Is there any way that
PadRight[a \[PlusMinus] b,2,""]
Returns
{a \[PlusMinus] b,""}
Instead of
a \[PlusMinus] b \[PlusMinus] ""
?
I believe that i need to somehow deactivate the operator properties of [PlusMinus].
Why do i need this?
I'm creating a program to display tables with physical quantities. To me, that means tables with entries like
(value of a) [PlusMinus] (uncertainty of a)
When i have several columns with different heights, i'm stuffing the shorter ones with "", so i can use Transpose the numeric part of the table.
If the column has more than one entrie, there's no problem:
PadRight[{a \[PlusMinus] b,c \[PlusMinus] d},4,""]
gives what i want:
{a \[PlusMinus] b,c \[PlusMinus] d,"",""}
It is when the column has only one entrie that my problem appears.
This is the code that constructs the body stuffed with "":
If[tested[Sbody],1,
body = PadRight[body, {Length[a], Max[Map[Length, body]]
With
tested[a__] :=
If[Length[DeleteDuplicates[Map[Dimensions, {a}]]] != 1, False,
True];
, a function that discovers if is arguments have the same dimension
and
a={Quantity1,Quantity2,...}
Where the quantities are the one's that i want on my table.
Thanks
First you need to be aware of that any expression in Mathematica is in the form of Head[Body]
where body may be empty, a single expression or a sequence of expressions separated by commas
Length operate on expressions, not necessarily lists
so
Length[PlusMinus[a,b]]
returns 2 since the body of the expression contains to expressions (atoms in this case) that are a and b
Read the documentation on PadRight. The second argument define the final length of the expression
so
PadRight[{a,b},4,c] results with a list of length 4 with the last two elements equal to
PadRight[{a,b},2,c] results with the original list since it is already of length 2
Therefore
PadRight[PlusMinus[a,b],2,anything] just returns the same PlusMinus[a,b] unchanged since it is already of length 2
so, youר first example is wrong. You are not able to get a result with head List using PadRight when you try to pad to an expression with head PlusMinus
There is no problem of executing
PadRight[PlusMinus[a,b],3,""]
but the result looks funny (at best) and logically meaningless, but if this is what you wanted in the first place you get it, and following my explanations above you can figure out why
HTH
best
yehuda
Related
I have to be honest that I don't quite understand Lua that well yet. I am trying to overwrite a local numeric value assigned to a set table address (is this the right term?).
The addresses are of the type:
project.models.stor1.inputs.T_in.default, project.models.stor2.inputs.T_in.default and so on with the stor number increasing.
I would like to do this in a for loop but cannot find the right expression to make the entire string be accepted by Lua as a table address (again, I hope this is the right term).
So far, I tried the following to concatenate the strings but without success in calling and then overwriting the value:
for k = 1,10,1 do
project.models.["stor"..k].inputs.T_in.default = 25
end
for k = 1,10,1 do
"project.models.stor"..j..".T_in.default" = 25
end
EDIT:
I think I found the solution as per https://www.lua.org/pil/2.5.html:
A common mistake for beginners is to confuse a.x with a[x]. The first form represents a["x"], that is, a table indexed by the string "x". The second form is a table indexed by the value of the variable x. See the difference:
for k = 1,10,1 do
project["models"]["stor"..k]["inputs"]["T_in"]["default"] = 25
end
You were almost close.
Lua supports this representation by providing a.name as syntactic sugar for a["name"].
Read more: https://www.lua.org/pil/2.5.html
You can use only one syntax in time.
Either tbl.key or tbl["key"].
The limitation of . is that you can only use constant strings in it (which are also valid variable names).
In square brackets [] you can evaluate runtime expressions.
Correct way to do it:
project.models["stor"..k].inputs.T_in.default = 25
The . in models.["stor"..k] is unnecessary and causes an error. The correct syntax is just models["stor"..k].
Consider this example
my #fib = (1,1, * + * … * > 200).rotor(2 => -1);
say #fib[0] ∈ #fib; # prints True
The first statement creates a Sequence of 2-element subsequences via the use of the rotor function. #fib will contain (1,1), (1,2) and so on. Quite obviously, the first element of a sequence is part of a sequence. Or is it?
my #fib = (1,1, * + * … * > 200).rotor(2 => -1);
say #fib[0], #fib[0].^name; # OUTPUT: «(1 1)List»
So the first element contains a list whose value is (1 1). OK, let's see
my $maybe-element = (1,1);
say $maybe-element, $maybe-element.^name; # OUTPUT: «(1 1)List»
say $maybe-element ∈ #fib; # OUTPUT: «False»
Wait, what? Let's see...
my $maybe-element = #fib[0];
say $maybe-element ∈ #fib; # OUTPUT: «True»
Hum. So it's not the container. But
say (1,1).List === (1,1).List; # OUTPUT: «False»
And
say (1,1).List == (1,1).List; # OUTPUT: «True»
So I guess ∈ is using object identity, and not equality. That being the case, how can we check, in sets or sequences of lists, if an independently generated list is included using this operator? Should we use another different strategy?
Maybe a subquestion is why the same literals generate completely different objects, but there's probably a good, and very likely security-related, answer for that.
So I guess ∈ is using object identity, and not equality.
That is correct.
That being the case, how can we check, in sets or sequences of lists, if an independently generated list is included using this operator?
You can use .grep or .first and the equality operator of your choice (presumably you want eqv here), or you can try to find a list-like value type. Off the top of my head, I don't know if one is built into Perl 6.
In order to extract the values (records) of an orddict as a sorted list, tried this:
-module(test).
-compile(export_all).
-record(node, {name="", cost=0}).
test() ->
List = orddict:append("A",#node{name="A",cost=1},
orddict:append("B",#node{name="B",cost=2},
orddict:new())),
lists:sort(fun({_,A},{_,B}) -> A#node.cost =< B#node.cost end,
orddict:to_list(List)).
The sort fails with exception error: {badrecord,node}.
What would be the correct syntax?
Solved:
The correct insertion method is orddict:store/2 instead of orddict:append/2. Then the pattern {_,A} matches for the comparison function.
The correct syntax is:
lists:sort(fun({_,[A]},{_,[B]}) -> A#node.cost =< B#node.cost end,
orddict:to_list(List)).
I not found note about this in documentation,but you can look in source code of module.
As #Pascal write in comments the reason is that orddict:append/3 is a function provided to append a value to an existing Key/Value pair where Value must be a list. In the use case, the key doesn't exist, so the pair is created and the Value append to an empty list.
Btw, you always can print and compare real and expected result.
io:format("~p~n",[orddict:to_list(List)])
For your example that is:
[{"A",[{node,"A",1}]},{"B",[{node,"B",2}]}]
I want to do a simple task but somehow I'm unable to do it. Assume that I have one column like:
a
z
e
r
t
How can I create a new column with the same value twice with the following result:
a
a
z
z
e
e
r
r
t
t
I've already tried to double my column and do something like :
=TRANSPOSE(SPLIT(JOIN(";",A:A,B:B),";"))
but it creates:
a
z
e
r
t
a
z
e
r
t
I get inspired by this answer so far.
Try this:
=SORT({A1:A5;A1:A5})
Here we use:
sort
{} to combine data
Accounting your comment, then you may use this formula:
=QUERY(SORT(ArrayFormula({row(A1:A5),A1:A5;row(A1:A5),A1:A5})),"select Col2")
The idea is to use additional column of data with number of row, then sort by row, then query to get only values.
And join→split method will do the same:
=TRANSPOSE(SPLIT(JOIN(",",ARRAYFORMULA(CONCAT(A1:A5&",",A1:A5))),","))
Here we use range only two times, so this is easier to use. Also see Concat + ArrayFormula sample.
Few hundreds rows is nothing :)
I created index from 1 to n, then pasted it twice and sorted by index. But it's obviously fancier to do it with a formula :)
Assuming Your list is in column A and (for now) the times of repeat are in C1 (can be changed to a number in the formula), then something simple like this will do (starting in B1):
=INDEX(A:A,(INT(ROW()-1)/$C$1)+1)
Simply copy down as you need it (will give just 0 after the last item). No sorting. No array. No sheets/excel problems. No heavy calculations.
I have a doubt making tables with format, I need to make tables, I know I can make it this way:
If for example my table is tabla("estudents",["name","age","id"]).
But I have a problem, I need to get the numbers of attributes of the table, then I'll set a length of 18 to each square and the length will be N..
print_table_name(C):- tabla(C,A), //I SEARCH MY TABLE
atom_codes(Name,C), //PASSING THE NAME TO ATOM
length(A,N), //I GET MY NUMBER OF ATRIBUTES
Length is 18*N, //Length WILL BE THE LENGTH OF THE TABLE
print_edge(N), //HERE I PRINT THE TOP EDGE
format('|~t~a~t~N|)|~n',Name), //HERE IS MY ERROR
print_edge(N). //HERE I PRINT THE BOTTOM EDGE
print_edge(0):- format('~n',[]).
print_edge(N):- format('+~`-t~18|+', []), M is N-1, print_edge(M), !.
format('|~t~a~t~N|)|~n',Name) here I can't pass N as a variable, then I dont know how I can do to format get the N, N is the length of the table..
It print this
+--------------------------------------------------------------------------+
|students
||
+--------------------------------------------------------------------------+
and if I put the length where is N, then it works.
+--------------------------------------------------------------------------+
| students |
+--------------------------------------------------------------------------+
The problem is that I don't know how to pass the variable N to format.
In this case you should pass N as an argument to the format/2 predicate. Replace the variable N in the format by the symbol * and put N in the argument list.
I'm not sure if you will get the desire effect, but at least it won't fail.
format('|~t~a~t~*|)|~n',[Name, N]).
Edit
Right now I can only test the solution in this limited online interpreter: I replace the ~t by the character dot ~46t to see the effect and this is the result:
?- format('|~46t~a~46t~*|)|~n',['students', 72]).
|...............................students................................)|
PS: Are you sure about the parenthesis between the two last vertical bar?