python

[Solved] How to print a work in string containing specific characters [closed]

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You could use the following regular expression: import re text = “fieldnameRelated_actions/fieldnamedatatypeRESOURCE_LIST![CDATA[nprod00123456]/value>value>![CDATA[nprod00765432]]/valuevaluesfield” print re.findall(r'(nprod\d+)’, text) Giving you: [‘nprod00123456’, ‘nprod00765432’] This works by finding any nprod in the text followed by one or more digits. Alternatively, without re, it might be possible as follows: print [‘nprod{}’.format(t.split(‘]’)[0]) for t in text.split(‘nprod’)[1:]] 1 solved How to print a … Read more

[Solved] Why are there so many formatting flavours in Python?

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This is what evolution looks like. “We need string concatenation”. Sure. “hello” + ”’ ”’ + ‘world’ or, if you like, ‘ ‘.join([‘hello’, ‘world’]) “But “found ” + str(var) + ” matches” is kind of clumsy. I hear Lisp has princ…?” Oh okay, here: ‘found %i matches’ % var (… Time passes… ) “I hear … Read more

[Solved] Python Error with iteration, for loop

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Please look at the formatting tips, it will help you display your code as you wish. that said, I think this is what you want. The following code will iterate through each character in your list string. If it finds a match, it will print success list = “9876554321” for char in list: if char … Read more

[Solved] Translate regular expression to python

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You can use the Regex pattern: (?:[A-Za-z.]*/PERSON\s*)+ [A-Za-z.]* matches zero or more of [A-Za-z.] /PERSON\s* matches /PERSON followed by zero or more whitespace The above is put in a non-captured group and the group is matched one or more time by the + token. Example: In [9]: re.search(r'(?:[A-Za-z.]*/PERSON\s*)+’, ‘Leo/PERSON Messi/PERSON hello’).group() Out[9]: ‘Leo/PERSON Messi/PERSON ‘ … Read more

[Solved] evaluate a python string expression using dictionary values

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At the end I found a solution to my problem: text = “‘my_home1’ in houses.split(‘,’) and ‘2018’ in iphone.split(‘,’) and 14 < mask” dict = {‘house’ : ‘my_home1,my_home2’, ‘iphone’ : ‘2015,2018’, ‘mask’ : ’15’ } expression = ‘false’ for key in dict.keys(): if isinstance(dict.get(key), str): text = re.sub(key, ‘\'{}\”.format(dict.get(key)), text) else: text = re.sub(key, dict.get(key), … Read more

[Solved] Python: Levenberg- Marquardt algorithm [closed]

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The requirement for the Levenberg Marquard algorithm is that you need to be able to calculate the jacoboan (derivative with respect to your parameter). If this is the case for your problem then yes. I guess that it is not. Perhaps the simplex algorithm is what you are looking for. solved Python: Levenberg- Marquardt algorithm … Read more

[Solved] One line recursive loop in python

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The list comprehension at the heart of this function is building up an implicit list that ultimately gets returned. We can unwind the function, remove the comprehension, and use an explicit list: def combine(n, k): “”” Given two integers n and k, return all possible combinations of k numbers out of 1 … n. “”” … Read more

[Solved] Why does return not return my Variable?

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I think you miss a return in the first function def get_input(high_run, low_run, run): pressed_key = input(”) if pressed_key == “n”: # Next Page return set_high_run_np(high_run, run), set_low_run_np(low_run, run) def set_high_run_np(high_run, run): if high_run != len(ldata): high_run = high_run + run return high_run def set_low_run_np(low_run, run): if low_run != len(ldata) – run: low_run = low_run … Read more

[Solved] WMI lib to start windows service remotely

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There is documentation regarding the library on github: https://github.com/tjguk/wmi/blob/master/docs/cookbook.rst I believe the above code is throwing an error because you are not specifying which service to start. Assuming you don’t know what services are available to you: import wmi c = wmi.WMI() # Pass connection credentials if needed # Below will output all possible service … Read more

[Solved] How to use re.findall to get the url string? [closed]

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re.findall(r”‘https://.*?'”, part_of_html) re.findall(pattern, string, flags=0) Return all non-overlapping matches of pattern in string, as a list of strings. The string is scanned left-to-right, and matches are returned in the order found. If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the … Read more