You can use the filter function to leave only those keywords contained in content: val match = keywords.filter { it in content } Here match is a List<String>. If you want to get an array in the result, you can add .toTypedArray() call. in operator in the expression it in content is the same as … Read more
Using a generator: function *intersperse(a, delim) { let first = true; for (const x of a) { if (!first) yield delim; first = false; yield x; } } console.log([…intersperse(array, ‘&’)]); Thanks to @Bergi for pointing out the useful generalization that the input could be any iterable. If you don’t like using generators, then [].concat(…a.map(e => … Read more
For the general case, where your search string can be in any column, you can do this: >>> rows, cols = np.where(t == ‘bar’) >>> t[rows] array([[‘2’, ‘3’, ‘4’, ‘bar’], [‘8’, ‘9’, ‘1’, ‘bar’]], dtype=”|S11″)
You have to use bracket notation: var obj = {}; obj[a[i]] = 0; x.push(obj); The result will be: x = [{left: 0}, {top: 0}]; Maybe instead of an array of objects, you just want one object with two properties: var x = {}; and x[a[i]] = 0; This will result in x = {left: 0, … Read more
The simplest thing would be: update table set array = array_append(array, ‘element’) where … or perhaps use the || operator: update table set array = array || ‘element’ where … Both of those are equivalent to the more common set n = n + 11 for numbers. Depending on the types involved, you might need … Read more
You can take Array.map(). This method returns an array with the elements from the callback returned. It expect that all elements return something. If not set, undefined will be returned. var students = [{ name: ‘Nick’, achievements: 158, points: 14730 }, { name: ‘Jordan’, achievements: ‘175’, points: ‘16375’ }, { name: ‘Ramon’, achievements: ’55’, points: … Read more
The extend() method appends to the existing array and returns None. In your case, you are creating an array — [4, 5, 6] — on the fly, extending it and then discarding it. The variable b ends up with the return value of None.
Sort within the same array variable Sort functions bellow are exactly the same, the only difference how short and expressive they are: Full declaration: myArr.sort { (lhs: EntryStruct, rhs: EntryStruct) -> Bool in // you can have additional code here return lhs.deadline < rhs.deadline } Shortened closure declaration: myArr.sort { (lhs:EntryStruct, rhs:EntryStruct) in return lhs.deadline … Read more
$testArray = [System.Collections.ArrayList]@() $tempArray = “123”, “321”, “453” foreach($item in $tempArray) { $arrayID = $testArray.Add($item) }
Why not simply do something like this? >>> a array([[ 0., 5., 2.], [ 0., 0., 3.], [ 0., 0., 0.]]) >>> b array([ 0., 0., 3.]) >>> a==b array([[ True, False, False], [ True, True, True], [ True, True, False]], dtype=bool) >>> np.all(a==b,axis=1) array([False, True, False], dtype=bool) >>> np.where(np.all(a==b,axis=1)) (array([1]),)