I want to create a function in which, first, it filters one element of a dataframe in Julia. Second, it tests if the element is "missing". If the answer is rue, it return the value "0.0". My issue is that the control evaluation "if" does not work and I don t know why. If the element is "String" the control evaluation works, however, the element is a 1-element Vector{SubString{String}}: after filtering; thus, the control evaluation does not work. I would like to know why and it is possible to turn the vector element into a string object.
Note: "isequal", '==', '===' do not work either.
For example:
example_ped = DataFrame(animal = collect(1:1:11),
sire = [fill(0,5); fill(4,3); fill(5,3)],
dam = [fill(0,4); fill(2,4); fill(3,3)])
CSV.write("ped_example.txt",example_ped, header=true,delim='\t')
pedi = CSV.read("ped_example.txt",delim = '\t', header=true, missingstrings=["0"], DataFrame)
pedi[!,1]=strip.(string.(pedi[!,1]))
pedi[!,2]=strip.(string.(pedi[!,2]))
pedi[!,3]=strip.(string.(pedi[!,3]))
Part of the function
function computAddRel!(ped,animal_1,animal_2)
elder,recent = animal_1 < animal_2 ? (animal_1,animal_2) : (animal_2,animal_1)
sireOfrecent = ped.sire[ped.animal.==recent]
damOfrecent = ped[ped.animal.==recent,"dam"]
if elder==recent
f_inbreed = (sireOfrecent=="missing" || damOfrecent=="missing") ? 0.0 : 0.5*computAddRel!(ped,sireOfrecent,damOfrecent)
adiv = 1.0 + f_inbreed
return adiv
end
end
if the animal_1 and animal_2 are equal to 5
julia> sireOfrecent = pedi.sire[pedi.animal.==recent]
1-element Vector{Union{Missing, Int64}}:
missing
However, the control evaluation is false
julia> sireOfrecent=="missing"
false
julia> isequal(sireOfrecent,"missing")
false
Thank in advance for your time.
You should write:
ismissing(only(sireOfrecent))
The meaning of this:
only checks if you picked exactly one row (if not - you will get an error, as then there is ambiguity; if yes - you extract out the element from an array)
ismissing is a function that you should use to check if some value is missing.
Here are some examples:
julia> x = [missing]
1-element Vector{Missing}:
missing
julia> only(x)
missing
julia> ismissing(only(x))
true
julia> only([1, 2])
ERROR: ArgumentError: Collection has multiple elements, must contain exactly 1 element
julia> ismissing(only([1]))
false
I would like to define a rule for proof by cases, to be used with proof (cases rule: <rule-name>). I managed to use the case_names and consumes parameters, but I did not manage to bind the schematic variable ?case, so that, inside a case of a proof using my rule, one can write show ?case .... How do I bind it?
Concretely: I have the Mizar-inspired notion of a trivial set, i.e. empty or singleton set. I would like to prove properties of trivial sets by empty vs. singleton case analysis. So far I have:
definition trivial where "trivial x = (x ⊆ {the_elem x})"
lemma trivial_cases [case_names empty singleton, consumes 1]:
assumes "trivial X"
assumes empty: "P {}"
and singleton: "⋀ x . X = {x} ⟹ P {x}"
shows "P X"
using assms unfolding trivial_def by (metis subset_singletonD)
and I can make use of this as follows:
notepad
begin
fix Q
fix X::"'a set"
have "trivial X" sorry
then have "Q X"
proof (cases rule: trivial_cases)
case empty
show "Q {}" sorry
next
case (singleton x)
show "Q {x}" sorry
qed
end
But I cannot use show ?case. If I try, it gives me the error message "Unbound schematic variable: ?case". print_cases inside the proof outputs the following:
cases:
empty:
let "?case" = "?P {}"
singleton:
fix x_
let "?case" = "?P {x_}"
assume "X = {x_}"
Which suggests that it doesn't work because ?P is not bound to trivial.
BTW: The full context in which I am using this can be seen at https://github.com/formare/auctions/blob/master/isabelle/Auction/SetUtils.thy.
As Joachim already mentioned, unlike induct, the cases method does not bind the schematic variable ?case. I would say the "canonical" way of conducting case analysis (as a proof method) conforms to this setup, since typically only different assumptions -- which taken together are exhaustive -- are considered, whereas the conclusion stays the same (abbreviated by ?thesis in Isabelle) throughout the different cases. I would set up your trivial_cases as follows:
lemma trivial_cases [consumes 1, case_names empty singleton]:
assumes "trivial X" and "X = {} ⟹ P" and "⋀x . X = {x} ⟹ P"
shows "P"
using assms by (auto simp: trivial_def)
Then you can use it like
notepad
begin
fix P and X :: "'a set"
have "trivial X" sorry
then have "P X"
proof (cases rule: trivial_cases)
case empty
then show ?thesis sorry
next
case (singleton x)
then show ?thesis sorry
qed
end
where the simplifier or explicit unfolding takes care of specializing X to {} and {x}, respectively.
Side Note: You can further tune trivial_cases by adding the attribtue cases pred: trivial. Then, whenever trivial ?X is the major assumption fed to cases, the rule trivial_cases is used implicitly, i.e., you can do the following
have "trivial X" sorry
then have "P X"
proof (cases)
in the above proof.
How can I define a cluster in Haskell using list comprehension?
I want to define a function for the cluster :
( a b c ) = [ a <- [1 .. 10],b<-[2 .. 10], c = (a, b)]
In your comment you gave the example [(1,2,1),(1,3,1),(1,4,1),(1,5,1),(1,6,1),(1,7,1)].
In that example, only the middle number changes, the other two are always 1. You can do this particular one with
ones = [(1,a,1)| a<-[1..7]]
However, you might want to vary the other ones. Let's have a look at how that works, but I'll use letters instead to make it clearer:
> [(1,a,b)| a<-[1..3],b<-['a'..'c']]
[(1,1,'a'),(1,1,'b'),(1,1,'c'),(1,2,'a'),(1,2,'b'),(1,2,'c'),(1,3,'a'),(1,3,'b'),(1,3,'c')]
You can see that the letters are varying more frequently than the numbers - the b<-[1..3] is like an outer loop, with c<-['a'..'c'] being the inner loop.
You could copy the c into the first of the three elements of the tuple:
> [(b,a,b)| a<-[1..3],b<-['a'..'b']]
[('a',1,'a'),('b',1,'b'),('a',2,'a'),('b',2,'b'),('a',3,'a'),('b',3,'b')]
Or give each its own varying input
> [(a,b,c)| a<-[1..2],b<-['a'..'b'],c<-[True,False]]
[(1,'a',True),(1,'a',False),(1,'b',True),(1,'b',False),(2,'a',True),(2,'a',False),(2,'b',True),(2,'b',False)]
The usage message for Set reminds us that multiple assignments can easily be made across two lists, without having to rip anything apart. For example:
Remove[x1, x2, y1, y2, z1, z2];
{x1, x2} = {a, b}
Performs the assignment and returns:
{a, b}
Thread, commonly used to generate lists of rules, can also be called explicitly to achieve the same outcome:
Thread[{y1, y2} = {a, b}]
Thread[{z1, z2} -> {a, b}]
Gives:
{a, b}
{z1 -> a, z2 -> b}
However, employing this approach to generate localized constants generates an error. Consider this trivial example function:
Remove[f];
f[x_] :=
With[{{x1, x2} = {a, b}},
x + x1 + x2
]
f[z]
Here the error message:
With::lvset: "Local variable specification {{x1,x2}={a,b}} contains
{x1,x2}={a,b}, which is an assignment to {x1,x2}; only assignments
to symbols are allowed."
The error message documentation (ref/message/With/lvw), says in the 'More Information' section that, "This message is generated when the first element in With is not a list of assignments to symbols." Given this explanation, I understand the mechanics of why my assignment failed. Nonetheless, I'm puzzled and wondering if this is necessary restriction by WRI, or a minor design oversight that should be reported.
So here's my question:
Can anyone shed some light on this behavior and/or offer a workaround? I experimented with trying to force Evaluation, without luck, and I'm not sure what else to try.
What you request is tricky. This is a job for macros, as already exposed by the others. I will explore a different possibility - to use the same symbols but put some wrappers around the code you want to write. The advantage of this technique is that the code is transformed "lexically" and at "compile-time", rather than at run-time (as in the other answers). This is generally both faster and easier to debug.
So, here is a function which would transform the With with your proposed syntax:
Clear[expandWith];
expandWith[heldCode_Hold] :=
Module[{with},
heldCode /. With -> with //. {
HoldPattern[with[{{} = {}, rest___}, body_]] :>
with[{rest}, body],
HoldPattern[
with[{
Set[{var_Symbol, otherVars___Symbol}, {val_, otherVals___}], rest___},
body_]] :>
with[{{otherVars} = {otherVals}, var = val, rest}, body]
} /. with -> With]
Note that this operates on held code. This has the advantage that we don't have to worry about possible evaluation o the code neither at the start nor when expandWith is finished. Here is how it works:
In[46]:= expandWith#Hold[With[{{x1,x2,x3}={a,b,c}},x+x1+x2+x3]]
Out[46]= Hold[With[{x3=c,x2=b,x1=a},x+x1+x2+x3]]
This is, however, not very convenient to use. Here is a convenience function to simplify this:
ew = Function[code, ReleaseHold#expandWith#Hold#code, HoldAll]
We can use it now as:
In[47]:= ew#With[{{x1,x2}={a,b}},x+x1+x2]
Out[47]= a+b+x
So, to make the expansion happen in the code, simply wrap ew around it. Here is your case for the function's definition:
Remove[f];
ew[f[x_] := With[{{x1, x2} = {a, b}}, x + x1 + x2]]
We now check and see that what we get is an expanded definition:
?f
Global`f
f[x_]:=With[{x2=b,x1=a},x+x1+x2]
The advantage of this approach is that you can wrap ew around an arbitrarily large chunk of your code. What happens is that first, expanded code is generated from it, as if you would write it yourself, and then that code gets executed. For the case of function's definitions, like f above, we cansay that the code generation happens at "compile-time", so you avoid any run-time overhead when usin the function later, which may be substantial if the function is called often.
Another advantage of this approach is its composability: you can come up with many syntax extensions, and for each of them write a function similar to ew. Then, provided that these custom code-transforming functions don't conlict with each other, you can simply compose (nest) them, to get a cumulative effect. In a sense, in this way you create a custom code generator which generates valid Mathematica code from some Mathematica expressions representing programs in your custom languuage, that you may create within Mathematica using these means.
EDIT
In writing expandWith, I used iterative rule application to avoid dealing with evaluation control, which can be a mess. However, for those interested, here is a version which does some explicit work with unevaluated pieces of code.
Clear[expandWithAlt];
expandWithAlt[heldCode_Hold] :=
Module[{myHold},
SetAttributes[myHold, HoldAll];
heldCode //. HoldPattern[With[{Set[{vars__}, {vals__}]}, body_]] :>
With[{eval =
(Thread[Unevaluated[Hold[vars] = Hold[vals]], Hold] /.
Hold[decl___] :> myHold[With[{decl}, body]])},
eval /; True] //. myHold[x_] :> x]
I find it considerably more complicated than the first one though.
The tricky issue is to keep the first argument of Set unevaluated.
Here is my suggestion (open to improvements of course):
SetAttributes[myWith, HoldAll];
myWith[{s : Set[a_List, b_List]}, body_] :=
ReleaseHold#
Hold[With][
Table[Hold[Set][Extract[Hold[s], {1, 1, i}, Hold],
Extract[Hold[s], {1, 2, i}]], {i, Length#b}], Hold#body]
x1 = 12;
Remove[f];
f[x_] := myWith[{{x1, x2} = {a, b}}, x + x1 + x2]
f[z]
results in
a+b+z
Inspired by halirutan below I think his solution, made slightly more safely, is equivalent to the above:
SetAttributes[myWith, HoldAll];
myWith[{Set[a : {__Symbol}, b_List]} /; Length[a] == Length[b],
body_] :=
ReleaseHold#
Hold[With][
Replace[Thread[Hold[a, b]], Hold[x_, y_] :> Hold[Set[x, y]], 1],
Hold#body]
The tutorial "LocalConstants" says
The way With[{x=Subscript[x, 0],...},body] works is to take body, and replace every
occurrence of x, etc. in it by Subscript[x, 0], etc. You can think of With as a
generalization of the /. operator, suitable for application to Mathematica code instead of
other expressions.
Referring to this explanation it seems obvious that something like
x + x1 + x2 /. {x1, x2} -> {a, b}
will not work as it might be expected in the With notation.
Let's assume you really want to hack around this. With[] has the attribute HoldAll, therefore everything you give as first parameter is not evaluated. To make such a vector-assignment work you would have to create
With[{x1=a, x2=b}, ...]
from the vector-notation. Unfortunately,
Thread[{a, b} = {1, 2}]
does not work because the argument to Thread is not held and the assignment is evaluated before Thread can do anything.
Lets fix this
SetAttributes[myThread, HoldFirst];
myThread[Set[a_, b_]] := mySet ### Transpose[{a, b}]
gives
In[31]:= myThread[{a, b, c} = {1, 2, 3}]
Out[31]= {mySet[a, 1], mySet[b, 2], mySet[c, 3]}
What looks promising at first, just moved the problem a bit away. To use this in With[] you have to replace at some point the mySet with the real Set. Exactly then, With[] does not see the list {a=1, b=2, c=3} but, since it has to be evaluated, the result of all assignments
In[32]:= With[
Evaluate[myThread[{a, b, c} = {1, 2, 3}] /. mySet :> Set], a + b + c]
During evaluation of In[32]:= With::lvw: Local
variable specification {1,2,3} contains 1, which is not an assignment to a symbol. >>
Out[32]= With[{1, 2, 3}, a + b + c]
There seems to be not easy way around this and there is a second question here: If there is a way around this restriction, is it as fast as With would be or do we lose the speed advantage compared to Module? And if speed is not so important, why not using Module or Block in the first place?
You could use Transpose to shorten Rolfs solution by 100 characters:
SetAttributes[myWith, HoldAll];
myWith[{Set[a_List, b_List]}, body_] :=
ReleaseHold[Hold[With][Hold[Set[#1, #2]] & ### Transpose[{a, b}],
Hold#body
]]
#Heike, yep the above breaks if either variable has already a value. What about this:
SetAttributes[myWith, HoldAll];
myWith[{Set[a_List, b_List]}, body_] :=
ReleaseHold#
Hold[With][Thread[Hold[a, b]] /. Hold[p__] :> Hold[Set[p]],
Hold#body]