Haskell AST Incomplete Location Information - parsing

I've been trying to build Haskell AST's so that I can find out which AST nodes exist on which lines of a source file. So far, I am using Language.Haskell.Parser and Language.Haskell.Syntax, which seem to work fairly well. I can generate a tree, then run through each part of it get out the line number using srcLine loc (where loc is a SrcLoc).
However, I am running into issues when the parsed file looks like this:
FILE 1
1| rangeLeq :: Integer -> NonnegRange
2| rangeLeq n =
3| Range BoundaryBelowAll (BoundaryAbove n)
The file could easily have been written like this:
FILE 2
1| rangeLeq :: Integer -> NonnegRange
2| rangeLeq n = Range BoundaryBelowAll (BoundaryAbove n)
The problem is that the parser sees these two as equivalent. It does not assign a SrcLoc to everything, only to certain parts of the AST. So what I end up with is the following output for both file 1 and file 2:
line 1: HsTypeSig
|--HsIdent (rangeLeq)
|--HsQualType
|--HsContext
|--HsTyFun
|--HsTyCon
|--HsUnQual
|--HsIdent (Integer)
|--HsTyCon
|--HsUnQual
|--HsIdent (NonnegRange)
line 2: HsMatch
|--HsIdent (rangeLeq)
|--HsPVar
|--HsIdent (n)
|--HsUnGuardedRhs
|--HsApp
|--HsApp
|--HsCon
|--HsUnQual
|--HsIdent (Range)
|--HsCon
|--HsUnQual
|--HsIdent (BoundaryBelowAll)
|--HsParen
|--HsApp
|--HsCon
|--HsUnQual
|--HsIdent (BoundaryAbove)
|--HsVar
|--HsUnQual
|--HsIdent (n)
So the issue here is that there is no location information on most of the nodes, so it sees the function definition (HsMatch) as a single line. In case it is unclear, The HsMatch represents rangeLeq n = Range BoundaryBelowAll (BoundaryAbove n) in the code. Since the only part of the AST that comes with a SrcLoc is the HsMatch itself, the parser is assuming that all parts of the HsMatch are on the same line.
tl;dr How can I get a correct parse, such that lines in the source file that are split unnecessarily still get tagged with the appropriate line? Id est, I want a parse that tags every single node with a SrcLoc, not just certain nodes.

haskell-src-exts has a location tag (SrcSpanInfo by default) on every subexpression; it should have enough detail to disambiguate your two example cases.

Related

How to reference a particular row for an existing variable in SPSS syntax?

I have 2 variables, one for raw p-values and another for adjusted p-values. I need to compute a new variable based on the values of these two variables. What I need to do isn't too complicated, but I have a hard time doing it in SPSS because I can't figure out how I can reference a particular row for an existing variable in SPSS syntax.
The first column lists raw p-values in ascending order. The next column lists adjusted p-values, but these adjusted p-values are still incomplete. I need to compare two adjacent p-values in the adjusted p-values column (e.g., row 1 and 2, row 2 and 3, row 3 and 4, and so forth), and take the p-values whichever is smaller in each of these comparisons and enter those p-values into the following column as values for a new variable.
However, that's not the end of the story. One more condition has to be met. That is, the new p-values have to be in the same order as the raw p-values. However, I cannot ensure this if I start the comparisons from the top row. You can see that (i') is greater than (h') and (g'), and (d') is greater than (c'), (b'), and (a') in the example below (picture).
In order to solve this issue, I would need to start the comparison of the adjusted p-values from the bottom. In addition, I would need to compare the adjusted p-values to the new p-values of one row below. One exception is that I can simply use the value of (a) as the value of (a') since the value of (a) should always be the greatest of all the p-values as a rule. Then, for (b') , I need to compare (b) and (a') and enter whichever is smaller as (b'). For (c'), I need to compare (c) and (b') and enter whichever is smaller as (c'), and so forth. By doing this way, (d') would be 0.911 and (i') would be 0.017.
Sorry for this long post, but I would really appreciate if I can get some help to do this task in SPSS.
Thank you in advance for your help.
Raw p-values | Adjusted p-values (Temporal)| New p-values (Final)
-------------|-----------------------------|---------------------
0.002 | 0.030 (i) | 0.025 (i')
0.003 | 0.025 (h) | 0.017 (h')
0.004 | 0.017 (g) | 0.017 (g')
0.005 | 0.028 (f) | 0.028 (f')
0.023 | 0.068 (e) | 0.068 (e')
0.450 | 1.061 (d) | 1.061 (d')
0.544 | 1.145 (c) | 0.911 (c')
0.850 | 0.911 (b) | 0.911 (b')
0.974 | 0.974 (a) | 0.974 (a')
Another tool that may be convenient is the SHIFT VALUES command. It can move one or more columns of data either forward or backward.
I wonder whether the purpose of this has to do with adjusting p values for multiple testing corrections as with Benjamin-Hochberg FDR or others similar. If that is the case, you might find the STATS PADJUST (Analyze > Descriptives > Calculate adjusted p values) extension command useful. It offers six adjustment methods. You can install it from the Utilities (pre-V24) or Extensions (V24+) menu.
To get you started, here are a few tools that can help you with this task:
The LAG function
you can compare values in this line and the previous one, for example, the following will compare the Pval in each line to the one in the previous one, and put the smaller of the two in the NewPval:
compute NewPVal=min(Pval, lag(Pval)).
If you want to do the same process only start from the bottom, you can easily sort your data in reverse order and do the same.
CREATE + LEAD
if you want to make comparisons to the next line instead of the previous line, you should first create a "lead" variable and then compare to it.
for example, the following syntax will create a new variable that for each line contains the value of Pval in the next line, and then chooses the smaller of the two for the NewPval:
create /LeadPval=LEAD(Pval 1).
compute NewPVal=min(Pval, LeadPval).
Using case numbers
You can use case numbers (line numbers) in calculations and in conditions. For example, the following syntax will let you make different calculations in the first line and the following ones:
if $casenum=1 NewPval=Pval.
if $casenum>1 NewPVal=min(Pval, lag(Pval)).

Cypher request to extract genealogy

I'm parsing a book within neo4j and I'd like to extract genealogy out of it I have sentences like :
"A begat B,C and D"
"X begat Y, and Y begat Z, ..."
and I store that as
(A:word)-[:subj]->(begat:word)-[:obj]-> (B:word)
(A:word)-[:subj]->(begat:word)-[:comp]-> (C:word)
(X:word)-[:subj]->(begat:word)-[:obj]-> (Y:word)
(Y:word)-[:subj]->(begat:word)-[:obj]-> (Z:word)
(X:word)-[:NNP]->(sentence:word)
(Y:word)-[:NNP]->(sentence:word)
(Z:word)-[:NNP]->(sentence:word)
(begat:word)-[:VBG]->(sentence:word)
How could I write my cypher request so that neo4j server visualization give me a tree instead of one "begat" node with all the other ones linking to it ? My genealogy is on several sentences and when linking word together I add the sentenceId to the relationship maybe we could use that.
The result would look like
A
______|_____
| | |
B C D
|
X
|
Y
|
Z
One more info the words are stored only once to avoid memory consumption.
Here is a sample of my data :
http://console.neo4j.org/r/xzsazf
Many thanks

Exception in thread "main" java.lang.IllegalArgumentException: Wrong number of attributes in the string + Mahout

I am trying to create a file descriptor using the command:
$ MAHOUT_HOME/core/target/mahout-core--job.jar org.apache.mahout.classifier.df.tools.Describe -p testdata/KDDTrain+.arff -f testdata/KDDTrain+.info -d N 3 C 2 N C 4 N C 8 N 2 C 19 N L
from the link:
https://mahout.apache.org/users/classification/partial-implementation.html on my data file but whatever file I take and change the number of attributes string N 3 C 2 N C 4 N C 8 N 2 C 19 N L .
I get the following exception:
Exception in thread "main" java.lang.IllegalArgumentException: Wrong number of attributes in the string
Please help!
There are a couple of reasons for which you might get an error like that...
Wrong Descriptor: Putting this for a sake of completeness. You must have already checked this one out. You have actually given a wrong descriptor for the data. Re-check the number and type of columns and then give them correctly to the descriptor.
Bad separator: Re-check the delimiter used in the data. That also might create some trouble. May be the data you have has some wrongly placed delimiter in some records. Make sure of that.
Special Characters: In my few experiments, I have noticed mahout does not enjoy if there are certain special characters, or data consists of characters of language other than English (unless of course, you tweak around the code). So make sure you have a way of handling them, and you should be good to go.
Anyways all these fight just so you can create a descriptor of the data. ATB.
Old question, but I had a more acute answer that I discovered after landing here with the same problem.
In this particular case, the problem I found was that the format of data file (from http://nsl.cs.unb.ca/NSL-KDD/) seems to have changed from the example as listed on the Mahout Random Forest example page.
The example lists a line format with the specifier
N 3 C 2 N C 4 N C 8 N 2 C 19 N L
but there's an extra element at the end of the lines; for example:
13,tcp,telnet,SF,118,2425,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,1,0.00,0.00,0.00,0.00,1.00,0.00,0.00,26,10,0.38,0.12,0.04,0.00,0.00,0.00,0.12,0.30,guess_passwd,2
which has one more field. Adding another number field (N) to the end of the specifier, as
N 3 C 2 N C 4 N C 8 N 2 C 19 N L N
I had luck using just the plain .txt file format instead of the .arff file format.

Binary to Integer -> Erlang

I have a binary M such that 34= will always be present and the rest may vary between any number of digits but will always be an integer.
M = [<<"34=21">>]
When I run this command I get an answer like
hd([X || <<"34=", X/binary >> <- M])
Answer -> <<"21">>
How can I get this to be an integer with the most care taken to make it as efficient as possible?
[<<"34=",X/binary>>] = M,
list_to_integer(binary_to_list(X)).
That yields the integer 21
As of R16B, the BIF binary_to_integer/1 can be used:
OTP-10300
Added four new bifs, erlang:binary_to_integer/1,2,
erlang:integer_to_binary/1, erlang:binary_to_float/1 and
erlang:float_to_binary/1,2. These bifs work similarly to how
their list counterparts work, except they operate on
binaries. In most cases converting from and to binaries is
faster than converting from and to lists.
These bifs are auto-imported into erlang source files and can
therefore be used without the erlang prefix.
So that would look like:
[<<"34=",X/binary>>] = M,
binary_to_integer(X).
A string representation of a number can be converted by N-48. For multi-digit numbers you can fold over the binary, multiplying by the power of the position of the digit:
-spec to_int(binary()) -> integer().
to_int(Bin) when is_binary(Bin) ->
to_int(Bin, {size(Bin), 0}).
to_int(_, {0, Acc}) ->
erlang:trunc(Acc);
to_int(<<N/integer, Tail/binary>>, {Pos, Acc}) when N >= 48, N =< 57 ->
to_int(Tail, {Pos-1, Acc + ((N-48) * math:pow(10, Pos-1))}).
The performance of this is around 100 times slower than using the list_to_integer(binary_to_list(X)) option.

How does the Erlang compiler handle pattern matching? What does it output?

I just asked a question about how the Erlang compiler implements pattern matching, and I got some great responses, one of which is the compiled bytecode (obtained with a parameter passed to the c() directive):
{function, match, 1, 2}.
{label,1}.
{func_info,{atom,match},{atom,match},1}.
{label,2}.
{test,is_tuple,{f,3},[{x,0}]}.
{test,test_arity,{f,3},[{x,0},2]}.
{get_tuple_element,{x,0},0,{x,1}}.
{test,is_eq_exact,{f,3},[{x,1},{atom,a}]}.
return.
{label,3}.
{badmatch,{x,0}}
Its all just plain Erlang tuples. I was expecting some cryptic binary thingy, guess not. I am asking this on impulse here (I could look at the compiler source but asking questions always ends up better with extra insight), how is this output translated in the binary level?
Say {test,is_tuple,{f,3},[{x,0}]} for example. I am assuming this is one instruction, called 'test'... anyway, so this output would essentially be the AST of the bytecode level language, from which the binary encoding is just a 1-1 translation?
This is all so exciting, I had no idea that I can this easily see what the Erlang compiler break things into.
ok so I dug into the compiler source code to find the answer, and to my surprise the asm file produced with the 'S' parameter to the compile:file() function is actually consulted in as is (file:consult()) and then the tuples are checked one by one for further action(line 661 - beam_consult_asm(St) -> - compile.erl). further on then there's a generated mapping table in there (compile folder of the erlang source) that shows what the serial number of each bytecode label is, and Im guessing this is used to generate the actual binary signature of the bytecode.
great stuff. but you just gotta love the consult() function, you can almost have a lispy type syntax for a random language and avoid the need for a parser/lexer fully and just consult source code into the compiler and do stuff with it... code as data data as code...
The compiler has a so-called pattern match compiler which will take a pattern and compile it down to what is essentially a series of branches, switches and such. The code for Erlang is in v3_kernel.erl in the compiler. It uses Simon Peyton Jones, "The Implementation of Functional
Programming Languages", available online at
http://research.microsoft.com/en-us/um/people/simonpj/papers/slpj-book-1987/
Another worthy paper is the one by Peter Sestoft,
http://www.itu.dk/~sestoft/papers/match.ps.gz
which derives a pattern match compiler by inspecting partial evaluation of a simpler system. It may be an easier read, especially if you know ML.
The basic idea is that if you have, say:
% 1
f(a, b) ->
% 2
f(a, c) ->
% 3
f(b, b) ->
% 4
f(b, c) ->
Suppose now we have a call f(X, Y). Say X = a. Then only 1 and 2 are applicable. So we check Y = b and then Y = c. If on the other hand X /= a then we know that we can skip 1 and 2 and begin testing 3 and 4. The key is that if something does not match it tells us something about where the match can continue as well as when we do match. It is a set of constraints which we can solve by testing.
Pattern match compilers seek to optimize the number of tests so there are as few as possible before we have conclusion. Statically typed language have some advantages here since they may know that:
-type foo() :: a | b | c.
and then if we have
-spec f(foo() -> any().
f(a) ->
f(b) ->
f(c) ->
and we did not match f(a), f(b) then f(c) must match. Erlang has to check and then fail if it doesn't match.

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