Consider using LISTAGG function in case you’re on 11g: select grp, listagg(name,’,’) within group( order by name ) from name_table group by grp sqlFiddle upd: In case you’re not, consider using analytics: select grp, ltrim(max(sys_connect_by_path (name, ‘,’ )), ‘,’) scbp from (select name, grp, row_number() over (partition by grp order by name) rn from tab … Read more
In XSLT 1.0, using Muenchian grouping. <xsl:stylesheet version=”1.0″ xmlns:xsl=”http://www.w3.org/1999/XSL/Transform”> <xsl:output method=”xml” indent=”yes” /> <xsl:key name=”division” match=”User” use=”@Division” /> <xsl:template match=”Users”> <AllUsers> <xsl:apply-templates select=”User[generate-id(.)=generate-id(key(‘division’,@Division)[1])]”/> </AllUsers> </xsl:template> <xsl:template match=”User”> <Division value=”{@Division}”> <xsl:for-each select=”key(‘division’, @Division)”> <User> <id><xsl:value-of select=”@id” /></id> <name><xsl:value-of select=”@name” /></name> </User> </xsl:for-each> </Division> </xsl:template> </xsl:stylesheet> In XSLT 2.0, use xsl:foreach-group <xsl:output method=”xml” indent=”yes” /> <xsl:template match=”Users”> … Read more
There’s no way to do it using standard Stream API. In general you cannot do it as it’s always possible that new item will appear in future which belongs to any of already created groups, so you cannot pass your group to downstream analysis until you process all the input. However if you know in … Read more
SELECT foo, count(bar) FROM mytable GROUP BY bar ORDER BY count(bar) DESC; The group by statement tells aggregate functions to group the result set by a column. In this case “group by bar” says count the number of fields in the bar column, grouped by the different “types” of bar. A better explanation is here: … Read more
Here’s a solution using reduce. Split the time into a date string, and then set a key for each date. If the key exists push it onto the array: Update added the array format version const data = [ {notes: ‘Game was played’, time: ‘2017-10-04T20:24:30+00:00’, sport: ‘hockey’, owner: ‘steve’, players: ’10’, game_id: 1}, { notes: … Read more
my $string = “var1=100 var2=90 var5=hello var3=\”a, b, c\” var7=test var3=hello”; while($string =~ /(?:^|\s+)(\S+)\s*=\s*(“[^”]*”|\S*)/g) { print “<$1> => <$2>\n”; } Prints: <var1> => <100> <var2> => <90> <var5> => <hello> <var3> => <“a, b, c”> <var7> => <test> <var3> => <hello> Explanation: Last piece first: the g flag at the end means that you can … Read more
defaultdict is how this is done. While for loops are largely essential, if statements aren’t. from collections import defaultdict groups = defaultdict(list) for obj in old_list: groups[obj.some_attr].append(obj) new_list = groups.values()
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
Reduce is great for situations like this. Given list below is your input data: const list = [{ ‘name’: ‘Display’, ‘group’: ‘Technical detals’, ‘id’: ’60’, ‘value’: ‘4’ }, { ‘name’: ‘Manufacturer’, ‘group’: ‘Manufacturer’, ‘id’: ’58’, ‘value’: ‘Apple’ }, { ‘name’: ‘OS’, ‘group’: ‘Technical detals’, ‘id’: ’37’, ‘value’: ‘Apple iOS’ } ]; const groups = list.reduce((groups, … Read more
Well, the brute force answer is: subList = [theList[n:n+N] for n in range(0, len(theList), N)] where N is the group size (3 in your case): >>> theList = list(range(10)) >>> N = 3 >>> subList = [theList[n:n+N] for n in range(0, len(theList), N)] >>> subList [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]] … Read more