Re: [personal kdb+] 'rank error

Community Support
1

Hi Rory,

You have to make the key of the dictionary into symbol format and also the rank error is due to the value d being all atoms and not list so these must be enlisted.

q)d:{count each group y xbar x}

q)x:1000?100

q)k:d[x;10]

q)k

50| 91

20| 101

30| 119

10| 118

70| 106

40| 99

80| 92

90| 84

60| 108

0 | 82

q)tab:flip(`$(string key k))!enlist each value k

q)tab

50 20 ?30 ?10 ?70 ?40 80 90 60 ?0

91 101 119 118 106 99 92 84 108 82

q)type tab

98h

Hope this helps.

Thanks Rory

2

Hi Guys,

Perhaps a slightly more terse solution would be;

q)tab:enlist (`$string key k)!value k
q)tab
10? 0?? 90 70? 40? 80 60? 50? 30? 20

104 100 99 101 100 89 100 109 109 89
q)type tab
98h

Kind regards,

Paul

3

Also if you change the ‘d’ function to:

d:{count each group `$string y xbar x}

you get a even shorter solution:

q)enlist d[x;10]
30 0  50  60  70 90  40  80  20 10

87 97 114 111 83 110 105 103 97 93

Regards,
Petru

4

Sorry - just as a final coda to this topic - I had neglected to sort the bins, so the function now looks like:

d:{h:count each group y xbar x; (`$(string asc key h))!(enlist each h[asc key h])}

Can anyone make this more compact / elegant? My approach looks a bit clunky to my eyes..

5

Hi,

This should do it.

q)x:1000?100
q)d:{h:count’[group y xbar x];(`$string a)!h a:asc key h}
q)enlist d[x;10]
0? 10 20 30 40? 50 60? 70? 80? 90

99 96 95 98 101 98 100 102 115 96

Regards,

Paul