rocuinneagain
Forum Replies Created
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A timespan can be used directly
q)select avg col by 0D00:00:00.005 xbar time from t time | col -----------------------------| ---- 2022.09.09D07:40:23.110000000| 3 2022.09.09D07:40:23.115000000| 11.5 2022.09.09D07:40:23.120000000| 18
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Yes that’s much cleaner – as you only ever compute one value
c2and only look back 1 step. this will do exactly as you need. -
Yes in that case an accumulator is needed. One method you could choose would be to pass a table through to accumulate while also allowing you to look back to previous rows:
q)update c2:1_@[;`c2]{y[`c2]:enlist $[(y[`c1][0]>last x[`c2]) or ((last x[`c])<last x[`c2]);y[`c1][0];last x`c2];x,y}/[enlist each {(1#0#x),x}update c2:0 from `c`c1#t] from t c c1 c2 ---------- 30 10 10 40 20 20 25 5 20 20 25 25 4 5 5 4 4 4 4.5 3 4 4.5 3.5 4 -
You can use a second q process help you to debug.
Set a Listening-port p and have the HTTP Post handler (.z.pp) print incoming message contents
p 5000 .z.pp:{show x;x}
Trying curl from command line
$ curl -H ‘Content-type: application/json’ -d ‘{“text”:”Hello World”}’ localhost:5000
The server q process prints:
” {“text”:”Hello World”}” `Host`User-Agent`Accept`Content-type`Content-Length!(“localhost:5000″;”curl/7.58.0″;”*/*”;”application/json”;”22″)
From a client q process using .Q.hp
q).Q.hp[“http://localhost:5000″;.h.ty`json] .j.j enlist[`text]!enlist”Hello World”
The server q process prints:
” {“text”:”Hello World”}” `Accept-Encoding`Connection`Host`Content-type`Content-length!(“gzip”;”close”;”localhost:5000″;”application/json”;”22″)
There are only slight differences in the headers.
You may need to consult documentation of the server you are connecting to to confirm if it has specific header requirements.
(Clients using KX Insights have access to kurl with more options than .Q.hp)
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Looking at your expected
c2this logic may be what you are looking for:q)update c2:fills ?[(c1>prev c1) or c<prev c1;c1;0N] from t c c1 c2 -------- 30 10 10 40 20 20 25 5 20 20 25 25 4 5 5 4 4 4
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changing
ctoprev[c]looks to be what was missingq)update c2:fills ?[(c1>prev c1) or prev[c]<prev c1;c1;0N] from t c c1 c2 -------- 30 10 10 40 20 20 25 5 20 20 25 25 4 5 5 4 4 4 1 3 4
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m'[x]is not parallel executionThere can be cases where the overhead of using multiple threads can make execution slower than single threaded. (.i.e performing many small operations)
.Q.fcis available to help in some of these cases.https://code.kx.com/q/ref/dotq/#qfc-parallel-on-cut
If you measure memory usage with
tsnote that it only sees usage in main thread. It does not sum usage from all threads. So you cannot compare results directly.q)\ts {til 10000000;x}'[til 1000] // \ts shows real memory usage 5395 134250848 q)\ts {til 10000000;x}':[til 1000] // \ts only sees memory usage in main thread - not the sum of threads 5612 33408q)\ts {x}'[til 100000] // Single core 8 4746288 q)\ts {x}':[til 100000] // Slower due to parallel overhead 13 4194864 q)\ts .Q.fc[{x}][til 100000] //Faster way to use multiple cores 1 3146512 -
You are seeing the same results as your
t2has an entry for every secondIf we swap
til 9for2*til 9we can create gaps and then you can see the differences wherewjwill bring in prevailing values butwj1will not.t1:([]sym:3#`ibm;time:07:01:01 07:01:03 07:01:05;price:100 101 105); a:101 103 103 104 104 107 108 107 108; b:98 99 102 103 103 104 106 106 107; t2:([]sym:`ibm; time:07:00:58+2*til 9; ask:a; bid:b); c:`sym`time; w:-2 1+:t1.time; o1:wj[w;c;t1;(t2;(max;`ask);(min;`bid))]; o2:wj1[w;c;t1;(t2;(max;`ask);(min;`bid))]; o1~o2; // 0b; q)wj[w;c;t1;(t2;(::;`ask);(::;`bid))] sym time price ask bid ------------------------------------------ ibm 07:01:01 100 101 103 103 98 99 102 ibm 07:01:03 101 103 103 104 99 102 103 ibm 07:01:05 105 103 104 104 102 103 103 q)wj1[w;c;t1;(t2;(::;`ask);(::;`bid))] sym time price ask bid ---------------------------------- ibm 07:01:01 100 103 103 99 102 ibm 07:01:03 101 103 104 102 103 ibm 07:01:05 105 104 104 103 103
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From: https://code.kx.com/q/kb/ssl/#tls-client-mode
By default, kdb+ will try to verify the servers certificate against a trusted source, using the certificates from
SSL_CA_CERT_FILEorSSL_CA_CERT_PATHto verify the servers certificate. If you don’t wish to verify a servers certificate, set -
In the basic kdb+ tick the RDB does do the write at EOD.
No data is lost, it will buffer in the outbound queue from the TP process.
As long as you have a safe memory overhead for this buffer then the system will continue to run.
Queries will also buffer while the RDB is performing the write.
More advanced architectures avoid the RDB being blocked, such as all the KX platforms. .e.g https://code.kx.com/insights/1.1/
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You could pass it in to a projection
q)f:{[xF] g:{[xG] xG + 2}; h:{[g;xH] g[xH] + 3}[g]; g[xF] + h[xF]} q)f[40] 87Or composition in this basic case
q)f:{[xF] g:{[xG] xG + 2}; h:+[3] g @; g[xF] + h[xF]} q)f[40] 87 -
Some ways:
q("{((),x)#y}", numpy.string_("iq"), numpy.string_("t")) q("{value"select ",string[x],"from ",string y}", "iq", "t") q("select %s from %s" % ("iq","t")) q("{?[y;();0b;{x!x}(),x]}", numpy.string_("iq"), numpy.string_("t"))- Using # https://code.kx.com/q/ref/take/
- Using value https://code.kx.com/q/ref/value/
- Python string formatting
- Functional select https://code.kx.com/q/basics/funsql/
Functional selects are the most powerful way https://code.kx.com/q/wp/parse-trees/
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See docs for more on how the library handles difference between symbols and strings:
https://qpython.readthedocs.io/en/latest/type-conversion.html#string-and-symbols
(Answer above updated to use
numpy.string_to pass symbols as needed) -
Do you have any details of your specific issue or error?
This request is vague. https://stackoverflow.com/help/how-to-ask
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You want to store it in a global variable in q?
Then you can use a lambda and
::orsetq('{`mydict set x}', qdict) q('{mydict::x}', qdict)