Is there a no-fuss serialization method for Haskell, similar to Erlang's term_to_binary/binary_to_term calls? Data.Binary seems unnecessarily complicated and raw. See this example where you are basically manually encoding terms to integers.
Use Data.Binary, and one of the deriving scripts that come with the package.
It's very simple to derive Binary instances, via the 'derive' or 'deriveM' functions provided in the tools set of Data.Binay.
derive :: (Data a) => a -> String
For any 'a' in Data, it derives a Binary instance for you as a String. There's a putStr version too, deriveM.
Example:
*Main> deriveM (undefined :: Drinks)
instance Binary Main.Drinks where
put (Beer a) = putWord8 0 >> put a
put Coffee = putWord8 1
put Tea = putWord8 2
put EnergyDrink = putWord8 3
put Water = putWord8 4
put Wine = putWord8 5
put Whisky = putWord8 6
get = do
tag_ <- getWord8
case tag_ of
0 -> get >>= \a -> return (Beer a)
1 -> return Coffee
2 -> return Tea
3 -> return EnergyDrink
4 -> return Water
5 -> return Wine
6 -> return Whisky
_ -> fail "no parse"
The example you cite is an example of what the machine generated output looks like -- yes, it is all bits at the lowest level! Don't write it by hand though -- use a tool to derive it for you via reflection.
Related
Can PICT (=Pairwise Independent Combinatorial Testing) handle/model independent parameters.
For example in following input a and b are independent, so they should not be combined.
Input in PICT:
a: 1, 2, 3, 4
b: 5, 6, 7, 8
//some line which models the independence: a independent of b
Output, that I would expect:
a b
1 5
2 6
3 7
4 8
This example, with only 2 parameters, of course normally would not make much sense, but it's illustrative.
The same could be applied to 3 parameters (a,b,c), where a is independent of b, but not c.
The main goal of declaring parameters as independent would be the reduce the number of tests.
I read the paper/user guide to PICT, but I didn't found any useful information.
I will answer my question by myself:
The solution is to define submodels and set the default order from 2 (=pairwise) to 1 (= no combination).
For example parameter a = {a_1, a_2, a_3} should be independent of
b = {b_1, b_2, b_3} and
c = {c_1, ..., c_4}.
Therefor I would expect 12 tests ((b x c) + a).
Resulting in the following input file:
a: 1,2,3
b: 1,2,3
c: 1,2,3,4
{b,c}#2
{b,c}#2 defines a submodel, consisting of b and c, which uses pairwise combination.
And running pict with the option: "/o:1".
In PICT you can define conditions and achieve your goal to not combine them with the rest. You can add additional option as N/A and have something like this;
If [A] = '1' Then [B] = 'N/A';
This is one possible option to handle this case.
I have a List of custom types that I would like to sort on one attribute which is of type Maybe Time.Posix. In reading the docs I've come to the conclusion I should use List.sortWith, as neither Maybe nor Time.Posix values are comparable. I've written a set of functions to prepare the values so they are comparable. The challenge that I'm facing, though, is pulling the values from the types in the list.
Here are the functions:
maybeCompare : (a -> a -> Order) -> Maybe a -> Maybe a -> Order
maybeCompare f a b =
case a of
Just some_a ->
case b of
Just some_b ->
f some_a some_b
Nothing ->
GT
Nothing ->
LT
posixCompare : Time.Posix -> Time.Posix -> Order
posixCompare a b = compare (posixToMillis(a)) (posixToMillis(b))
posMay = maybeCompare (posixCompare)
Which I combine and use like this:
List.sortWith (posMay .time) objList
On data that looks like this:
obj1 = {id=1,time= Just time1}
obj2 = {id=2,time= Just time2}
obj3 = {id=3,time= Just time3}
obj4 = {id=4,time= Just time4}
obj5 = {id=5,time= Nothing}
objList = obj1 :: obj2 :: obj3 :: obj4 :: obj5 :: []
Now, this approach works for a list like this List (Maybe Time.Posix). By which I mean I get the output I expect, the list is sorted on the Posix time with the Nothing values in the desired location.
However, for a List of types where Maybe Time.Posix is one of the values I get this error (one of many, but I think this is the source):
List.sortWith (posMay .time) objList
^^^^^
This .time field access function has type:
{ b | time : a } -> a
But `posMay` needs the 1st argument to be:
Maybe Time.Posix
Is there are way to make the types of my functions align to sort this kind of data? Or, should I rethink my approach?
I've created a working example at https://ellie-app.com/8dp2qD6fDzBa1
Your posMay function is of the type Maybe a -> Maybe a -> Order, so it's not expecting .time, which is a function of type {id:Int,time:Maybe Posix} -> Maybe Posix.
Instead, you can create a different function which shims between posMay and List.sortWith, which would look like this: List.sortWith (\a b -> posMay a.time b.time)
If you want to be able to pass a getter function to posMay, you could rewrite it to accept that function:
posMay getter a b =
maybeCompare posixCompare (getter a) (getter b)
Then, you would use it like this: List.sortWith (posMay .time) (or List.sortWith (posMay identity) for a List (Maybe Posix). A version that works like this is at https://ellie-app.com/8dp7gm3qthka1
It seems that if I write a set to a file, it's not in a format where it can be read back in easily as a set. Here's an example:
#lang racket
(let ([out (open-output-file "test.rkt" #:exists 'replace)])
(write (set 1 2 3 4 5) out)
(close-output-port out))
This makes a file with #<set: 1 3 5 2 4>, which the reader complains about. There is a related unanswered question on the mailing list here.
The way I'm getting around it right now is by printing literally the string "(set " to a file, then all the integers with spaces, then a closing ")". Super ugly and I would like to use the reader if possible.
You can use the Racket serialization library to do this. Here's an example:
Welcome to Racket v6.4.0.7.
-> (require racket/serialize)
-> (with-output-to-file "/tmp/set.rktd"
(lambda () (write (serialize (set 1 2 3)))))
-> (with-input-from-file "/tmp/set.rktd"
(lambda () (deserialize (read))))
(set 1 3 2)
Note that a serialized value is just a special kind of s-expression, so you can manipulate it like other values (like store it in a database, write it to disk, send it over a network, etc.):
-> (serialize (set 1 2 3))
'((3)
1
(((lib "racket/private/set-types.rkt")
.
deserialize-info:immutable-custom-set-v0))
0
()
()
(0 #f (h - (equal) (1 . #t) (3 . #t) (2 . #t))))
would anyone be able to provide me with a working example of dataframe zipping in C#? I am bit lost in the operation.
Thanks!
The frame.Zip operation is the same thing as zipAlign in the more documented F# API, so have a look at zipAlign in this section of the documentation.
Given a frame df1:
A
1 -> 1
2 -> 2
And a frame df2:
A
2 -> 2
3 -> 3
When you call df1.Zip(df2, (int a, int b) -> a + b), you get:
A
1 -> <missing>
2 -> 4
3 -> <missing>
That is, for cells where both frames contain a value, a + b is calculated. For all other cells, you get a missing value. Note that you need type annotations in the lambda function - this has to match the type of values in the frame (for not matching types, the function just returns the values from the first frame unchanged).
I read in a book on non-deterministic mapping there is mapping from Q*∑ to 2Q for M=(Q,∑,trans,q0,F)
where Q is a set of states.
But I am not able to understand how it's 2Q;
if there are 3 states a, b, c, how does it map to 8 states?
I always found that the easiest way to think about these (since the set of states is finite) is as having each of those subsets be an encoding of a base-2 number that ranges from 0 (all bits zero) to 2|Q|-1 (all bits one), where there are as many bits in the number as there are members in the state set, Q. Then, you can just take one of these numbers and map it into a subset by using whether a particular bit in the number is set. Easy!
Here's a worked example where Q = {a,b,c}. In this case, |Q| is 3 (there are three elements) and so 23 is 8. That means we get this if we say that the leading bit is for element a, the next bit is for b, and the trailing bit for c:
0 = 000 = {}
1 = 001 = {c}
2 = 010 = {b}
3 = 011 = {b,c}
4 = 100 = {a}
5 = 101 = {a,c}
6 = 110 = {a,b}
7 = 111 = {a,b,c}
See? That initial three states has been transformed into 8, and we have a natural numbering of them that we could use to create the labels of those states if we chose.
Now, to the interpretations of this within a non-deterministic context. Basically, the non-determinism means that we're uncertain about what state we're in. We represent this by using a pseudo-state that is the set of “real” states that we might be in; if we have total non-determinism then we are in the pseudo-state where all real-states are possible (i.e., {a,b,c}) whereas the pseudo-state where no real-states are possible (i.e., {}) is the converse (and really ought to be impossible to reach in the transition system). In a real system, you're usually not dealing with either of those extremes.
The logic of how you convert the deterministic transition system into a non-deterministic one is rather more complex than I want to go into here. (I had to read a substantial PhD thesis to learn it so it's definitely more than an SO answer's worth!)
2Q means the set of all subsets of Q. For each state q and each letter x from sigma, there is a subset of Q states to which you can go from q with letter x. So yeah, if there are three states abc the set 2Q consists of 8 elements {{}, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}}. It doesn't map to 8 states, it maps to one of these 8 sets. HTH