Given ohlcv where each column has the same date: 2024.03.06
q)ohlcv
date ticker open close high low volume time entry
-----------------------------------------------------------------------------------------
2024.03.06 IBM 174.3 150.5 174.3 144.1 17110 2024.03.06D09:15:00.000000000 0
2024.03.06 IBM 153.95 158.7 158.7 153.95 775 2024.03.06D09:16:00.000000000 0
2024.03.06 IBM 160 157.05 160 157 8455 2024.03.06D09:17:00.000000000 0
2024.03.06 IBM 157.85 157.1 157.85 155.8 14334 2024.03.06D09:18:00.000000000 0
2024.03.06 IBM 157.05 157.75 157.8 156.2 11478 2024.03.06D09:19:00.000000000 0
2024.03.06 IBM 157.5 158 158 157.15 12446 2024.03.06D09:20:00.000000000 0
2024.03.06 IBM 158 160.2 161 157.6 29500 2024.03.06D09:21:00.000000000 0
2024.03.06 IBM 160.55 161.25 161.25 159.7 33726 2024.03.06D09:22:00.000000000 1
2024.03.06 IBM 162 164.4 164.7 161.55 44981 2024.03.06D09:23:00.000000000 0
...
2024.03.06 IBM 154.75 154.4 155 154.05 14003 2024.03.06D15:29:00.000000000 0 / last row
q)distinct ohlcv`date
,2024.03.06
q) count ohlcv
375
q)meta ohlcv
c | t f a
---------------| -----
date | d
ticker | s
open | f
close | f
high | f
low | f
volume | f
time | p
entry | b
why does the following select by date from ohlcv statement yield only one row, which, as it turns out is the last row? My understaning was that we should get all 375 rows since they all have the same date.
q)select by date from ohlcv
date | ticker open close high low volume time entry
----------| ---------------------------------------------------------------------------
2024.03.06 IBM 154.75 154.4 155 154.05 14003 2024.03.06D15:29:00.000000000 0 / last row
q) count ohlcv
1
>Solution :
The kdb Q language is only slightly related to SQL. You need to do some more reading. select by is the same as SQL’s group by, so it aggregates all the rows based on the grouped column.
https://code.kx.com/q/ref/select/#select-phrase