commiting to branch which is itself a checkout branch - bazaar

I have repo located at A, of which I did a checkout to B, which I checked out to C (so C is bound to B and B to A).
I did some changes in C, did bzr commit on B, but I do not seem to see changes in B automatically.
Is this an expected behavior or did I screw something up without knowing?

Checkouts of checkouts are not supported, and Bazaar will generally refuse to do bind to a branch that is already bound to something else.
How did you create C, is it actually bound to B ? "bzr info C" should tell you.

Related

How to get all reviewers using GitHub REST API?

When I make REST GET to my repository by this path
https://api.github.com/repos/{owner}/{repo}/pulls/{pull_number}/requested_reviewers
I get not all reviewers assigned to PR.
I receive only those who did nothing with review. For example, I have PR with reviewers A, B, C and D.
After creation PR A requested some changes, C approved pull request, B and D did nothing. If I get requested_reviewers three will be only B and D, but I need all of them.
The way to get A and C is
https://api.github.com/repos/{owner}/{repo}/pulls/{pull_number}/reviews
You will get all active reviews, and state will show is it APPROVED or CHANGES_REQUESTED

Yarn transitive dependency replacement (not version)

I have these dependencies:
A -> B -> C -> D
D has a vulnerability and there's no version out with the fix (the PR is out there but the repo is dead).
I'd like to do something like:
"resolutions": {
"D": "myDwithTheFix#1.0.0"
}
Is there a way to do something like that?
If not I have to create my own myA -> myB -> myC -> myD ...
If you want to know, D is underscore.string and its ReDOS.
EDIT:
I think my question only makes sense if it is combined with some redirection feature from a bundler, like webpack aliases.
Because if I replace D for myDwithTheFix, I'd need to replace or redirect also the references from C to point to myDwithTheFix instead of D.

Optaplanner:Add Dynamic visits without changing the already created visits

I am saving the best solution into the DB, and we display that on the web page. I am looking for some solution where a user can add more visits, but that should not change already published trips.
I have checked the documentation and found ProblemFactChange can be used, but only when the solver is already running.
In my case solver is already terminated and the solution is also published. Now I want to add more visits to the vehicle without modifying the existing visits of the Vehicle. Is this possible with Optaplanner? if yes any example of documentation would be very helpful.
You can use PlanningPin annotation for avoiding unwanted changes.
Optaplanner - Pinned planning entities
If you're not looking for pinning (see Ismail's excellent answer), take a look at the OptaPlanner School Timetabling example, which allows adding lessons between solver runs. The lessons simply get stored in the database and then get loaded when the solver starts.
The difficulty with VRP is the chained model complexity (we're working on an alternative): If you add a visit X between A and B, then make sure that afterwards A.next = X, B.previous = X, X.previous = A, X.next = B and X.vehicle = A.vehicle. Not the mention the arrival times etc.
My suggestion would be to resolve what is left after the changes have been introduced. Let's say you are you visited half of your destinations (A -> B -> C) but not yet (C - > D -> E) when two new possible destinations (D' and E') are introduced. Would not this be the same thing as you are starting in C and trying plan for D, D', E and E'? The solution needs to be updated on the progress though so the remainder + changes can be input to the next solution.
Just my two cent.

Forward every third email to a given person

I need something (script, rule) that would forward every third email that comes to my account to some other email address. But I need this to be able to run even when the Outlook/PC is closed/ turned off.
Is it possible to do that?
It's third message to one given person. So every message counts as "third"
- first go to person A, second to person B and third to person C, fourth to person A, fifth to person B etc etc...
Well if I could manage to create the script being server side I would imagine doing something like:
do 3 rules:
Received message = Send message to person A, delete it, turn rule 2
on and turn itself off
Received message = Send message to person B,
delete it, turn rule 3 on and turn itself off
Received message =
Send message to person C, delete it, turn rule 1 on and turn itself
off
This should work, I just don't know if it is possible to make it server-side so Outlook doesn't have to be running.
This would only be possible with a client side rule, which will not run when Outlook is not running. So, no, you cannot do that.

Hadoop Pig - Replace strings in a relation with their corresponding values in a map

I have a relation called conversations_grouped made up of bags of tuples of varying sizes, like so:
DUMP conversations_grouped:
...
({(L194),(L195),(L196),(L197)})
({(L198),(L199)})
({(L200),(L201),(L202),(L203)})
({(L204),(L205),(L206)})
({(L207),(L208)})
({(L271),(L272),(L273),(L274),(L275)})
({(L276),(L277)})
({(L280),(L281)})
({(L363),(L364)})
({(L365),(L366)})
({(L666256),(L666257)})
({(L666369),(L666370),(L666371),(L666372)})
({(L666520),(L666521),(L666522)})
Each L[0-9]+ is a tag corresponding to a string. For example, L194 might be "Hello, how are you doing?" and L195 might be "fine, how are you?". This correspondence is maintained by a map called line_map. Here's a sample:
DUMP line_map;
...
([L666324#Do you think she might be interested in someone?])
([L666264#Well that's typical of Her Majesty's army. Appoint an engineer to do a soldier's work.])
([L666263#Um. There are rumours that my Lord Chelmsford intends to make Durnford Second in Command.])
([L666262#Lighting COGHILL' 5 cigar: Our good Colonel Dumford scored quite a coup with the Sikali Horse.])
([L666522#So far only their scouts. But we have had reports of a small Impi farther north, over there. ])
([L666521#And I assure you, you do not In fact I'd be obliged for your best advice. What have your scouts seen?])
([L666520#Well I assure you, Sir, I have no desire to create difficulties. 45])
([L666372#I think Chelmsford wants a good man on the border Why he fears a flanking attack and requires a steady Commander in reserve.])
([L666371#Lord Chelmsford seems to want me to stay back with my Basutos.])
([L666370#I'm to take the Sikali with the main column to the river])
([L666369#Your orders, Mr Vereker?])
([L666257#Good ones, yes, Mr Vereker. Gentlemen who can ride and shoot])
([L666256#Colonel Durnford... William Vereker. I hear you 've been seeking Officers?])
What I'm trying to do now is parse through each line and replace the L[0-9]+ tags with their corresponding text from line_map. Is it possible to make references to line_map from within a Pig FOREACH statement, or is there something else I have to do?
The first issue with this is that in a map the key must be a quoted string. So you can't use a schema value to access the map. E.G. This will not work.
C: {foo: chararray, M: [value:chararray]}
D = FOREACH C GENERATE M#foo ;
The solution that comes to mind is to FLATTEN conversations_grouped. Then do a join between conversations_grouped and line_map on the L[0-9]+ tag. You'll probably want to project out some of the extra fields (like the L[0-9]+ tag after the join) to make the next step faster. After that you'll have to regroup the data, and massage it into the correct format.
This won't work unless each bag has it's own unique ID for the regrouping, but if each of the L[0-9]+ tags appear in only one bag (conversation) you can use this to create a unique id.
-- A is dumped conversations_grouped
B = FOREACH A {
-- Pulls out an element from the bag to use as the id
id = LIMIT tags 1 ;
-- Flattens B into id, tag form. Each group of tags will have the same id.
GENERATE FLATTEN(id), FLATTEN(tags) ;
}
The schema and output for B is:
B: {id: chararray,tags::tag: chararray}
(L194,L194)
(L194,L195)
(L194,L196)
(L194,L197)
(L198,L198)
(L198,L199)
(L200,L200)
(L200,L201)
(L200,L202)
(L200,L203)
(L204,L204)
(L204,L205)
(L204,L206)
(L207,L207)
(L207,L208)
(L271,L271)
(L271,L272)
(L271,L273)
(L271,L274)
(L271,L275)
(L276,L276)
(L276,L277)
(L280,L280)
(L280,L281)
(L363,L363)
(L363,L364)
(L365,L365)
(L365,L366)
(L666256,L666256)
(L666256,L666257)
(L666369,L666369)
(L666369,L666370)
(L666369,L666371)
(L666369,L666372)
(L666520,L666520)
(L666520,L666521)
(L666520,L666522)
Assuming that the tags are unique, the rest is done like:
-- A2 is line_map, loaded in tag/message pairs instead of a map
-- Joins conversations_grouped and line_map on tag
C = FOREACH (JOIN B by tags::tag, A2 by tag)
-- This generate removes the tag
GENERATE id, message ;
-- Regroups C on the id created in B
D = FOREACH (GROUP C BY id)
-- This step limits the output to just messages
GENERATE C.(message) AS messages ;
Schema and output from D:
D: {messages: {(A2::message: chararray)}}
({(Colonel Durnford... William Vereker. I hear you 've been seeking Officers?),(Good ones, yes, Mr Vereker. Gentlemen who can ride and shoot)})
({(Your orders, Mr Vereker?),(I'm to take the Sikali with the main column to the river),(Lord Chelmsford seems to want me to stay back with my Basutos.),(I think Chelmsford wants a good man on the border Why he fears a flanking attack and requires a steady Commander in reserve.)})
({(Well I assure you, Sir, I have no desire to create difficulties. 45),(And I assure you, you do not In fact I'd be obliged for your best advice. What have your scouts seen?),(So far only their scouts. But we have had reports of a small Impi farther north, over there. )})
NOTE: If at worst, (the L[0-9]+ tags aren't unique) you can give each line of the input file(s) a sequential, integer id before you load it into pig.
UPDATE: If you are using pig 0.11, then you can also use the RANK operator.