Currently, I have a filter that is a string, so it's ordering the options like this:
The order of the options is incorrect, as you can see $1M is right after $15k and before $200k.
Without prefixing the values with something "A." , "B.", "C.", etc... how do I change the order of the filter options?
I think you can first create a dimension like the below:
dimension: sort_arr_bands {
type:number
hidden: yes
sql: CASE
WHEN ${arr_bands} = '<=$1M' THEN 1
WHEN ${arr_bands} = '<=$2M' THEN 2
WHEN ${arr_bands} = '<=$3M' THEN 3
ELSE 4
END;;
description: "This dimension is used to force sort the arr_bands dimension."
}
Then add to your main/original dimension:
dimension: arr_bands {
type: string
sql: ${TABLE}.arr_bands ;;
order_by_field: sort_arr_bands
}
Related
I have set up a log-based alert in Microsoft Azure. The deployment of the alerts done via ARM template.
Where you can input your query and set threshold like below.
"triggerThresholdOperator": {
"value": "GreaterThan"
},
"triggerThreshold": {
"value": 0
},
"frequencyInMinutes": {
"value":15
},
"timeWindowInMinutes": {
"value": 15
},
"severityLevel": {
"value": "0"
},
"appInsightsQuery": {
"value": "exceptions\r\n| where A_ != '2000' \r\n| where A_ != '4000' \r\n| where A_ != '3000' "
}
As far as I understand we can only set threshold once ON an entire query.
Questions: I have multiple statements in my query which I am excluding since it's just a noise. But now I want to set a threshold on value 3000 to 5 and also want to set a time-window to 30 in the same query. meaning only exclude 3000 when it occurs 5 times in the last 30 minutes(when query get run).
exceptions
| where A_ != '2000'
| where A_ != '4000'
| where A_ != '3000'
I am pretty sure that I can't set a threshold like this in the query and the only workaround is to create a new alert just for value 3000 and set a threshold in ARM template. I haven't found any heavy threshold/time filters in Aure. Is there any way I can set multiple thresholds and time filters in a single query? which is again getting checked by different threshold and time filetrs in the ARM template.
Thanks.
I don't fully understand your question.
But for your time window question you could do something like
exceptions
| summarize count() by A_, bin(TimeGenerated, 30m)
That way you will get a count of A_ in blocks of 30 minutes.
Another way would be to do:
let Materialized = materialize(
exceptions
| summarize Count=count(A_) by bin(TimeGenerated, 30m)
);
Materialized | where Count == 10
But then again it all depends on what you would like to achieve
You can easily set that in the query and fire based on the aggregate result.
exceptions
| where timestamp > ago(30m)
| summarize count2000 = countif(A_ == '2000'), count3000 = countif(A_ == '3000'), count4000 = countif(A_ == '4000')
| where count2000 > 5 or count3000 > 3 or count4000 > 4
If the number of results is greater than one than the aggregate condition applies.
I want to create a function that can take a dictionary of dictionaries such as the following
information = {
"sample information": {
"ID": 169888,
"name": "ttH",
"number of events": 124883,
"cross section": 0.055519,
"k factor": 1.0201,
"generator": "pythia8",
"variables": {
"trk_n": 147,
"zappo_n": 9001
}
}
}
and then print it in a neat way such as the following, with alignment of keys and values using whitespace:
sample information:
ID: 169888
name: ttH
number of events: 124883
cross section: 0.055519
k factor: 1.0201
generator: pythia8
variables:
trk_n: 147
zappo_n: 9001
My attempt at the function is the following:
def printDictionary(
dictionary = None,
indentation = ''
):
for key, value in dictionary.iteritems():
if isinstance(value, dict):
print("{indentation}{key}:".format(
indentation = indentation,
key = key
))
printDictionary(
dictionary = value,
indentation = indentation + ' '
)
else:
print(indentation + "{key}: {value}".format(
key = key,
value = value
))
It produces the output like the following:
sample information:
name: ttH
generator: pythia8
cross section: 0.055519
variables:
zappo_n: 9001
trk_n: 147
number of events: 124883
k factor: 1.0201
ID: 169888
As is shown, it successfully prints the dictionary of dictionaries recursively, however is does not align the values into a neat column. What would be some reasonable way of doing this for dictionaries of arbitrary depth?
Try using the pprint module. Instead of writing your own function, you can do this:
import pprint
pprint.pprint(my_dict)
Be aware that this will print characters such as { and } around your dictionary and [] around your lists, but if you can ignore them, pprint() will take care of all the nesting and indentation for you.
I have a file .txt with 3 columns: ID-polygon-1, ID-polygon-2 and distance.
When I import my file into Netlogo, I obtain 3 lists [[list1][list2][list3]] which corresponds with the 3 columns.
I used table:from-list list to create a table with the content of 3 lists.
I obtain {{table: [[1 1] [67 518] [815 127]]}} (The table displays the first two lines of my dataset).
For example, I would like to get the value of distance (list3) between ID-polygon-1 = 1 (list1) and ID-polygon-2 = 67 (list1), that is, 815.
How can I use table:get table key when I have need of 2 keys (ID-polygon-1 and ID-polygon-2) ?
Thanks very much your help.
Using table:from-list will not help you there: it expects "a list of two element lists, or pairs" where the "the first element in the pair is the key and the second element is the value." That's not what you have in your original list.
Furthermore, NetLogo tables (and associative arrays in general) cannot have two keys. They are always just key-value pairs. Nothing prevents the value from being another table, however, and in your case, that is what you need: a table of tables!
There is no primitive to build that directly, however. You will need to build it yourself:
extensions [ table ]
globals [ t ]
to setup
let lists [
[ 1 1 ] ; ID-polygon-1 column
[ 67 518 ] ; ID-polygon-2 column
[ 815 127 ] ; distance column
]
set t table:make
foreach n-values length first lists [ ? ] [
let id1 item ? (item 0 lists)
let id2 item ? (item 1 lists)
let dist item ? (item 2 lists)
if not table:has-key? t id1 [
table:put t id1 table:make
]
table:put (table:get t id1) id2 dist
]
end
Here is what you get when you print the resulting table:
{{table: [[1 {{table: [[67 815] [518 127]]}}]]}}
And here is a small reporter to make it convenient to get a distance from the table:
to-report get-dist [ id1 id2 ]
report table:get (table:get t id1) id2
end
Using get-dist 1 67 will give the 815 result you were looking for.
I have a problem nesting the result tags in each other the right way.
The result should look like this:
aimed result
<categoryA>
<position>...</position>
<position>...</position>
...
</categoryA>
<categoryB>
<position>...</position>
<position>...</position>
...
</categoryB>
currently I have only managed to get the right results for the positions, the categoryA and B are 1 hierarchic layer higher than the positions. the positions should be nested in the categories. The categories can be referenced by let $y := $d/Bilanz/Aktiva/* (respectively $d$d/Bilanz/Aktiva/LangfristigesVermoegen and $d$d/Bilanz/Aktiva/KurzfristigesVermoegen).
Here is my query:
query
let $d := doc('http://etutor.dke.uni-linz.ac.at/etutor/XML?id=5001')/Bilanzen
let $a02 := $d/Bilanz[#jahr='2002']/Aktiva/*
let $a03 := $d/Bilanz[#jahr='2003']/Aktiva/*
for $n02 in $a02//* , $n03 in $a03//*
(:
where name($n02) = name($n03)
where node-name($n02) = node-name($n03)
:)
where name($n02) = name($n03)
return <position name="{node-name($n02)}">
<j2002>{data($n02/#summe)}</j2002>
<j2003>{data($n03/#summe)}</j2003>
<diff>{data($n03/#summe) - data($n02/#summe)}</diff>
</position>
xml
<Bilanzen>
<Bilanz jahr="2002">
<Aktiva>
<LangfristigesVermoegen>
<Sachanlagen summe="1486575.8"/>
<ImmateriellesVermoegen summe="67767.2"/>
<AssoziierteUnternehmen summe="190826.3"/>
<AndereBeteiligungen summe="507692.7"/>
<Uebrige summe="92916.4"/>
</LangfristigesVermoegen>
<KurzfristigesVermoegen>
<Vorraete summe="78830.9"/>
<Forderungen summe="198210.3"/>
<Finanzmittel summe="181102.0"/>
</KurzfristigesVermoegen>
</Aktiva>
<Passiva>
<Eigenkapital>
<Grundkapital summe="91072.4"/>
<Kapitalruecklagen summe="186789.5"/>
<Gewinnruecklagen summe="798176.2"/>
<Bewertungsruecklagen summe="-34922.4"/>
<Waehrungsumrechnung summe="0"/>
<EigeneAktien summe="0"/>
</Eigenkapital>
<AnteileGesellschafter summe="23613.1"/>
<LangfristigeVerb>
<Finanzverbindlichkeiten summe="680007.1"/>
<Steuern summe="36555.8"/>
<Rueckstellungen summe="429286.1"/>
<Baukostenzuschuesse summe="169246.0"/>
<Uebrige summe="36166.9"/>
</LangfristigeVerb>
<KurzfristigeVerb>
<Finanzverbindlichkeiten summe="14614.6"/>
<Steuern summe="65247.6"/>
<Lieferanten summe="94939.2"/>
<Rueckstellungen summe="123664.8"/>
<Uebrige summe="89464.8"/>
</KurzfristigeVerb>
</Passiva>
</Bilanz>
<Bilanz jahr="2003">
<Aktiva>
<LangfristigesVermoegen>
<Sachanlagen summe="1590313.7"/>
<ImmateriellesVermoegen summe="69693.2"/>
<AssoziierteUnternehmen summe="198224.7"/>
<AndereBeteiligungen summe="418489.3"/>
<Uebrige summe="104566.7"/>
</LangfristigesVermoegen>
<KurzfristigesVermoegen>
<Vorraete summe="20609.8"/>
<Forderungen summe="289458.5"/>
<Finanzmittel summe="302445.9"/>
</KurzfristigesVermoegen>
</Aktiva>
<Passiva>
<Eigenkapital>
<Grundkapital summe="91072.4"/>
<Kapitalruecklagen summe="186789.5"/>
<Gewinnruecklagen summe="875723.4"/>
<Bewertungsruecklagen summe="-15459.5"/>
<Waehrungsumrechnung summe="-633.7"/>
<EigeneAktien summe="0"/>
</Eigenkapital>
<AnteileGesellschafter summe="22669.8"/>
<LangfristigeVerb>
<Finanzverbindlichkeiten summe="733990.2"/>
<Steuern summe="68156.8"/>
<Rueckstellungen summe="395997.2"/>
<Baukostenzuschuesse summe="177338.5"/>
<Uebrige summe="38064.9"/>
</LangfristigeVerb>
<KurzfristigeVerb>
<Finanzverbindlichkeiten summe="6634.7"/>
<Steuern summe="97119.1"/>
<Lieferanten summe="89606.0"/>
<Rueckstellungen summe="128237.5"/>
<Uebrige summe="98495.2"/>
</KurzfristigeVerb>
</Passiva>
</Bilanz>
</Bilanzen>
I would really appreciate some help, i have no clue at all. Thank you.
If I understand you correctly, you want the information about LangfristigesVermoegen (and its children) to be grouped in the output under element categoryA, and the information about Kurzfristigesvermoegen to be grouped under categoryB.
So you will want first of all to do something to generate the categoryA and categoryB elements. For example,
let $d := doc(...)/Bilanzen
return (
<categoryA>{ ... children of category A here ... }</categoryA>,
<categoryB>{ ... children of category B here ... }</categoryB>
)
The positions in each category can be generated using code similar to what you've now got, except that instead of iterating over
for $n02 in $a02//* , $n03 in $a03//*
you will need to iterate over $a02[self::LangfristigesVermoegen]/* for category A, and over $a02[self::KurzfristigesVermoegen]/* for category B (and similarly, of course, for $n02 and $n03).
If the set of categories is not static and you just want to group things in the output using the same grouping elements present in the input, then you'll want an outer structure something like this:
for $assetclass1 in $anno2002/*
let $assetclass2 := $anno2003/*[name() = name($assetclass1)]
return
(element {name($assetclass1)} {
for $old in $assetclass1/*,
$new in $assetclass2/*
where name($old) eq name($new)
return <position name="{node-name($old)}">
<j2002>{data($old/#summe)}</j2002>
<j2003>{data($new/#summe)}</j2003>
<diff>{data($new/#summe) - data($old/#summe)}</diff>
</position>
})
I am very new to lua and my plan is to create a table. This table (I call it test) has 200 entries - each entry has the same subentries (In this example the subentries money and age):
This is a sort of pseudocode:
table test = {
Entry 1: money=5 age=32
Entry 2: money=-5 age=14
...
Entry 200: money=999 age=72
}
How can I write this in lua ? Is there a possibility ? The other way would be, that I write each subentry as a single table:
table money = { }
table age = { }
But for me, this isn't a nice way, so maybe you can help me.
Edit:
This question Table inside a table is related, but I cannot write this 200x.
Try this syntax:
test = {
{ money = 5, age = 32 },
{ money = -5, age = 14 },
...
{ money = 999, age = 72 }
}
Examples of use:
-- money of the second entry:
print(test[2].money) -- prints "-5"
-- age of the last entry:
print(test[200].age) -- prints "72"
You can also turn the problem on it's side, and have 2 sequences in test: money and age where each entry has the same index in both arrays.
test = {
money ={1000,100,0,50},
age={40,30,20,25}
}
This will have better performance since you only have the overhead of 3 tables instead of n+1 tables, where n is the number of entries.
Anyway you have to enter your data one way or another. What you'd typically do is make use some easily parsed format like CSV, XML, ... and convert that to a table. Like this:
s=[[
1000 40
100 30
0 20
50 25]]
test ={ money={},age={}}
n=1
for balance,age in s:gmatch('([%d.]+)%s+([%d.]+)') do
test.money[n],test.age[n]=balance,age
n=n+1
end
You mean you do not want to write "money" and "age" 200x?
There are several solutions but you could write something like:
local test0 = {
5, 32,
-5, 14,
...
}
local test = {}
for i=1,#test0/2 do
test[i] = {money = test0[2*i-1], age = test0[2*i]}
end
Otherwise you could always use metatables and create a class that behaves exactly like you want.