I've found this code for getting articles by tag and display them as a list with links in xWiki, but I want it sorted by date.
Has anyone a suggestion for me?
{{velocity}}
#set ($list = $xwiki.tag.getDocumentsWithTag('myTag'))
#foreach($reference in $list)
#set ($document = $xwiki.getDocument($reference))
#set ($label = $document.getTitle())
[[$label>>$reference]]
#end
{{/velocity}}
Thanks in advance!
Sorting in velocity can hit one of 2 performance penalties:
Actually sorting in velocity, either with a sorting algorithm -> unnecesarrily compicated
Loading all the document results into memory (a collection) and
sorting that collection with the sort/collection tool -> you risk quickly running out of memory if the result is larger than you expected.
The easiest alternative, given that there is XWiki running behind it, would be to do an XWQL query for the XWiki.TagClass objects stored inside the documents and do the sorting at the database level. At this point, in velocity, you only need to display the results:
{{velocity}}
#foreach ($docStringRef in $services.query.xwql("from doc.object(XWiki.TagClass) tagsObj where 'conference' member of tagsObj.tags order by doc.creationDate DESC").setLimit(10).execute())
#set ($document = $xwiki.getDocument($docStringRef))
[[$document.title>>$docStringRef]]
#end
{{/velocity}}
For future use/reference, the list of available Velocity tools in XWiki might also be useful https://extensions.xwiki.org/xwiki/bin/view/Extension/Velocity%20Module#HVelocityTools since they can help with common operations, including sorting (that I mentioned at point 2. above)
Related
I have 25K tif files (please don't ask why) that I want to organize into stacks on image J. Basically for each region of interest (ROI), there are 50 images which breaks down into 25 z-planes for two channels. I want everything in a single stack. And I'd like to batch process the whole folder without opening 50 images 500 times at a time. I've attached a picture of what the file names look like:
Folder organization
r01c01f01p01-ch1.tif - the first 10 characters are unique ID to each ROI, then plane number (p01) then channel - ch1 or ch2, then file extension
Here's what I have so far (which I cobbled together based on other macros so this may not make sense...).This is using the ImageJ macros language.
//Processing loop to process each file in the folder.
for (i=0; i<list.length; i++) {
showProgress(i+1, list.length);
if (endsWith(list[i], ".tif")) { // skip the subfolder (I create a subfolder earlier in the macros)
print("-- Processing file: " + list[i] + " --");
open(dir+list[i]);
imageTitle= getTitle();
newTitle = substring(imageTitle, 0, lengthOf(imageTitle)-10); // r01c01f01p, cutting off plane number and then the rest to just get the ROI ID
//This is where I'm stuck:
// find all files containing newTitle and open them (which would be 50 at a time), then run the following macros on them
run("Images to Stack", "name=Ch1 title=[] use");
run("Duplicate...", "title=Ch2 duplicate");
selectWindow("Ch1");
run("Slice Remover", "first=1 last=50 increment=2");
selectWindow("Ch2");
run("Slice Remover", "first=2 last=50 increment=2");
run("Merge Channels...", "c1=Ch1 c2=Ch2 create");
saveAs("tiff", dirNew + newTitle + "_Stack.tif");
//Close(All)?
}
}
print("-- Done --");
showStatus("Finished.");
setBatchMode(false); // Exit batch mode
run("Collect Garbage");
Thank you!
You could do something like:
for (plane=1; plane<51; plane++) {
open(newTitle+plane+"-ch1.tif");
open(newTitle+place+"-ch2.tif");
}
Which would take care of the opening. I would be inclined to have a loop prior to this which would collate the number of unique "newTitle"'s, as your current setup would end up doing something like opening the first item, assembling the combined TIF, and then repeat the process 25K times if I understand it correctly.
Given that you know the number of unique "r01c01f01p" values, in principle you could do a set of stacked loops akin to:
newTitleArray = newArray();
for (r=1; r<50; r++) {
titleBit = "r0" + toString(r);
for (c=1; c<501; c++) {
titleBit = titleBit + "f0"...
Alternatively, you could set up a loop where you check for unique "r01c01f01p" values and add them to an array. In any case, you'd replace the for "list" loop with the for "newTitleArray" loop, and then continue onto the opener I listed above, instead of your existing one.
If I am understanding correctly, it seems like you might do well to stack by channel first, then merge the two. I am not 100% sure, but I think you could potentially use a macro I have already created to do that. It was originally meant to batch process terabytes of 5D data, so it should be very comfortable handling your volume of images. It is not exactly what you are looking for, but should be super easy to modify (I went a little overboard with the commenting in the code), and I think the only thing it does that you might rather it not is produce max projects from the inputs. I'll throw a link here and look for your reply. If it's of interest, let me know and we can work to make it suit your needs together :-) Otherwise, if you could provide a little more detail about where you're getting stuck and/or where I may have misunderstood, I will do my very best to help!
https://github.com/evanjkiely/FIJIMacros
How do you do something like gun.get({startkey, endkey}) ?
Previously: https://github.com/amark/gun/issues/479
#qwe123wsx #sebastianmacias apologies for the delay! Originally posted at: https://github.com/amark/gun/issues/479
The wire spec has a protocol for this but it isn't implemented yet. It looks something like this:
gun.on('out', {get: {'#': {'>': 'a', '<': 'b'}}});
However this doesn't work yet. I would recommend instead:
(1) Pagination behavior is very different from one app to another and will be hard for us to create a "one-size-fits-all" solution, so it would be highly helpful if you could implement your own* pagination and make it available as a user-module, then we can learn from your experience (what worked, what didn't) and make the best solution part of core.
(2) Your app will probably work fine without pagination in the meanwhile, while it can be built (it is targeted for after 1.0), and then as your app becomes more popular, it should be fairly easy to add in without much refactor, once you need it and it is available.
... * How to build your own?
Lots of good articles on this, best one I've seen yet is from Neo4j on how to do it in a graph database (which applies to gun as well) https://graphaware.com/neo4j/2014/08/20/graphaware-neo4j-timetree.html .
Another rough idea is you model your data based on pagination or time. So rather than having ALL tweets go into user's tweet table, instead, the user's tweet table is a table of DAYS (or weeks), and then you put the tweet inside the week table. Now when you load the data, you can scan/skip based off of week very easily while it being super bandwidth efficient.
Rough PSEUDO code:
function onTweetSend(tweet){
gun.get('user').get('alice').get('tweets').get(Date.uniqueYear() + Date.uniqueWeek()).set(tweet)
}
function paginateUserTweet(howMany, cb){
var range = convertToArrayOfUniqueWeekNamesFromToday(howMany);
var all = [];
range.forEach(function(week){
gun.get('user').get('alice').get('tweets').get(week).load(function(tweets){
all.push(tweets);
if(all.length < range.length){ return }
all = flattenArray(all);
cb(all);
});
});
}
Now we can use https://gun.eco/docs/RAD#lex
gun.get(...).get({'.': {'>': startkey, '<': endkey}, '%': 50000}).map().once(...)
I need to get the available options for a certain question in Watson conversation api?
For example I have a conversation app and in some cases Y need to give the users a list to select an option from it.
So I am searching for a way to get the available reply options for a certain question.
I can't answer to the NPM part, but you can get a list of the top 10 possible answers by setting alternate_intents to true. For example.
{
"context":{
"conversation_id":"cbbea7b5-6971-4437-99e0-a82927607079",
"system":{
"dialog_stack":["root"
],
"dialog_turn_counter":1,
"dialog_request_counter":1
}
},
"alternate_intents":true,
"input":{
"text":"Is it hot outside?"
}
}
This will return at most the top ten answers. If there is a limited number of intents it will only show them.
Part of your JSON response will have something like this:
"intents":[{
"intent":"temperature",
"confidence":0.9822100598134365
},
{
"intent":"conditions",
"confidence":0.017789940186563623
}
This won't get you the output text though from the node. So you will need to have your answer store elsewhere to cross reference.
Also be aware that just because it is in the list, doesn't mean it's a valid answer to give the end user. The confidence level needs to be taken into account.
The confidence level also does not work like a normal confidence. You need to determine your upper and lower bounds. I detail this briefly here.
Unlike earlier versions of WEA, the confidence is relative to the
number of intents you have. So the quickest way to find the lowest
confidence is to send a really ambiguous word.
These are the results I get for determining temperature or conditions.
treehouse = conditions / 0.5940327076534431
goldfish = conditions / 0.5940327076534431
music = conditions / 0.5940327076534431
See a pattern?🙂 So the low confidence level I will set at 0.6. Next
is to determine the higher confidence range. You can do this by mixing
intents within the same question text. It may take a few goes to get a
reasonable result.
These are results from trying this (C = Conditions, T = Temperature).
hot rain = T/0.7710267712183176, C/0.22897322878168241
windy desert = C/0.8597747113239446, T/0.14022528867605547
ice wind = C/0.5940327076534431, T/0.405967292346557
I purposely left out high confidence ones. In this I am going to go
with 0.8 as the high confidence level.
Category Name
|
Geograpy (8)
Study Db (18)
i am implement my own advance search in alfresco. i need to read all files which related with particular category.
example:
if there is 20 file under geograpy, lucene query should read particular document under search key word "banana".
Further explanation -
I am using search.lib.js to search. I would like to analyze the result to find out to which category the documents belong to. For example I would like to know how many documents belong to the category under Languages and the subcategories. I experimented with the Classification API but I don't get the result I want. Any Idea how to go through the result to get the category name of each document?
is there any simple method like node.properties["cm:creator"]?
thanks
janaka
I think you should specify more your question:
Are you using cm:content or a customized content?
Are you going to search the keyword inside the content of the file? or are you going to search the keyword in a specific metadata(s)?
Do you want to create a webscript (java or javascript)?
One thing to take in consideration:
if you use +PATH:"cm:generalclassifiable/...." for the categorization in your lucene queries, the performance will be slow (following my experince)
You can use for example the next query to find all nodes at any depth below /cm:Languages:
var results = search.luceneSearch("+PATH:\"cm:generalclassifiable/cm:Languages//*\");
Take a look to this url: https://wiki.alfresco.com/wiki/Search#Path_Queries
Once you have all the elements, you can loop all, and get to which category below. Of course you need to create some counter per each category/subcategory:
for(i = 0; i < results.length; i++){
var node = results[i];
var categoryNodeRef = node.properties["cm:categories"];
var categoryDesc = categoryNodeRef.properties["cm:description"];
var categoryName = categoryNodeRef.properties["cm:name"];
}
This is not exactly the solution, but can be a useful idea to start.
Sorry if it's not what you're asking for, I have just arrived from my holidays.
I have a small index with ~1000 documents with only two fields:
- id (string)
- content (text_general)
I noticed that when I do MLT search by id for similar content, the original document(which id is the searched id) have a score 5.241327.
There is 1:1 duplicated document and for the duplicated content it is returning score = 1.5258181. Why? Why it is not 5.241327 when it is 100% duplicate.
Another question is can I in any way to get similarity documents by content by passing some text in the query.
Example:
/mlt/?q=content:Some encoded long text&mlt.fl=content
I am trying to check if there is similar content uploaded and the check must be performed at new content upload time.
It might be worth to try some different parameters. I also use MLT on only one field, I use the following parameters:
'mlt.boost': 'true',
'mlt.fl': 'my_field_name',
'mlt.maxqt': 1000,
'mlt.mindf': '0',
'mlt.mintf': '0',
'qt': 'mlt',
'rows': '10'
See http://wiki.apache.org/solr/MoreLikeThis for an explanation of the parameters. I think with a small index mindf might be important and I see the default mintf (term frequency) is 2, so I assume an ID is only one term, so this is probably ignored!
First, how does Solr More-Like-This works?
A regular Solr query is conducted (e.g. "?q=content:Some encoded long text&.....".
For each document returned by the above query, More-Like-This conduct More like this query...
So, the first result set "response", is just like any Solr query results set.
The More-Like-This appears below and start with something like that (Json format):
"moreLikeThis":{
"57375":{"numFound":18155,"start":0,"docs":["
For an explanation about More Like This algorithm, please read that:
http://blog.brattland.no/node/18
and: http://cephas.net/blog/2008/03/30/how-morelikethis-works-in-lucene/
If you didn't solved the problem yet, please let me know and I will guide you through.