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How can I download a specific Maven artifact in one command line?

...lns="http://maven.apache.org/PLUGIN_REGISTRY/1.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> <plugins> <plugin> <groupId>org.apache.maven.plugins</groupId> <artifactId>maven-dependency-plugin</artifactId> <useVersion>...
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Array Size (Length) in C#

... is a rectangular multi-dimensional array (for example, int[,] b = new int[3, 5];) b.Rank will give the number of dimensions (2) and b.GetLength(dimensionIndex) will get the length of any given dimension (0-based indexing for the dimensions - so b.GetLength(0) is 3 and b.GetLength(1) is 5). S...
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Python str vs unicode types

... ways into a sequence of binary data represented via str. Note: In Python 3, unicode was renamed to str and there is a new bytes type for a plain sequence of bytes. Some differences that you can see: >>> len(u'à') # a single code point 1 >>> len('à') # by default utf-8 -&gt...
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How to select Python version in PyCharm?

... | edited Dec 30 '14 at 21:56 Eric Leschinski 114k4949 gold badges368368 silver badges313313 bronze badges ...
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UnicodeDecodeError: 'ascii' codec can't decode byte 0xd1 in position 2: ordinal not in range(128)

... 153 Unicode is not equal to UTF-8. The latter is just an encoding for the former. You are doing it ...
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How to get commit history for just one branch?

... Adam Rosenfield 347k9090 gold badges477477 silver badges564564 bronze badges answered Jun 7 '13 at 0:06 alexalex ...
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Reading a binary file with python

... | edited Jan 3 '12 at 11:15 answered Jan 3 '12 at 10:46 ...
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How to efficiently compare two unordered lists (not sets) in Python?

... 253 O(n): The Counter() method is best (if your objects are hashable): def compare(s, t): retu...
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Join a list of strings in python and wrap each string in quotation marks

...clear with the quotes – jamylak May 3 '17 at 2:47 1 @jamlak ok, repr just seemed safer to me inca...
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Given an array of numbers, return array of products of all other numbers (no division)

... 1, a[0], a[0]*a[1], a[0]*a[1]*a[2], } { a[1]*a[2]*a[3], a[2]*a[3], a[3], 1, } Both of which can be done in O(n) by starting at the left and right edges respectively. Then multiplying the two arrays element by element gives the required result My...