable names are almost invariably (see what I did there?) a bad idea. Suppose you already have a variable called X_list and then you process … Read more
[ad_1] The problem with your code is simple. When count equals 3, you want to add a newline after printing out the last digit, otherwise the next set of 4 numbers will be printed on the same line. So, in summary, this is what you do: if count == 3: print(i) count = 0 One … Read more
[ad_1] If you mean that you have several fields like 98 but they all contain a title, you could do this: titles = list() for k in my_dict[“countries”].keys(): if my_dict[“countries”][k].has_key(“title”): titles.append(my_dict[“countries”][k][“title”]) or, as suggested in the comments try: titles = [item[‘title’] for item in dictName[‘countries’]] except KeyError: print(“no countries/title”) [ad_2] solved How to get specific … Read more
[ad_1] Did you google the error before asking here?Anyways try this BoostInstallationHowto#LD_LIBRARY_PATH.Please google before you ask here.Hope this helps you. 1 [ad_2] solved PyCuda Error in Execution
[ad_1] If you want all combinations of [0, 1] for a length of 7 then you can take the cartesian product of [0, 1] repeated 7 times, e.g.: In []: import itertools as it list(it.product([0, 1], repeat=7)) Out[]: [(0, 0, 0, 0, 0, 0, 0), (0, 0, 0, 0, 0, 0, 1), (0, 0, 0, … Read more
[ad_1] Call stron the number and str.join the result of that: i = 4316 print(“+”.join(str(i))) 4+3+1+6 [ad_2] solved Python – How to show addition of integers digits
[ad_1] How to write a code in python that is a recursive function that returns a string with each character in its argument repeated? [ad_2] solved How to write a code in python that is a recursive function that returns a string with each character in its argument repeated?
[ad_1] Not much to go on here, but after a for statement such as this, you need a line after this and 4 spaces before your next statement. A lot of programming languages use { and } to signify the start and stop of a block of code, Python uses indentation (4 spaces) to do … Read more
[ad_1] Use defaultdict() to collect parameters per url path: from collections import defaultdict from urllib import quote from urlparse import parse_qsl, urlparse urls = defaultdict(list) with open(‘links.txt’) as f: for url in f: parsed_url = urlparse(url.strip()) params = parse_qsl(parsed_url.query, keep_blank_values=True) for key, value in params: urls[parsed_url.path].append(“%s=%s” % (key, quote(value))) # printing results for url, params … Read more
[ad_1] Change the 9 to a 10: if temp2[4] == ‘-‘ and temp2[10] != ‘-‘: temp3 = temp2[:10] + ‘-‘ + temp2[10:] When you call a string its str[from:to], so temp2[:9] is equivalent to temp2[0:9] and it would only return the characters from 0-9 instead of the required 0-10 0 [ad_2] solved Python String unwanted … Read more
[ad_1] I think the recommended answer would be something along these lines: hello=”Hello” world=’World’ hello_world=hello + ‘ ‘ + world print=hello_world Run output: Hello World [ad_2] solved What’s wrong here? (concatenation) in Python