I am using the rythmengine.
The output I want is: x=prefix:value without any whitespaces.
#def test(String param){
#{param="prefix:"+param} #param
}
1
2 x=#test("value")
3
output is: 2 x= prefix:value
#def test(String param){
#{param="prefix:"+param}#param
}
1
2 x=#test("value")
3
output is: 2 x=param
This seems like a bug. Does anyone know a solution or workaround?
Try the following code on http://fiddle.rythmengine.org/#/editor:
#def test(String param){
#{param="prefix:"+param}##param
}
1
2 x=#test("value")
3
Related
I am trying to compile this piece of CL code using Rational Series but keep getting error.
This is my CL code:
PGM
DCLF FILE(LAB4DF)
SNDRCVF RCDFMT(RECORD1) /* send, recieve file */
DOWHILE (&IN03 = '0')
SELECT
WHEN (&USERINPUT = '1' *OR &USERINPUT = '01') CALLSUBR OPTION1
OTHERWISE DO
*IN03 = '1'
ENDDO
ENDSELECT
ENDDO
SUBR OPTION1
DSPLIBL
ENDSUBR
ENDPGM
And this is my DSPF code
A R RECORD1
A 1 38'LAB 4'
A 3 3'Please select one of the following-
A options:'
A 6 11'3. Maximum Invalid Signon Attempt-
A s allowed'
A 8 11'5. Run Instructor''s Insurance Pr-
A ogram'
A 5 11'2. Signed on User''s Message Queu-
A e'
A 1 3'Yathavan Parameshwaran'
A 7 11'4. Initial number of active jobs -
A for storage allocation'
A 4 11'1. Previous sign on by signed on -
A user'
A 14 11'F3 = Exit'
A 14 31'F21 = Command Line'
A 2 70TIME
A 1 72DATE
A 9 11'Option: '
A USERINPUT 2 B 9 19
A 91 DSPATR(RI)
A 92 DSPATR(PC)
A MSGTXT1 70 O 11 11
A MSGTXT2 70 O 12 11
Is there a problem with my CL code or DSPF code?
You forgot to say what error you were getting. It's always important to put all the information about error messages into your questions.
There are two errors.
&IN03 is not defined
Your assignment to *IN03 should be to &IN03, but that's not how you do an assignment in CLP
If you want to be able to press F3, you have to code something like CA03(03) in the "Functions" for the record format.
To assign a variable in CL, code
CHGVAR name value
Looking at the documentation here, I suspect you need to add RCDFMT to your DCLF spec like so:
DCLF FILE(LAB4DF) RCDFMT(RECORD1)
SNDRCVF RCDFMT(RECORD1) /* send, recieve file */
If you really do only have 1 record format in your display file, then you can also omit the RCDFMT from both commands like so:
DCLF FILE(LAB4DF)
SNDRCVF /* send, recieve file */
Basically, I have a graph coloring program where each node with an edge to another node has to be different colors. Here, is my code:
node(1..4).
edge(1,2).
edge(2,3).
edge(3,4).
edge(4,1).
edge(2,4).
color(1..3).
{ assign(N,C) : color(C) } = 1 :- node(N).
1 { assign(N,1) : color(1) } 1 :- node(N). %line in question
:- edge(N,M), assign(N,C), assign(M,C).
How would I tell the program to only assign color 1, once? The line labeled %line in question is the line giving me problems. Here is another solution I tried that didn't work:
node(1..4).
edge(1,2).
edge(2,3).
edge(3,4).
edge(4,1).
edge(2,4).
color(1..3).
{ assign(N,C) : color(C) } = 1 :- node(N).
:- edge(N,M), assign(N,C), assign(M,C).
vtx(Node, Color) :- node(Node), color(Color).
1 { vtx(N, 1) : color(1) } 1 :- node(N).
#show vtx/2.
If anyone could help me out it would be much appreciated.
In this simple case of restricting a single color to be used once, you can write the a single constraint
:- assign(N, 1), assign(M, 1), node(N), node(M), N!=M.
Actually, the line you marked as in question :
1 { assign(N,1) : color(1) } 1 :- node(N). %line in question
can be translated as
If N is a node, we will (and we must) assign color(1) to node(N) and only assign once, i.e. If node(i) is true, we will have exactly one node(i, 1).
Therefore, with this rule and your facts node(1..4), you will immediately get assign(1,1), assign(2,1), assign(3,1), assign(4,1). This is defninitely unsatisfiable under color problem (with the last constraint).
Back to your requirement:
How would I tell the program to only assign color 1, once?
The problem here is the constraint you set in the line: "color 1 is assigned only once" applies to each node(i), i=1,2,3,4 instead of all nodes.
To make it clearer, you might as well consider that this line would be instantiated as:
1 { assign(1,1) : color(1) } 1 :- node(1).
1 { assign(2,1) : color(1) } 1 :- node(2).
1 { assign(3,1) : color(1) } 1 :- node(3).
1 { assign(4,1) : color(1) } 1 :- node(4).
With node(1..4) all true, we will have assign(1,1), assign(2,1), assign(3,1), assign(4,1).
What you want is assign(N, 1) appears once and only once in the answer, thus in your rule, this should be true with no premiere condition.
Therefore, change the problem line into:
{ assign(N,1): node(N), color(1) } = 1. %problem line changed
You will get the proper assignment:
clingo version 5.4.0
Reading from test.lp
Solving...
Answer: 1
assign(2,2) assign(1,3) assign(3,3) assign(4,1)
Answer: 2
assign(1,2) assign(2,3) assign(3,2) assign(4,1)
Answer: 3
assign(2,1) assign(1,3) assign(3,3) assign(4,2)
Answer: 4
assign(2,1) assign(1,2) assign(3,2) assign(4,3)
SATISFIABLE
Intuitively, this line means the assign(N, 1) should be in answer set under no condition, as long as N is a node. This will count all nodes instead of every single one.
I'm trying to plot individual sequences by means of function seqIplot() in TraMineR. These individual sequences represent work trajectories, completed by former school's graduates via a WEB questionnaire.
Using argument "sortv", I'd like to sort my sequences according to the order of the levels of one covariate, the year of graduation, named "PROMO".
"PROMO" is a factor variable contained in a data frame named "covariates.seq", gathering covariates together:
str(covariates.seq)
'data.frame': 733 obs. of 6 variables:
$ ID_SQ : Factor w/ 733 levels "1","2","3","5",..: 1 2 3 4 5 6
7 8 9 10 ...
$ SEXE : Factor w/ 2 levels "Féminin","Masculin": 1 1 1 1 2 1
1 2 2 1 ...
$ PROMO : Factor w/ 6 levels "1997","1998",..: 1 2 2 4 4 3 2 2
2 2 ...
$ DEPARTEMENT : Factor w/ 10 levels "BC","GCU","GE",..: 1 4 7 8 7 9
9 7 7 4 ...
$ NIVEAU_ADMISSION: Factor w/ 2 levels "En Premier Cycle",..: NA 1 1 1 1
1 NA 1 1 1 ...
$ FILIERE_SECTION : Factor w/ 4 levels "Cursus Classique",..: NA 4 2 NA
1 1 NA NA 4 3 ..
I'm also using "SEXE", the graduates' gender, as a grouping variable. To plot the individual sequences so, my command is as follows:
seqIplot(sequences, group = covariates.seq$SEXE,
sortv = covariates.seq$PROMO,
cex.axis = 0.7, cex.legend = 0.7)
I expected that, by using a process time axis (with the year of graduation as sequence-dependent origin), sorting the sequences according to the order of the levels of "PROMO" would give a plot with groups of sequences from the longest (for the older graduates) to the shortest (for the younger graduates).
But I've got an issue: in the output plot, the sequences don't appear to be correctly sorted according to the levels of "PROMO". Indeed, by using "sortv = covariates.seq$PROMO" as in the command above, the plot doesn't show groups of sequences from the longest to the shortest, as expected. It looks like the plot obtained without using the argument "sortv" (see Figures below).
Without using argument "sortv"
Using "sortv = covariates.seq$PROMO"
Note that I have 733 individual sequences in my object "sequences", created as follows:
labs <- c("En poste","Au chômage (d'au moins 6 mois)", "Autre situation
(d'au moins 6 mois)","En poursuite d'études (thèse ou hors
thèse)", "En reprise d'études / formation (d'au moins 6 mois)")
codes <- c("En poste", "Au chômage", "Autre situation", "En poursuite
d'études", "En reprise d'études / formation")
sequences <- seqdef(situations, alphabet = labs, states = codes, left =
NA, right = "DEL", missing = NA,
cnames = as.character(seq(0,7400/365,1/365)),
xtstep = 365)
The values of the covariates are sorted in the same order as the individual sequences. The covariate "PROMO" doesn't contain any missing value.
Something's going wrong, but what?
Thank you in advance for your help,
Best,
Arnaud.
Using a factor as sortv argument in seqIplot works fine as illustrated by the example below:
sdc <- c("aabbccdd","bbbccc","aaaddd","abcabcab")
sd <- seqdecomp(sdc, sep="")
seq <- seqdef(sd)
fac <- factor(c("2000","2001","2001","2000"))
par(mfrow=c(1,3))
seqIplot(seq, with.legend=FALSE)
seqIplot(seq, sortv=fac, with.legend=FALSE)
seqlegend(seq)
I have the following definition in my *profile.ps1 file:
if(Test-Path $env:M2_HOME){
function mvn{
$cmd = "$env:M2_HOME\bin\mvn.bat"
& $cmd $args
}
}
When I define a function using this function in powershell like:
function d { mvn help:describe $args }
using like:
d -Dplugin=jar
everything is fine as opposed to defining the latter as:
function d { mvn help:describe -Dplugin=$args }
using like:
d jar
Is there some builtin to handle this corner case?
It looks like you just need to make sure you're passing the arguments as strings and ensure they're evaluated first:
function mvn{
$cmd = "$env:M2_HOME\bin\mvn.bat"
& $cmd $args
}
function d { mvn "help:describe" "-Dplugin=$($args)" }
for get argument through call function you should use like this
function test {
write-host $args[0]
write-host $args[1]
}
test stackoverflow powershell
output
stackoverflow
powershell
stackoverflow is first argument passed to function and powershell is second argument passed to function
for every argument passed to function
function test {
foreach ($a in $args){
write-host "output:$args"
}
}
test 1 2 3 4 5 6 7 8
output:
test 1 2 3 4 5 6 7 8
output:1 2 3 4 5 6 7 8
output:1 2 3 4 5 6 7 8
output:1 2 3 4 5 6 7 8
output:1 2 3 4 5 6 7 8
output:1 2 3 4 5 6 7 8
output:1 2 3 4 5 6 7 8
output:1 2 3 4 5 6 7 8
output:1 2 3 4 5 6 7 8
for your function
if(Test-Path $env:M2_HOME){
function mvn{
$cmd = "$env:M2_HOME\bin\mvn.bat"
foreach ($arg in $args) {
& $cmd $args}
}
}
Someone seems to have made a similar observation:
The reason seems -D being a special character in powershell more or less marking where exactly each particluar special option string determined by -D ends, at least this modification to my calling function then works for me:
function x {mvn help:describe `-Dplugin=$args}
Sure it would be nice to handle such occasions in the hosting function somehow (mvn definition in my *profile.ps1), but that solutions rather seems out of scope of my question.
I'm quite new with Matlab and I've been searching, unsucessfully, for the following issue: I have an unstructure txt file, with several rows I don't need, but there are a number of rows inside that file that have an structured format. I've been researching how to "load" the file to edit it, but cannot find anything.
Since i don't know if I was clear, let me show you the content in the file:
8782 PROJCS["UTM-39",GEOGC.......
1 676135.67755473056 2673731.9365976951 -15 0
2 663999.99999999302 2717629.9999999981 -14.00231124135486 3
3 709999.99999999162 2707679.2185399458 -10 2
4 679972.20003752434 2674637.5679516452 0.070000000000000007 1
5 676124.87132483651 2674327.3183533219 -18.94794942571912 0
6 682614.20527054626 2671000.0000000549 -1.6383425512446661 0
...........
8780 682247.4593014461 2676571.1515358146 0.1541080392180566 0
8781 695426.98657108378 2698111.6168302582 -8.5039945992245904 0
8782 674723.80100125563 2675133.5486935056 -19.920312922947179 0
16997 3 21
1 2147 658 590
2 1855 2529 5623
.........
I'd appreciate if someone can just tell me if there is the possibility to open the file to later load only the rows starting with 1 to the one starting with 8782. First row and all the others are not important.
I know than manually copy and paste to a new file would be a solution, but I'd like to know about the possibility to read the file and edit it for other ideas I have.
Thanks!
% Now lines{i} is the string of the i'th line.
lines = strsplit(fileread('filename'), '\n')
% Now elements{i}{j} is the j'th field of the i'th line.
elements = arrayfun(#(x){strsplit(x{1}, ' ')}, lines)
% Remove the first row:
elements(1) = []
% Take the first several rows:
n_rows = 8782
elements = elements(1:n_rows)
Or if the number of rows you need to take is not fixed, you can replace the last two statements above by:
firsts = arrayfun(#(x)str2num(x{1}{1}), elements)
n_rows = find((firsts(2:end) - firsts(1:end-1)) ~= 1, 1, 'first')
elements = elements(1:n_rows)