I am having trouble figuring out how to get the length of a matrix within a matrix within a matrix (nested depth of 3). So what the code is doing in short is... looks to see if the publisher is already in the array, then it either adds a new column in the array with a new publisher and the corresponding system, or adds the new system to the existing array publisher
output[k][1] is the publisher array
output[k][2][l] is the system
where the first [] is the amount of different publishers
and the second [] is the amount of different systems within the same publisher
So how would I find out what the length of the third deep array is?
function reviewPubCount()
local output = {}
local k = 0
for i = 1, #keys do
if string.find(tostring(keys[i]), '_') then
key = Split(tostring(keys[i]), '_')
for j = 1, #reviewer_code do
if key[1] == reviewer_code[j] and key[1] ~= '' then
k = k + 1
output[k] = {}
-- output[k] = reviewer_code[j]
for l = 1, k do
if output[l][1] == reviewer_code[j] then
ltable = output[l][2]
temp = table.getn(ltable)
output[l][2][temp+1] = key[2]
else
output[k][1] = reviewer_code[j]
output[k][2][1] = key[2]
end
end
end
end
end
end
return output
end
The code has been fixed here for future reference: http://codepad.org/3di3BOD2#output
You should be able to replace table.getn(t) with #t (it's deprecated in Lua 5.1 and removed in Lua 5.2); instead of this:
ltable = output[l][2]
temp = table.getn(ltable)
output[l][2][temp+1] = key[2]
try this:
output[l][2][#output[l][2]+1] = key[2]
or this:
table.insert(output[l][2], key[2])
Related
In the Help Documentation of Scilab 6.0.2, I can read the following instruction on the Overloading entry, regarding the last operation code "iext" showed in this entry's table:
"The 6 char code may be used for some complex insertion algorithm like x.b(2) = 33 where b field is not defined in the structure x. The insertion is automatically decomposed into temp = x.b; temp(2) = 33; x.b = temp. The 6 char code is used for the first step of this algorithm. The 6 overloading function is very similar to the e's one."
But I can't find a complete example on how to use this "char 6 code" to overload a function. I'm trying to use it, without success. Does anyone have an example on how to do this?
The code bellow creates a normal "mlist" as a example. Which needs overloading functions
A = rand(5,3)
names = ["colA" "colB" "colC"]
units = ["ft" "in" "lb"]
M = mlist(["Mlog" "names" "units" names],names,units,A(:,1),A(:,2),A(:,3))
Following are the overload functions:
//define display
function %Mlog_p(M)
n = size(M.names,"*")
formatStr = strcat(repmat("%10s ",1,n)) + "\n"
formatNum = strcat(repmat("%0.10f ",1,n)) + "\n"
mprintf(formatStr,M.names)
mprintf(formatStr,M.units)
disp([M(M.names(1)),M(M.names(2)),M(M.names(3))])
end
//define extraction operation
function [Mat]=%Mlog_e(varargin)
M = varargin($)
cols = [1:size(M.names,"*")] // This will also work
cols = cols(varargin($-1)) // when varargin($-1) = 1:1:$
Mat = []
if length(varargin)==3 then
for i = M.names(cols)
Mat = [Mat M(i)(varargin(1))]
end
else
for i=1:size(M.names(cols),"*")
Mat(i).name = M.names(cols(i))
Mat(i).unit = M.units(cols(i))
Mat(i).data = M(:,cols(i))
end
end
endfunction
//define insertion operations (a regular matrix into a Mlog matrix)
function ML=%s_i_Mlog(i,j,V,M)
names = M.names
units = M.units
A = M(:,:) // uses function above
A(i,j) = V
ML = mlist(["Mlog" "names" "units" names],names,units,A(:,1),A(:,2),A(:,3))
endfunction
//insertion operation with structures (the subject of the question)
function temp = %Mlog_6(j,M)
temp = M(j) // uses function %Mlog_e
endfunction
function M = %st_i_Mlog(j,st,M)
A = M(:,:) // uses function %Mlog_e
M.names(j) = st.name // uses function above
M.units(j) = st.unit // uses function above
A(:,j) = st.data // uses function above
names = M.names
units = M.units
M = mlist(["Mlog" "names" "units" names],names,units,A(:,1),A(:,2),A(:,3))
endfunction
The first overload (displays mlist) will show the matrix in the form of the following table:
--> M
M =
colA colB colC
ft in lb
0.4720517 0.6719395 0.5628382
0.0623731 0.1360619 0.5531093
0.0854401 0.2119744 0.0768984
0.0134564 0.4015942 0.5360758
0.3543002 0.4036219 0.0900212
The next overloads (extraction and insertion) Will allow the table to be access as a simple matrix M(i,j).
The extraction function Will also allow M to be access by column, which returns a structure, for instance:
--> M(2)
ans =
name: "colB"
unit: "in"
data: [5x1 constant]
The last two functions are the overloads mentioned in the question. They allow the column metadata to be changed in a structure form.
--> M(2).name = "length"
M =
colA length colC
ft in lb
0.4720517 0.6719395 0.5628382
0.0623731 0.1360619 0.5531093
0.0854401 0.2119744 0.0768984
0.0134564 0.4015942 0.5360758
0.3543002 0.4036219 0.0900212
I would like to iterate over a list and occasionally delete items of said list. Below a toy example:
function delete_item!(myarray, item)
deleteat!(myarray, findin(myarray, [item]))
end
n = 1000
myarray = [i for i = 1:n];
for a in myarray
if a%2 == 0
delete_item!(myarray, a)
end
end
However I get error:
BoundsError: attempt to access 500-element Array{Int64,1} at index [502]
How can I fix it (as efficiently as possible)?
Additional information. The above seems like a silly example, in my original problem I have a list of agents which interact. Therefore I am not sure if iterating over a copy would be the best solution. For example:
#creating my agent
mutable struct agent <: Any
id::Int
end
function delete_item!(myarray::Array{agent, 1}, item::agent)
deleteat!(myarray, findin(myarray, [item]))
end
#having my list of agents
n = 1000
myarray = agent[agent(i) for i = 1:n];
#trying to remove agents from list while having them interact
for a in myarray
#agent does stuff
if a.id%2 == 0 #if something happens remove
delete_item!(myarray, a)
end
end
Unfortunately there is no single answer to this question as most efficient approach depends on the logic of the whole model (in particular do other agents' actions depend on the fact that some entry is actually deleted from an array).
In most cases the following approach should be the simplest (I am leaving findin which is inefficient but I understand that you may have duplicates in myarray in general):
n = 1000
myarray = [i for i = 1:n];
keep = trues(n)
for (i, a) in enumerate(myarray)
keep[i] || continue # do not process an agent that is marked for deletion
if a%2 == 0 # here application logic might also need to check keep in some cases
keep[findin(myarray, [a])] = false
end
end
myarray = myarray[keep]
If for some reason you really need to delete elements of myarray in each iteration here is how you can do it:
n = 1000
myarray = [i for i = 1:n];
i = 1
while i <= length(myarray)
a = myarray[i]
if a%2 == 0
todelete = findin(myarray, [a])
i -= count(x -> x < i, todelete) # if myarray has duplicates of a you have to move the counter back
deleteat!(myarray, todelete)
else
i += 1
end
end
In general the code you give will not be very fast (e.g. if you know myarray does not contain duplicates it can be much simpler - and I guess you can).
EDIT: Here is how you can implement both versions if you know you do not have duplicates (you can simply use agent's index - observe that we can also avoid unnecessary checks):
n = 1000
myarray = [i for i = 1:n];
keep = trues(n)
for (i, a) in enumerate(myarray)
if a%2 == 0 # here application logic might also need to check keep in some cases
keep[i] = false
end
end
myarray = myarray[keep]
If for some reason you really need to delete elements of myarray in each iteration here is how you can do it:
n = 1000
myarray = [i for i = 1:n];
i = 1
while i <= length(myarray)
a = myarray[i]
if a%2 == 0
deleteat!(myarray, i)
else
i += 1
end
end
I am stuck at creating a matrix of a matrix (vector in this case)
What I have so far
index = zeros(size(A)) // This is some matrix but isn't important to the question
indexIndex = 1;
for rows=1:length(R)
for columns=1:length(K)
if(A(rows,columns)==x)
V=[rows columns]; // I create a vector holding the row + column
index(indexIndex) = V(1,2) // I want to store all these vectors
indexIndex = indexIndex + 1
end
end
end
I have tried various ways of getting the information out of V (such as V(1:2)) but nothing seems to work correctly.
In other words, I'm trying to get an array of points.
Thanks in advance
I do not understand your question exactly. What is the size of A? What is x, K and R? But under some assumptions,
Using list
You could use a list
// Create some matrix A
A = zeros(8,8)
//initialize the list
index = list();
// Get the dimensions of A
rows = size(A,1);
cols = size(A,2);
x = 0;
for row=1:rows
for col=1:cols
if(A(row,col)==x)
// Create a vector holding row and col
V=[row col];
// Append it to list using $ (last index) + 1
index($+1) = V
end
end
end
Single indexed matrices
Another approach would be to make use of the fact an multi-dimensional matrix can also be indexed by a single value.
For instance create a random matrix named a:
-->a = rand(3,3)
a =
0.6212882 0.5211472 0.0881335
0.3454984 0.2870401 0.4498763
0.7064868 0.6502795 0.7227253
Access the first value:
-->a(1)
ans =
0.6212882
-->a(1,1)
ans =
0.6212882
Access the second value:
-->a(2)
ans =
0.3454984
-->a(2,1)
ans =
0.3454984
So that proves how the single indexing works. Now to apply it to your problem and knocking out a for-loop.
// Create some matrix A
A = zeros(8,8)
//initialize the array of indices
index = [];
// Get the dimensions of A
rows = size(A,1);
cols = size(A,2);
x = 0;
for i=1:length(A)
if(A(i)==x)
// Append it to list using $ (last index) + 1
index($+1) = i;
end
end
Without for-loop
If you just need the values that adhere to a certain condition you could also do something like this
values = A(A==x);
Be carefull when comparing doubles, these are not always (un)equal when you expect.
Say I have this dictionary in Lua
places = {dest1 = 10, dest2 = 20, dest3 = 30}
In my program I check if the dictionary has met my size limit in this case 3, how do I push the oldest key/value pair out of the dictionary and add a new one?
places["newdest"] = 50
--places should now look like this, dest3 pushed off and newdest added and dictionary has kept its size
places = {newdest = 50, dest1 = 10, dest2 = 20}
It's not too difficult to do this, if you really needed it, and it's easily reusable as well.
local function ld_next(t, i) -- This is an ordered iterator, oldest first.
if i <= #t then
return i + 1, t[i], t[t[i]]
end
end
local limited_dict = {__newindex = function(t,k,v)
if #t == t[0] then -- Pop the last entry.
t[table.remove(t, 1)] = nil
end
table.insert(t, k)
rawset(t, k, v)
end, __pairs = function(t)
return ld_next, t, 1
end}
local t = setmetatable({[0] = 3}, limited_dict)
t['dest1'] = 10
t['dest2'] = 20
t['dest3'] = 30
t['dest4'] = 50
for i, k, v in pairs(t) do print(k, v) end
dest2 20
dest3 30
dest4 50
The order is stored in the numeric indices, with the 0th index indicating the limit of unique keys that the table can have.
Given that dictionary keys do not save their entered position, I wrote something that should be able to help you accomplish what you want, regardless.
function push_old(t, k, v)
local z = fifo[1]
t[z] = nil
t[k] = v
table.insert(fifo, k)
table.remove(fifo, 1)
end
You would need to create the fifo table first, based on the order you entered the keys (for instance, fifo = {"dest3", "dest2", "dest1"}, based on your post, from first entered to last entered), then use:
push_old(places, "newdest", 50)
and the function will do the work. Happy holidays!
What i want is to retrieve quantity in database from piece and covert it to dozen. Then input as dozen and convert back to pieces and save to database again.
when I input data eg. 10.3, it should convert to 123 piece for me ((10 * 12) + 3). My code work well without my "If clause" but only when data was "single" type. It made error when I input integer number, so I added "If.." statement to check it first which is now the output was correct for Integer but incorrect when I input single number.
I have this code..
Function DzToPcs(val)
'If CLng(val) = val then <-- not work
'if Fix(val) <> val then <-- work but the output was not correct when input single type number.
if Int(vInt) = vInt then <-- work but the output was not correct when input single type number.
DztoPcs = val * 12
else
strInt = Cstr(val)
a = Split(strInt,".")
dz = a(0)
pcs = a(1)
getdz = Cint(dz)
getpcs = Cint(pcs)
DztoPcs = (getdz * 12) + getpcs
end if
I'm not sure what's wrong with your if statements (my VBScript is a little rusty), but you could try this alternative:
Function DzToPcs(val)
strInt = Cstr(val)
a = Split(strInt,".")
dz = a(0)
if UBound(a) > 0 then
pcs = a(1)
getdz = Cint(dz)
getpcs = Cint(pcs)
DztoPcs = (getdz * 12) + getpcs
else
DztoPcs = dz * 12
end if
end function