flatten

I need to copy all *.doc files (but not folders whose names match *.doc ) from a network folder \\server\source (including files in all nested folders) to a local folder C:\destination without preserving the nested folders hierarchy (i.e. all files should go directly into C:\destination and no nested folders should be created in C:\destination ). In case there are several files with the same name from different subfolders of \\server\source , only the first one should be copied and never overwritten then — all conflicting files found later should be skipped (there could be many cases like this

I have a javascript object that I need to flatten into a string so that I can pass as querystring, how would I do that? i.e: { cost: 12345, insertBy: 'testUser' } would become cost=12345&insertBy=testUser I can't use jQuery AJAX call for this call, I know we can use that and pass the object in as data but not in this case. Using jQuery to flatten to object would be okay though. Thank you. You want jQuery.param : var str = $.param({ cost: 12345, insertBy: 'testUser' }); // "cost=12345&insertBy=testUser" Note that this is the function used internally by jQuery to serialize objects passed as the

I have a list of objects where objects can be lists or scalars. I want an flattened list with only scalars. Eg: L = [35,53,[525,6743],64,63,[743,754,757]] outputList = [35,53,525,6743,64,63,743,754,757] P.S. The answers in this question does not work for heterogeneous lists. Flattening a shallow list in Python Here is a relatively simple recursive version which will flatten any depth of list l = [35,53,[525,6743],64,63,[743,754,757]] def flatten(xs): result = [] if isinstance(xs, (list, tuple)): for x in xs: result.extend(flatten(x)) else: result.append(xs) return result print flatten(l) it

How to simply flatten array in jQuery? I have: [1, 2, [3, 4], [5, 6], 7] And I want: [1, 2, 3, 4, 5, 6, 7] You can use jQuery.map , which is the way to go if you have the jQuery Library already loaded. $.map( [1, 2, [3, 4], [5, 6], 7], function(n){ return n; }); Returns [1, 2, 3, 4, 5, 6, 7] Use the power of JavaScript: var a = [[1, 2], 3, [4, 5]]; console.log( Array.prototype.concat.apply([], a) ); //will output [1, 2, 3, 4, 5] Here's how you could use jquery to flatten deeply nested arrays: $.map([1, 2, [3, 4], [5, [6, [7, 8]]]], function recurs(n) { return ($.isArray(n) ? $.map(n, recurs):

Have in mind that the JSON structure is not known before hand i.e. it is completely arbitrary, we only know that it is JSON format. For example, The following JSON { "Port": { "@alias": "defaultHttp", "Enabled": "true", "Number": "10092", "Protocol": "http", "KeepAliveTimeout": "20000", "ThreadPool": { "@enabled": "false", "Max": "150", "ThreadPriority": "5" }, "ExtendedProperties": { "Property": [ { "@name": "connectionTimeout", "$": "20000" } ] } } } Should be deserialized into Map-like structure having keys like (not all of the above included for brevity): port[0].alias port[0].enabled port

What's the best way to convert a List of Lists in scala (2.9)? I have a list: List[List[A]] which I want to convert into List[A] How can that be achieved recursively? Or is there any other better way? List has the flatten method. Why not use it? List(List(1,2), List(3,4)).flatten > List(1,2,3,4) Given the above example, I'm not sure you need recursion. Looks like you want List.flatten instead. e.g. scala> List(1,2,3) res0: List[Int] = List(1, 2, 3) scala> List(4,5,6) res1: List[Int] = List(4, 5, 6) scala> List(res0,res1) res2: List[List[Int]] = List(List(1, 2, 3), List(4, 5, 6)) scala> res2