After reading this topic and after experimenting a bit, I am trying to understand how the Lua length operator works when a table contains nil values.
Before I started to investigate, I thought that the length was simply the number of consecutive non-nil elements, starting at index 1:
print(#{nil}) -- 0
print(#{"o"}) -- 1
print(#{"o",nil}) -- 1
print(#{"o","o"}) -- 2
print(#{"o","o",nil}) -- 2
That looks pretty simple, right?
But my headache started when I accidentally added an element after a nil-terminated table:
print(#{"o",nil,"o"})
My guess was that it should probably print 1 because it would stop counting when the first nil is found. Or maybe it should print 2 if the length operator is greedy enough to look for non-nil elements after the first nil. But the above code prints 3.
So I’ve ran several other tests to see what happens:
-- nil before the end
print(#{nil,"o"}) -- 2
print(#{nil,"o","o"}) -- 3
print(#{"o",nil,"o"}) -- 3
-- several nil elements
print(#{"o",nil,nil}) -- 1
print(#{nil,"o",nil}) -- 0
print(#{nil,nil,"o"}) -- 3
I should mention that repl.it currently uses Lua 5.1.5 which is rather old, but if you test with the Lua demo, which currently uses Lua 5.3.5, you’ll get the same results.
By looking at those results and by looking at this answer, I assume that:
if the last element is not nil, the length operator returns the full size of the table, including nil entries if any
if the last element is nil, it counts the number of consecutive non-nil and stops counting at the first nil
Are those assumptions correct?
Can we predict a 100% well-defined behavior when a table contains one or several nil values?
The Lua documentation states that the length of a table is only defined if the table is a sequence. Does that mean that the length operator has undefined behavior for non-sequences?
Apart from the length operator, can nil values cause any trouble in a table?
We can predict some behaviour, but it is not standardised, and as such you should never rely on it. It's quite possible that the behaviour may change within this major version of Lua.
Should you ever need to fill a table with nil values, I suggest wrapping the table and replace holes with a unique placeholder value (eg. NIL={}; if v==nil then t[k]=NIL end, this is quite cheap to test against and safe.).
That said...
As there is even a difference in the result of # depending on how the table is defined, you'll have to distinguish between statically defined (constant) tables and dynamic defined (muted) tables.
Static table definitions:
#{nil,nil,nil,nil,nil, 1} -- 6
#{3, 2, nil, 1} -- 4
#{nil,nil,nil, 1, 1,nil} -- 0
#{nil,nil, 1, 1, 1,nil} -- 5
#{nil, 1, 1, 1, 1,nil} -- 5
#{nil,nil,nil,nil, 1,nil} -- 0
#{nil,nil, 1,nil, 1,nil,nil} -- 5
#{nil,nil,nil, 1,nil,nil, 1,nil} -- 4
Using this kind of definition, as long as the last value is non-nil, you will get a length equal to the position of the last value. If the last value is nil, Lua starts a (non-linear) search from the tail until it finds the first non-nil value.
Dynamic data definition
local x={}; x[5]=1;print(#x) -- 0
local x={}; x[1]=1;x[2]=1;x[3]=1;x[5]=1;print(#x) -- 3
local x={}; x[1]=1;x[2]=1;x[4]=1;x[5]=1;print(#x) -- 5
#{[5]=1} -- 0
local x={nil,nil,nil,1};x[5]=1;print(#x) -- 0
As soon as the table was changed once, the operator works the other way (that includes static definitions with []). If the first element is nil, # always returns 0, but if not it starts a search that I did not investigate further (I guess you can check the sources, though I don't think it's a standard binary search), until it finds a nil value that is preceded by a non-nil value.
As said before, relying on this behaviour is not a good idea, and invites lots of issues down the road. Though if you want to make a nasty unmaintainable program to mess with a colleague, that's a sure way to do it.
When a table is a sequence (all numeric keys start at 1 and there are no nil gaps), # is defined to be precisely the count of those elements.
For non-sequence tables, it is a bit more complicated. Lua 5.2 seems to leave the result as undefined. For 5.1 and 5.3, the result of the operation is a border.
A border in a table is any positive index that contains a non-nil value followed by nil, or 0 if the first element is nil. # is defined to return any value that satifies these conditions.
Looking at it from another perspective, since tables contain an "array" part and a "map" part, Lua has no way of knowing where the "map" indices start. For example, you can create a table with 1000 values and then set the first 999 of them to nil; that could leave you with a table of "size" 1000. However, you can also start with an empty table and set the 1000th element, having a table of "size" 0 but still structurally equivalent to the first one. The result of # is then simply the first valid value the internal algorithm finds.
The length operator produces undefined behaviour for tables that aren't sequences (i.e. tables with nil elements in the middle of the array). This means that even if the Lua implementation always behaves in a certain way, you shouldn't rely on that behaviour, as it may change in future versions of Lua, or in different implementations like LuaJIT.
You can use nils in tables - there is nothing wrong with that - just don't use the length operator on a table which might have nils before non-nil values.
The post you linked to contains more details about how the actual algorithm works. It mentions counting elements with a "binsearch", i.e. a binary search. This is not the same as just counting the elements one by one - if there are nils in the table, then depending on their exact position, the binary search algorithm may treat them as the end of the table, or may just ignore them.
To sum up, the algorithm is harder to predict than you were assuming, and even though it is technically possible to predict what will happen in any given case, you shouldn't rely on that behaviour as it is liable to change.
I am working with a third party device which has some implementation of Lua, and communicates in BACnet. The documentation is pretty janky, not providing any sort of help for any more advanced programming ideas. It's simply, "This is how you set variables...". So, I am trying to just figure it out, and hoping you all can help.
I need to set a long list of variables to certain values. I have a userdata 'ME', with a bunch of variables named MVXX (e.g. - MV21, MV98, MV56, etc).
(This is all kind of background for BACnet.) Variables in BACnet all have 17 'priorities', i.e., every BACnet variable is actually a sort of list of 17 values, with priority 16 being the default. So, typically, if I were to say ME.MV12 = 23, that would set MV12's priority-16 to the desired value of 23.
However, I need to set priority 17. I can do this in the provided Lua implementation, by saying ME.MV12_PV[17] = 23. I can set any of the priorities I want by indexing that PV. (Corollaries - what is PV? What is the underscore? How do I get to these objects? Or are they just interpreted from Lua to some function in C on the backend?)
All this being said, I need to make that variable name dynamic, so that i can set whichever value I need to set, based on some other code. I have made several attempts.
This tells me the object(MV12_PV[17]) does not exist:
x = 12
ME["MV" .. x .. "_PV[17]"] = 23
But this works fine, setting priority 16 to 23:
x = 12
ME["MV" .. x] = 23
I was trying to attempt some sort of what I think is called an evaluation, or eval. But, this just prints out function followed by some random 8 digit number:
x = 12
test = assert(loadstring("MV" .. x .. "_PV[17] = 23"))
print(test)
Any help? Apologies if I am unclear - tbh, I am so far behind the 8-ball I am pretty much grabbing at straws.
Underscores can be part of Lua identifiers (variable and function names). They are just part of the variable name (like letters are) and aren't a special Lua operator like [ and ] are.
In the expression ME.MV12_PV[17] we have ME being an object with a bunch of fields, ME.MV12_PV being an array stored in the "MV12_PV" field of that object and ME.MV12_PV[17] is the 17th slot in that array.
If you want to access fields dynamically, the thing to know is that accessing a field with dot notation in Lua is equivalent to using bracket notation and passing in the field name as a string:
-- The following are all equivalent:
x.foo
x["foo"]
local fieldname = "foo"
x[fieldname]
So in your case you might want to try doing something like this:
local n = 12
ME["MV"..n.."_PV"][17] = 23
BACnet "Commmandable" Objects (e.g. Binary Output, Analog Output, and o[tionally Binary Value, Analog Value and a handful of others) actually have 16 priorities (1-16). The "17th" you are referring to may be the "Relinquish Default", a value that is used if all 16 priorities are set to NULL or "Relinquished".
Perhaps your system will allow you to write to a BACnet Property called "Relinquish Default".
def digits(n):
total=0
for i in range(0,n):
if n/(10**(i))<1 and n/(10**(i-1))=>1:
total+=i
else:
total+=0
return total
I want to find the number of digits in 13 so I do the below
print digits(13)
it gives me $\0$ for every number I input into the function.
there's nothing wrong with what I've written as far as I can see:
if a number has say 4 digits say 1234 then dividing by 10^4 will make it less than 1: 0.1234 and dividing by 10^3 will make it 1.234
and by 10^3 will make it 1.234>1. when i satisfies BOTH conditions you know you have the correct number of digits.
what's failing here? Please can you advise me on the specific method I've tried
and not a different one?
Remember for every n there can only be one i which satisfies that condition.
so when you add i to the total there will only be i added so total returning total will give you i
your loop makes no sense at all. It goes from 0 to exact number - not what you want.
It looks like python, so grab a solution that uses string:
def digits(n):
return len(str(int(n))) # make sure that it's integer, than conver to string and return number of characters == number of digits
EDIT:
If you REALLY want to use a loop to count number of digits, you can do this this way:
def digits(n):
i = 0
while (n > 1):
n = n / 10
++i
return i
EDIT2:
since you really want to make your solution work, here is your problem. Provided, that you call your function like digits(5), 5 is of type integer, so your division is integer-based. That means, that 6/100 = 0, not 0.06.
def digits(n):
for i in range(0,n):
if n/float(10**(i))<1 and n/float(10**(i-1))=>1:
return i # we don't need to check anything else, this is the solution
return null # we don't the answer. This should not happen, but still, nice to put it here. Throwing an exception would be even better
I fixed it. Thanks for your input though :)
def digits(n):
for i in range(0,n):
if n/(10**(i))<1 and n/(10**(i-1))>=1:
return i
I have an array named xTain of size nDatax1
I initialize it as
xTrain = torch.linspace(-1,1,nData)
To access the array, the author uses xTrain[{{i}}]
can you please explain this notation? Why not simply xTrain[i] ?
Please refer the author's code here on Pg No 21- https://www.cs.ox.ac.uk/people/nando.defreitas/machinelearning/lecture4.pdf
As an additional note-
xTrain=torch.linespace(-1,1,10)
when I do
th> print(xTrain[1])
-1
th> print(xTrain[{1}])
-1
th> print(xTrain[{{1}}])
-1
[torch.DoubleTensor of size 1]
Why does it also print [torch.DoubleTensor of size 1] in 3rd case. My guess is in first 2 case its returning a scalar value at that location and in 3rd case a DoubleTensor
Good place to start is the Lua manual, it's syntax and explessions. You can see, what is meaning of {...} in Lua:
{...} -- creates a list with all vararg parameters
So in short your {1} creates a list with single value 1. Repeating it once more you got a list containing list containing single number 1.
If the xTrain would be simple table, it would probably fail, because it is hard to index using lists, but Lua supports metatables so the actual value is not used for indexing the table, but passed to some function which takes care of the lists.
Also reading further about the Tensor class, which is returned from the torch.linespace() function is a good place to see. The indexing using "array access" is explained in the section [Tensor] [{ dim1,dim2,... }] or [{ {dim1s,dim1e}, {dim2s,dim2e} }]
Could someone please tell me how to write a custom function in Open Office Basic to be used in Open Office Calc and that returns an array of values. An example of one such built-in function is MINVERSE. I need to write a custom function that populates a range of cells in much the same way.
Help would be much appreciated.
Yay, I just figured it out: all you do is return an array from your macro, BUT you also have to press Ctrl+Shift+Enter when typing in the cell formula to call your function (which is also the case when working with other arrays in calc). Here's an example:
Function MakeArray
Dim ret(2,2)
ret(0,0) = 1
ret(1,0) = 2
ret(0,1) = 3
ret(1,1) = 4
MakeArray = ret
End Function
FWIW, damjan's MakeArray function returns a Variant containing an array, I think. (The type returned by MakeArray is unspecified, so it defaults to Variant. A Variant is a container with a descriptive header, apparently cast as needed by the interpreter.)
Almost, but not quite, the same thing as returning an array. According to http://www.cpearson.com/excel/passingandreturningarrays.htm, Microsoft did not introduce the ability to return an array until 2000. His example [ LoadNumbers(Low As Long, High As Long) As Long()] does not compile in OO, flagging a syntax error on the parens following Long. It appears that OO's Basic emulates the pre-2k VBA.