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Ruby Gem install Json fails on Mavericks and Xcode 5.1 - unknown argument: '-multiply_definedsuppres

... 10 Answers 10 Active ...
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How to parse Excel (XLS) file in Javascript/HTML5

...he function. <script src="https://cdnjs.cloudflare.com/ajax/libs/xlsx/0.8.0/jszip.js"></script> <script src="https://cdnjs.cloudflare.com/ajax/libs/xlsx/0.8.0/xlsx.js"></script> <script> var ExcelToJSON = function() { this.parseExcel = function(file) { var read...
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Capitalize only first character of string and leave others alone? (Rails)

... that the only letter changed is the first one. new_string = string.slice(0,1).capitalize + string.slice(1..-1) Update: irb(main):001:0> string = "i'm from New York..." => "i'm from New York..." irb(main):002:0> new_string = string.slice(0,1).capitalize + string.slice(1..-1) => "I'm ...
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How to get the index of a maximum element in a numpy array along one axis

... >>> a.argmax(axis=0) array([1, 1, 0]) share | improve this answer | follow | ...
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Creating a new column based on if-elif-else condition

... 140 To formalize some of the approaches laid out above: Create a function that operates on the rows...
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Less aggressive compilation with CSS3 calc

The Less compilers that I'm using ( OrangeBits and dotless 1.3.0.5 ) are aggressively translating 4 Answers ...
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Join strings with a delimiter only if strings are not null or empty

...ple: [...].filter(x => typeof x === 'string' && x.length > 0) will only keep non-empty strings in the list. -- (obsolete jquery answer) var address = "foo"; var city; var state = "bar"; var zip; text = $.grep([address, city, state, zip], Boolean).join(", "); // foo, bar ...
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How do I apply a perspective transform to a UIView?

... 330 As Ben said, you'll need to work with the UIView's layer, using a CATransform3D to perform the l...
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How to center a “position: absolute” element

...may use: position: absolute; margin-left: auto; margin-right: auto; left: 0; right: 0; text-align: center; share | improve this answer | follow | ...
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Flatten nested dictionaries, compressing keys

...t(items) >>> flatten({'a': 1, 'c': {'a': 2, 'b': {'x': 5, 'y' : 10}}, 'd': [1, 2, 3]}) {'a': 1, 'c_a': 2, 'c_b_x': 5, 'd': [1, 2, 3], 'c_b_y': 10} share | improve this answer | ...