I know that there is no guarantee concerning table element order when iterating over all table elements using pairs(). The table elements could be returned in any order.
But what about initializing a table, e.g. consider the following code:
function func(x)
print(x)
return(x)
end
t = {func(0), x = func(1), y = func(2), [0] = func(3), func(4), [1000] = func(5)}
The test shows that func() is called in the order the table elements are initialized but is this guaranteed? I don't seem to find anything about this in the Lua reference but I'm sure there must be some official explanation on this.
The order of evaluation in table constructors and in function arguments is not defined to allow room for optimization by the compiler.
You shouldn't rely on the order of evaluation anyway.
This is not specific to Lua. Most languages don't specify order of evaluation in lists of expressions.
Related
I'm modding a game. I'd like to optimize my code if possible for a frequently called function. The function will look into a dictionary table (consisting of estimated 10-100 entries). I'm considering 2 patterns a) direct reference and b) lookup with ipairs:
PATTERN A
tableA = { ["moduleName.propertyName"] = { some stuff } } -- the key is a string with dot inside, hence the quotation marks
result = tableA["moduleName.propertyName"]
PATTERN B
function lookup(type)
local result
for i, obj in ipairs(tableB) do
if obj.type == "moduleName.propertyName" then
result = obj
break
end
end
return result
end
***
tableB = {
[1] = {
type = "moduleName.propertyName",
... some stuff ...
}
}
result = lookup("moduleName.propertyName")
Which pattern should be faster on average? I'd expect the 'native' referencing to be faster (it is certainly much neater), but maybe this is a silly assumption? I'm able to sort (to some extent) tableB in a order of frequency of the lookups whereas (as I understand it) tableA will have in Lua random internal order by default even if I declare the keys in proper order.
A lookup table will always be faster than searching a table every time.
For 100 elements that's one indexing operation compared to up to 100 loop cycles, iterator calls, conditional statements...
It is questionable though if you would experience a difference in your application with so little elements.
So if you build that data structure for this purpose only, go with a look-up table right away.
If you already have this data structure for other purposes and you just want to look something up once, traverse the table with a loop.
If you have this structure already and you need to look values up more than once, build a look up table for that purpose.
I have a table of strings like this:
{
"1",
"1.5",
"3.13",
"1.2.5.7",
"2.5",
"1.3.5",
"2.2.5.7.10",
"1.17",
"1.10.5",
"2.3.14.9",
"3.5.21.9.3",
"4"
}
And would like to sort that like this:
{
"1",
"1.2.5.7",
"1.3.5",
"1.5",
"1.10.5",
"1.17",
"2.2.5.7.10",
"2.3.14.9",
"2.5",
"3.5.21.9.3",
"3.13",
"4"
}
How do I sort this in Lua? I know that table.sort() will be used, I just don't know the function (second parameter) to use for comparison.
Given your requirements, you probably want something like natural sort order. I described several possible solution as well as their impact on the results in a blog post.
The simplest solution may look like this (below), but there are 5 different solutions listed with different complexity and the results:
function alphanumsort(o)
local function padnum(d) return ("%03d%s"):format(#d, d) end
table.sort(o, function(a,b)
return tostring(a):gsub("%d+",padnum) < tostring(b):gsub("%d+",padnum) end)
return o
end
table.sort sorts ascending by default. You don't have to provide a second parameter then. As you're sorting strings Lua will compare the strings character by character. Hence you must implement a sorting function that tells Lua which comes first.
I just don't know the function (second parameter) to use for
comparison.
That's why people wrote the Lua Reference Manual
table.sort (list [, comp])
Sorts the list elements in a given order, in-place, from list1 to
list[#list]. If comp is given, then it must be a function that
receives two list elements and returns true when the first element
must come before the second in the final order, so that, after the
sort, i <= j implies not comp(list[j],list[i]). If comp is not given,
then the standard Lua operator < is used instead.
The comp function must define a consistent order; more formally, the
function must define a strict weak order. (A weak order is similar to
a total order, but it can equate different elements for comparison
purposes.)
The sort algorithm is not stable: Different elements considered equal
by the given order may have their relative positions changed by the
sort.
Think about how you would do it with pen an paper. You would compare each number segment. As soon as a segment is smaller than the other you know this number comes first.
So a solution would probably require you to get those segments for the strings, convert them to numbers so you can compare their values...
The reference manual has this to say about the table.move function, introduced in Lua 5.3:
table.move (a1, f, e, t [,a2])
Moves elements from table a1 to table a2, performing the equivalent to the following multiple assignment: a2[t],··· = a1[f],···,a1[e]. The default for a2 is a1. The destination range can overlap with the source range. The number of elements to be moved must fit in a Lua integer.
This description leaves a lot to be desired. I'm hoping for a general, canonical explanation of the function that goes into more detail than the reference manual. (Oddly, I could not find such an explanation anywhere on the web, perhaps because the function is fairly new.)
Particular points I am still confused on after reading the reference manual's explanation a few times:
When it says "move", that means the items are being removed from their original location, correct? Do the indices of items above the removed items shift down to fill the gaps? If so, and we're moving within the same table, does t point to the original location before anything starts moving?
Is there some significance to the choice of index letters f, e, and t?
There is no similar function in any other language I know. What's an example of how I might use this? Since it's one of only seven table functions, I presume it's quite useful.
Moves elements from table a1 to table a2, performing the equivalent to the following multiple assignment a2[t],··· = a1[f],···,a1[e]
Maybe they could have added the information this is done using consecutive integer values from f to e.
If you know Lua a bit more you'll know that a Lua table has no order. So the only way to make that code work is to use consecutive integer keys. Especially as the documentation mentions a source range.
Giving the equivalent syntax is the most unambiguous way of describing a function.
If you know the very basic concept of multiple assignment in Lua (see 3.3.3. Assignment) , you know what this function does.
table.move(a1, 1, 4, 6, a2) would copy a1[1], a1[2], a1[3], a1[4] into a2[6], a2[7], a2[8], a2[9]
The most common usecase is probably to get a subset of a list.
local values = {1,45,1,44,123,2354,321,745,1231}
old syntax:
local subset = {}
for i = 3, 7 do
table.insert(subset, values[i])
end
new:
local subset = table.move(values, 5, 7, 1, {})
Or maybe you quickly want to remove the last 3 values from a table?
local a = {1,2,3,4,5,6,7}
table.move({}, 1,3,#a-2, a)
Our task is create a table, and read values to the table using a loop. Print the values after the process is complete. - Create a table. - Read the number of values to be read to the table. - Read the values to the table using a loop. - Print the values in the table using another loop. for this we had written code as
local table = {}
for value in ipairs(table) do
io.read()
end
for value in ipairs(table) do
print(value)
end
not sure where we went wrong please help us. Our exception is
Input (stdin)
3
11
22
abc
Your Output (stdout)
~ no output ~
Expected Output
11
22
abc
Correct Code is
local table1 = {}
local x = io.read()
for line in io.lines() do
table.insert(table1, line)
end
for K, value in ipairs(table1) do
print(value)
end
Let's walk through this step-by-step.
Create a table.
Though the syntax is correct, table is a reserved pre-defined global name in Lua, and thus cannot should not be declared a variable name to avoid future issues. Instead, you'll need to want to use a different name. If you're insistent on using the word table, you'll have to distinguish it from the function global table. The easiest way to do this is change it to Table, as Lua is a case-sensitive language. Therefore, your table creation should look something like:
local Table = {}
Read values to the table using a loop.
Though Table is now established as a table, your for loop is only iterating through an empty table. It seems your goal is to iterate through the io.read() instead. But io.read() is probably not what you want here, though you can utilize a repeat loop if you wish to use io.read() via table.insert. However, repeat requires a condition that must be met for it to terminate, such as the length of the table reaching a certain amount (in your example, it would be until (#Table == 4)). Since this is a task you are given, I will not provide an example, but allow you to research this method and use it to your advantage.
Print the values after the process is complete.
You are on the right track with your printing loop. However, it must be noted that iterating through a table always returns two results, an index and a value. In your code, you would only return the index number, so your output would simply return:
1
2
3
4
If you are wanting the actual values, you'll need a placeholder for the index. Oftentimes, the placeholder for an unneeded variable in Lua is the underscore (_). Modify your for loop to account for the index, and you should be set.
Try modifying your code with the suggestions I've given and see if you can figure out how to achieve your end result.
Edited:
Thanks, Piglet, for corrections on the insight! I'd forgotten table itself wasn't a function, and wasn't reserved, but still bad form to use it as a variable name whether local or global. At least, it's how I was taught, but your comment is correct!
I have the following function for a hacky project:
% The Record variable is some known record with an associated table.
Query = qlc:q([Existing ||
Existing <- mnesia:table(Table),
ExistingFields = record_to_fields(Existing),
RecordFields = record_to_fields(Record),
ExistingFields == RecordFields
]).
The function record_to_fields/1 simply drops the record name and ID from the tuple so that I can compare the fields themselves. If anyone wants context, it's because I pre-generate a unique ID for a record before attempting to insert it into Mnesia, and I want to make sure that a record with identical fields (but different ID) does not exist.
This results in the following (redacted for clarity) stack trace:
{aborted, {{case_clause, {stuff}},
[{db, '-my_func/2-fun-1-',8, ...
Which points to the line where I declare Query, however there is no case clause in sight. What is causing this error?
(Will answer myself, but I appreciate a comment that could explain how I could achieve what I want)
EDIT: this wouldn't be necessary if I could simply mark certain fields as unique, and Mnesia had a dedicated insert/1 or create/1 function.
For your example, I think your solution is clearer anyway (although it seems you can pull the record_to_fields(Record) portion outside the comprehension so it isn't getting calculated over and over.)
Yes, list comprehensions can only have generators and assignments. But you can cheat a little by writing an assignment as a one-element generator. For instance, you can re-write your expression as this:
RecordFields = record_to_fields(Record),
Query = qlc:q([Existing ||
Existing <- mnesia:table(Table),
ExistingFields <- [record_to_fields(Existing)],
ExistingFields == RecordFields
]).
As it turns out, the QLC DSL does not allow assignments, only generators and filters; as per the documentation (emphasis mine):
Syntactically QLCs have the same parts as ordinary list
comprehensions:
[Expression || Qualifier1, Qualifier2, ...]
Expression (the template)
is any Erlang expression. Qualifiers are either filters or generators.
Filters are Erlang expressions returning boolean(). Generators have
the form Pattern <- ListExpression, where ListExpression is an
expression evaluating to a query handle or a list.
Which means we cannot variable assignments within a QLC query.
Thus my only option, insofar as I know, is to simply write out the query as:
Query = qlc:q([Existing ||
Existing <- mnesia:table(Table),
record_to_fields(Existing) == record_to_fields(Record)
]).