I want to create a tex table showing the different coefficients of an x variable when regressed on different y variables. My code is as follows:
local vars yvar1 yvar2 yvar3 etc. [there are over 100]
foreach var of local vars{
reghdfe `var' xvar, vce(cluster i.country) a(i.country)
scalar `var'_coefficient = _b[xvar]
}
Therefore, I have a list of scalars which correspond to the coefficients that I want to include in a tex table with the following format:
Variable
X Coefficient
yvar1
0.5
yvar2
-0.4
yvar3
0.7
...
...
I have tried a few ways of creating a table like this but so far to no avail. I was wondering if anyone has any ideas of how I could create such a table?
Thank you very much!
Sounds like you can use the following code. However, does not seem efficient since you'd need to input the variable names into the frmttable command. You can do this automatically with outreg2. However, you would have 100 columns. There is probably a more efficient way to present your output than a table with 100 roes.
bcuse wage2, clear
local vars hours married south sibs meduc lwage
local count = 0
foreach var of local vars {
reghdfe `var' lwage, vce(cluster urban) a(urban)
scalar `var'_coefficient = e(b)[1,1]
local count = `count' + 1
}
matrix coefficients = J(`count',1,.)
local count = 1
foreach var of local vars {
mat coefficients[`count',1] = `var'_coefficient
local count = `count' + 1
}
frmttable, statmat(coefficients) sdec(2) rtitle( "var1" \ "var2" \ "var3" \ "var4" \ "var5" \ "var6") ctitle("variables", "coeff.")
Related
my problem is that he uses all 3 spawn points instead of only 1 of the 3 that I put there to choose from
YY_Pos = {}
YY_Pos[1] = {m=3017299663, x=1700, y=13154, z=2450}
YY_Pos[2] = {m=3017299663, x=1775, y=12413, z=2436}
YY_Pos[3] = {m=3017299663, x=1775, y=12413, z=2500}
function YY_tmhz_1_OnCreatureDisappear(MapID, InstanceID, Creatur ID, x, y, z)
for i = 1,3 do
local Index_pos = math.random(1,3)
local CreatureID = map.MapCreateCreature(YY_Pos[i].m, InstanceID, 1534207, YY_Pos[i].x, YY_Pos[i].y, YY_Pos[i].z)
end
end
You create a random value Index_pos but you never use it.
Instead you use the loop counter variable i to index your coordinate table.
Also note that the code as is will not use 1 of three coordinates but 3 random coordinates.
You would have to move the math.random call out of the loop if you want to do something 3 times for the same random index.
As you can tell I'm a beginner in lua. I am trying to understand a function I'm stuck at what the following code segment does?
It is used in the following code snippet in the last line:
function classify(txt_dir, img_dir, cls_list)
local acc = 0.0
local total = 0.0
local fea_img = {}
local fea_txt = {}
for fname in io.lines(cls_list) do
local imgpath = img_dir .. '/' .. fname .. '.t7'
local txtpath = txt_dir .. '/' .. fname .. '.t7'
fea_img[#fea_img + 1] = extract_img(imgpath)
fea_txt[#fea_txt + 1] = extract_txt(txtpath)
end
for i = 1,#fea_img do
-- loop over individual images.
for k = 1,fea_img[i]:size(1) do
local best_match = 1
local best_score = -math.huge
for j = 1,#fea_txt do
local cur_score = torch.dot(fea_img[i][{k,{}}], fea_txt[j])
From my understanding, fea_img is a lua table. Is the line fea_img[i][{k,{}}] some sort of slicing for the value for the key 'i' in the table fea_img?
I tried searching for more examples and found this being used here too (last line):
for i = 1,nsamples,batchsize do
-- indices
local lasti = math.min(i+batchsize-1,nsamples)
local m = lasti - i + 1
-- k-means step, on minibatch
local batch = x[{ {i,lasti},{} }]
Any help on this would be really appreciated. Thank you!
In lua you can access a specific index on a table in multiple ways. Like these two examples
local myValue = someTable.theIndex
-- or
local myOtherValue = someTable[2]
So the construct you see here is to access some values from a (nested) table.
Also in lua you can use anything except nil as a index, so even tables are possible.
The line
fea_img[i][{k,{}}]
Can be extended to this:
local index1 = i -- i in this case is your loop variable
local index2 = { k , { } } -- This creates a table with 2 values, the first one will be the vaule of the var k, the second one is an empty table
local value1 = fea_img[index1] -- This will get you a table
local value2 = value1[index2] -- This will get the same as: fea_img[i][{k,{}}]
Correction and Addition:
As Nicol Bolas already said in the comments: The index must be an exact match. Which means it literally has to be the same table, which is not the case for the presented code from you. Either you dropped code you thought is unnecessary or fea_img has some some kind of metatable on it.
In the case of
local k = 2
local table1 = {k, { } }
local table2 = {k, { } }
table2 and table1 do have the exact same content. But they are not the same table. Which will lead to nil always being retrieved if one is used to store data in a table and the other is used to get it back.
Syntactically, t[k] is indexing a table with a key. Normally, if there is a record in the table with the key k, its value is returned. Nothing more, nothing less.
If fea_img[i] was a normal table, {k,{}} would always return nil, since table indices are resolved based on their identity ({k,{}} is always a new table). Based on your code, I have to conclude that the elements of fea_img (i.e. what extract_img returns) are not normal tables.
In Lua, you can override the indexing operation using a metatable. If you index a value that has a metatable with __index, it will be used if there is no matching record in the table:
local t = {}
setmetatable(t, {
__index = function(t, k)
return k
end
})
print(t[{}])
This table has a metatable associated with it, which is used in the indexing operation. In this case __index returns the key, but whatever library you are using might provide more complex behaviour.
This is specific to the library you are using, not something related to the Lua syntax.
I have a logic problem for an iOS app but I don't want to solve it using brute-force.
I have a set of integers, the values are not unique:
[3,4,1,7,1,2,5,6,3,4........]
How can I get a subset from it with these 3 conditions:
I can only pick a defined amount of values.
The sum of the picked elements are equal to a value.
The selection must be random, so if there's more than one solution to the value, it will not always return the same.
Thanks in advance!
This is the subset sum problem, it is a known NP-Complete problem, and thus there is no known efficient (polynomial) solution to it.
However, if you are dealing with only relatively low integers - there is a pseudo polynomial time solution using Dynamic Programming.
The idea is to build a matrix bottom-up that follows the next recursive formulas:
D(x,i) = false x<0
D(0,i) = true
D(x,0) = false x != 0
D(x,i) = D(x,i-1) OR D(x-arr[i],i-1)
The idea is to mimic an exhaustive search - at each point you "guess" if the element is chosen or not.
To get the actual subset, you need to trace back your matrix. You iterate from D(SUM,n), (assuming the value is true) - you do the following (after the matrix is already filled up):
if D(x-arr[i-1],i-1) == true:
add arr[i] to the set
modify x <- x - arr[i-1]
modify i <- i-1
else // that means D(x,i-1) must be true
just modify i <- i-1
To get a random subset at each time, if both D(x-arr[i-1],i-1) == true AND D(x,i-1) == true choose randomly which course of action to take.
Python Code (If you don't know python read it as pseudo-code, it is very easy to follow).
arr = [1,2,4,5]
n = len(arr)
SUM = 6
#pre processing:
D = [[True] * (n+1)]
for x in range(1,SUM+1):
D.append([False]*(n+1))
#DP solution to populate D:
for x in range(1,SUM+1):
for i in range(1,n+1):
D[x][i] = D[x][i-1]
if x >= arr[i-1]:
D[x][i] = D[x][i] or D[x-arr[i-1]][i-1]
print D
#get a random solution:
if D[SUM][n] == False:
print 'no solution'
else:
sol = []
x = SUM
i = n
while x != 0:
possibleVals = []
if D[x][i-1] == True:
possibleVals.append(x)
if x >= arr[i-1] and D[x-arr[i-1]][i-1] == True:
possibleVals.append(x-arr[i-1])
#by here possibleVals contains 1/2 solutions, depending on how many choices we have.
#chose randomly one of them
from random import randint
r = possibleVals[randint(0,len(possibleVals)-1)]
#if decided to add element:
if r != x:
sol.append(x-r)
#modify i and x accordingly
x = r
i = i-1
print sol
P.S.
The above give you random choice, but NOT with uniform distribution of the permutations.
To achieve uniform distribution, you need to count the number of possible choices to build each number.
The formulas will be:
D(x,i) = 0 x<0
D(0,i) = 1
D(x,0) = 0 x != 0
D(x,i) = D(x,i-1) + D(x-arr[i],i-1)
And when generating the permutation, you do the same logic, but you decide to add the element i in probability D(x-arr[i],i-1) / D(x,i)
I'm writing a game using Corona SDK in lua language. I'm having a hard time coming up with a logic for a system like this;
I have different items. I want some items to have 1/1000 chance of being chosen (a unique item), I want some to have 1/10, some 2/10 etc.
I was thinking of populating a table and picking a random item. For example I'd add 100 of "X" item to the table and than 1 "Y" item. So by choosing randomly from [0,101] I kind of achieve what I want but I was wondering if there were any other ways of doing it.
items = {
Cat = { probability = 100/1000 }, -- i.e. 1/10
Dog = { probability = 200/1000 }, -- i.e. 2/10
Ant = { probability = 699/1000 },
Unicorn = { probability = 1/1000 },
}
function getRandomItem()
local p = math.random()
local cumulativeProbability = 0
for name, item in pairs(items) do
cumulativeProbability = cumulativeProbability + item.probability
if p <= cumulativeProbability then
return name, item
end
end
end
You want the probabilities to add up to 1. So if you increase the probability of an item (or add an item), you'll want to subtract from other items. That's why I wrote 1/10 as 100/1000: it's easier to see how things are distributed and to update them when you have a common denominator.
You can confirm you're getting the distribution you expect like this:
local count = { }
local iterations = 1000000
for i=1,iterations do
local name = getRandomItem()
count[name] = (count[name] or 0) + 1
end
for name, count in pairs(count) do
print(name, count/iterations)
end
I believe this answer is a lot easier to work with - albeit slightly slower in execution.
local chancesTbl = {
-- You can fill these with any non-negative integer you want
-- No need to make sure they sum up to anything specific
["a"] = 2,
["b"] = 1,
["c"] = 3
}
local function GetWeightedRandomKey()
local sum = 0
for _, chance in pairs(chancesTbl) do
sum = sum + chance
end
local rand = math.random(sum)
local winningKey
for key, chance in pairs(chancesTbl) do
winningKey = key
rand = rand - chance
if rand <= 0 then break end
end
return winningKey
end
I have six variables, that these variables have different position on the screen, I wanna put different images in these variables, hence i have an Array with the images.
misImagenes = {[1] = "rec/ro.png",[2] ="rec/az.png",[3] ="rec/ros.png",[4] ="rec/ne.png",[5] ="rec/ve.png",[6] ="rec/am.png"}
I put the elements of this Array into another Array into that have 2 different randoms, like this:
randoms = {[1] = misImagenes[math.random(1,6)],[2] = misImagenes[math.random(1,6)] }
So, I wanna put this randoms of random form, hence, i create an random of the randoms.
randomRan = randoms[math.random(1,2)]
I put the randomRan into the 6 variables, but the images of the variables are always equals.
uno = display.newImageRect(randomRan,340,280)
dos = display.newImageRect(randomRan,340,280)
tres = display.newImageRect(randomRan,340,280)
cuatro = display.newImageRect(randomRan,340,280)
cinco = display.newImageRect(randomRan,340,280)
seis = display.newImageRect(randomRan,340,280)
This variables have the randomRan, but the images are alway equals, i need that the images are differents, 2 differents images in random variables.
Thanks
It looks like what you want to do is commonly called shuffling and filtering an array.
Once you assign randoms[math.random(1,2)] to the randomRan variable, it is going to stay the same no matter what. It isn't like randomRan is going to be random each time it's used. However, if it were a function call, like randomRan(), then that would be a different case, depending on what the function did. A variable, once assigned to, generally stays the same unless changed.
math.randomseed(os.time()) -- Make sure to seed the random number generator.
local function shuffle(t)
local n = #t
while n >= 2 do
-- n is now the last pertinent index
local k = math.random(n) -- 1 <= k <= n
-- Quick swap
t[n], t[k] = t[k], t[n]
n = n - 1
end
return t
end
local misImagenes = {"rec/ro.png", "rec/az.png", "rec/ros.png", "rec/ne.png", "rec/ve.png", "rec/am.png"}
local randomImages = {}
-- Make a copy of misImagenes for randomImages.
for i, v in ipairs(misImagenes) do randomImages[i] = v end
-- Shuffle the new array. This will randomize the order of its contents.
shuffle(randomImages)
-- Since you want only two unique images for a total of six rectangles,
-- we'll have to duplicate and overwrite the other four, randomly.
for i = 1+2, 6 do
randomImages[i] = randomImages[math.random(1, 2)]
end
-- Now to filter the array with newImageRect.
for i=1, #randomImages do
randomImages[i] = display.newImageRect(randomImages[i], 340, 280)
end
-- randomImages now contains all of your randomized image rectangles.
The shuffle algorithm was borrowed from here to show an example of how this could work.
If you are doing
var1 = randomRan
var2 = randomRan
then var1 and var2 will have the same value - randomRan does not get recomputed each time its evaluated. YIf that is what you want, you can repeat the expression you used to initialize randomRan:
var1 = randoms[math.random(1,2)]
var2 = randoms[math.random(1,2)]
and if you want to avoid retyping that ocmplex expression, you can encapsulate it in a function:
--Return a random image
local function randomRan()
return randoms[math.random(1,2)]
end
var1 = randomRan()
var2 = randomRan()