You can define grouping in JasperReports. JasperReports calculates the total for you, there is comfortable way to add groups and totals. Here an overview what you need to do in iReport. To add the group modify your query to order by pledgeType – JasperReports requires the data sorted according to your grouping. right click on … Read more
Use itertools.groupby: import itertools import operator data=[(1, ‘A’, ‘foo’), (2, ‘A’, ‘bar’), (100, ‘A’, ‘foo-bar’), (‘xx’, ‘B’, ‘foobar’), (‘yy’, ‘B’, ‘foo’), (1000, ‘C’, ‘py’), (200, ‘C’, ‘foo’), ] for key,group in itertools.groupby(data,operator.itemgetter(1)): print(list(group)) yields [(1, ‘A’, ‘foo’), (2, ‘A’, ‘bar’), (100, ‘A’, ‘foo-bar’)] [(‘xx’, ‘B’, ‘foobar’), (‘yy’, ‘B’, ‘foo’)] [(1000, ‘C’, ‘py’), (200, ‘C’, ‘foo’)] … Read more
You can use Array.prototype.reduce() in combination with Array.prototype.forEach(). var array = [ [0, 1, 3], [2, 4, 6], [5, 5, 7], [10, 0, 3] ], result = array.reduce(function (r, a) { a.forEach(function (b, i) { r[i] = (r[i] || 0) + b; }); return r; }, []); document.write(‘<pre>’ + JSON.stringify(result, 0, 4) + ‘</pre>’); Update, … Read more
Such a list of lists could be constructed using a list comprehension: In [17]: seq=[1,2,3,4,5,6,7,8] In [18]: [seq[i:i+3] for i in range(0,len(seq),3)] Out[18]: [[1, 2, 3], [4, 5, 6], [7, 8]] There is also the grouper idiom: In [19]: import itertools In [20]: list(itertools.izip_longest(*[iter(seq)]*3)) Out[20]: [(1, 2, 3), (4, 5, 6), (7, 8, None)] but … Read more
The theory (by theory I mean SQL Standard) says that WHERE restricts the result set before returning rows and HAVING restricts the result set after bringing all the rows. So WHERE is faster. On SQL Standard compliant DBMSs in this regard, only use HAVING where you cannot put the condition on a WHERE (like computed … Read more
Suppose the bottom left coordinates and the upper right coordinates are respectively [0, 0] and [100, 100]. From MongoDB 3.2 you can use the $slice operator to return a subset of an array which is what you want. db.collection.aggregate([ { “$match”: { “loc”: { “$geoWithin”: { “$box”: [ [0, 0], [100, 100] ] } }} … Read more
You’d have to do this manually using a loop. Something like this should work: $result = array(); foreach( $input as $row) { $id = array_shift( $row); foreach( $row as $key => $value) { $result[ $id ][ $key ] = ( isset( $result[ $id ][ $key ]) ? $result[ $id ][ $key ] + $value : … Read more
The documentation of accumarray says: Note If the subscripts in subs are not sorted, fun should not depend on the order of the values in its input data. And your subs is not sorted (at least not in ascending order). If you rewrite the code so that subs is sorted and vals is also rearranged … Read more
You could use a generic function: function _group_by($array, $key) { $return = array(); foreach($array as $val) { $return[$val[$key]][] = $val; } return $return; } I added some sample code to test <?php $list= [ [ ‘No’ => 101, ‘Paper_id’ => ‘WE3P-1’, ‘Title’ => “a1”, ‘Author’ => ‘ABC’, ‘Aff_list’ => “University of South Florida, Tampa, United … Read more