I have a condition which follows this logic Av(B^C) which is equivalent to (AvB) ^ (AvC).
Where:
A = NAF> 2
B= (COUNT_INT + COUNT_NOINT < 25)
C= (NAF> 1)
This is the condition given in the specs:
NAF> 2 OR (( COUNT_INT + COUNT_NOINT < 25) AND (NAF> 1))
This is how I coded it in sas, but its not producing the correct results. I appreciate any suggestions. Thanks.
Actually this is where it fails when I code up the array :
if (sum(count_int_arr[i],count_nonint_arr[i]) lt 25 ) and
( naf_arr[i] gt 2) or (naf_arr[i] gt 1) then check_naf_arr[i] = "bad";
The way you coded it is not what you gave in the initial specs. Your code is equivalent to (A AND B) OR C, whereas you want A OR (B AND C).
This should give you correct results:
DATA myData;
/* (...) */
DO i = 1 TO someValue;
IF (naf_arr[i] > 2) OR (SUM(count_int_arr[i], count_nonint_arr[i]) < 25 AND naf_arr[i] > 1)
THEN check_naf_arr[i] = "bad";
END;
/* (...) */
run;
Related
I was looking at some code which came with the following short-hand statement
score = ((initialPlayer == player) ? CAPTURE_SCORE : -CAPTURE_SCORE) + recursive_solver(lastMap, initialPlayer, findOpponent(player), 1 + rounds);
I think that I understand the first portion of the code,
if(initialPlayer == player){
score = CAPTURE_SCORE;
}
else
score = -CAPTURE_SCORE;
but im confused to how the +recursive_solver function is added to this, any help would be greatly appreciated :)
As stated above I tried to write out the statement in a longer from thats easier for me to read. My best guess is that the recursive solver function is then added to the score of the if-else statement?
if(initialPlayer == player)
score = CAPTURE_SCORE + recursive_solver(lastMap, initialPlayer, findOpponent(player), 1 + rounds);
else
score = -CAPTURE_SCORE + recursive_solver(lastMap, initialPlayer, findOpponent(player), 1 + rounds);
Explanation:
A = (C ? B : D) + E;
If C is true: A = (B) + E;
If C is false: A = (D) + E;
In sum
if (C)
A = B + E;
else // if (!C)
A = D + E;
score = ((initialPlayer == player) ? CAPTURE_SCORE : -CAPTURE_SCORE) + recursive_solver(lastMap, initialPlayer, findOpponent(player), 1 + rounds);
Equals to
if(initialPlayer == player) {
score = CAPTURE_SCORE;
} else {
score = -CAPTURE_SCORE;
}
score += recursive_solver(lastMap, initialPlayer, findOpponent(player), 1 + rounds);
I need help with looping in Mata. I have to write a code for Beta coefficients for OLS in Mata using a loop. I am not sure how to call for the variables and create the code. Here is what I have so far.
foreach j of local X {
if { //for X'X
matrix XX = [mata:XX = cross(X,1 , X,1)]
XX
}
else {
mata:Xy = cross(X,1 , y,0)
Xy
}
I am getting an error message "invalid syntax".
I'm not sure what you need the loop for. Perhaps you can provide more information about that. However the following example may help you implement OLS in mata.
Load example data from bcuse:
ssc install bcuse
clear
bcuse bwght
mata
x = st_data(., ("male", "parity","lfaminc","packs"))
cons = J(rows(x), 1, 1)
X = (x, cons)
y = st_data(., ("lbwght"))
beta_hat = (invsym(X'*X))*(X'*y)
e_hat = y - X * beta_hat
s2 = (1 / (rows(X) - cols(X))) * (e_hat' * e_hat)
B = J(cols(X), cols(X), 0)
n = rows(X)
for (i=1; i<=n; i++) {
B =B+(e_hat[i,1]*X[i,.])'*(e_hat[i,1]*X[i,.])
}
V_robust = (n/(n-cols(X)))*invsym(X'*X)*B*invsym(X'*X)
se_robust = sqrt(diagonal(V_robust))
V_ols = s2 * invsym(X'*X)
se_ols = sqrt(diagonal(V_ols))
beta_hat
se_robust
end
This is far from the only way to implement OLS using mata. See the Stata Blog for another example using quadcross, I like my example because it preserves a little more of the matrix algebra in the code.
I'm trying to implemenet an assignment problem. I have the following problem when trying to multiply two variables in linear programming (using glpk gusek) in my goal function:
minimize PATH_COST: sum{k in Rodzaj_Transportu}(sum{z in numery_Zlecen}Koszty_Suma[k,z])*y[k,z]; #y is a binary variable; Koszty_Suma is total cost for ordez z and car type k
The following error is arising: "model.mod:47: multiplication of linear forms not allowed".
Code (.dat file):
data;
set numery_Zlecen := 1, 2, 3; #order numbers
set Miasta := '*some data: *' #cities.
#order numer (from city to city)
set Zlecenie[1] := Warszawa Paris;
set Zlecenie[2] := Berlin Praha;
set Zlecenie[3] := Praha Amsterdam;
#number of packages for transport for a particular order
param Ilosc_Wyrobow :=
1 10
2 50
3 110;
param Godziny_Pracy := 9; #number of working hours during the day
param Pojemnosc_Samochodu := 35; #capacity of the car (how many packages it can take)
param Srednia_Predkosc := 80; #average car speed
param Spalenie_Paliwa := 0.25; #fuel combustion
param Wynagrodzenie_za_Godzine := 20; #salary for one working hour
param Cena_Noclegu := 100; #price of accommodation
param Dystans: '*some data: *' #km between cities.
param Koszt_Paliwa : '*some data: *' #fuel consumption depends on country.
end;
Code (.mod file):
#INDEXY
#=====================================================================
set Miasta; #i,j
set numery_Zlecen; #z
set Zlecenie{numery_Zlecen} dimen 2; #p,q
set Rodzaj_Transportu; #k
#PARAMETRY
#=====================================================================
param Dystans {Miasta,Miasta};
param Ilosc_Wyrobow{numery_Zlecen};
param Godziny_Pracy >= 0;
param Pojemnosc_Samochodu {Rodzaj_Transportu}>= 0;
param Srednia_Predkosc >=0;
param Spalenie_Paliwa >=0;
param Koszt_Paliwa {Miasta,Miasta};
param Wynagrodzenie_za_Godzine >= 0;
param Cena_Noclegu >= 0;
#ZMIENE
#=====================================================================
var x{Miasta,Miasta,numery_Zlecen} <= 1, >= 0; #variable x equal 1 when we're going the path from city A to city B; otherwise it equals 0
var y{Rodzaj_Transportu,numery_Zlecen} binary <=1, >=0; #variable that shows what types of car/s we are using for order (can be 0 or 1)
var Koszty_Suma{Rodzaj_Transportu,numery_Zlecen}; #total costs
var Koszty_Transportu{numery_Zlecen}; #transport costs
var Koszty_Odpoczynku{numery_Zlecen}; #rest costs
var Koszty_Wynagrodzenia{numery_Zlecen}; #salary costs
#FUNKCjA CELU
#=====================================================================
minimize PATH_COST: sum{k in Rodzaj_Transportu}(sum{z in numery_Zlecen}Koszty_Suma[k,z])*y[k,z];
#OGRANICZENIA (constraints)
#=====================================================================
s.t. SOURCE{z in numery_Zlecen, (p,q) in Zlecenie[z], i in Miasta: i = p && p != q}:
sum {j in Miasta} (x[i ,j ,z ]) - sum {j in Miasta}( x[j ,i ,z ]) = 1;
s.t. INTERNAL {z in numery_Zlecen, (p,q) in Zlecenie[z],i in Miasta: i != p && i != q && p != q }:
sum {j in Miasta} (x[i ,j ,z ]) - sum {j in Miasta}( x[j ,i ,z ]) = 0;
s.t. OGR_KM_DZIEN{z in numery_Zlecen,(p,q) in Zlecenie[z], j in Miasta, i in Miasta: i != q}:
if (Dystans[i,j] > (Godziny_Pracy*Srednia_Predkosc)) and i != q then x[i,j,z] = 0;
s.t. OGR_KOSZTY_SUMA{z in numery_Zlecen, k in Rodzaj_Transportu}:
Koszty_Suma[k,z] = (Koszty_Transportu[z] + Koszty_Odpoczynku[z] + Koszty_Wynagrodzenia[z])*ceil(Ilosc_Wyrobow[z]/Pojemnosc_Samochodu[k]);
s.t. OGR_KOSZTY_TRANSPORTU{z in numery_Zlecen}:
Koszty_Transportu[z] = (sum{i in Miasta} (sum{j in Miasta} ( Dystans[i,j]*x[i,j, z]*Koszt_Paliwa[i,j] ) ))*Spalenie_Paliwa;
s.t. OGR_KOSZTY_ODPOCZYNKU{z in numery_Zlecen}:
Koszty_Odpoczynku[z] =
(sum{i in Miasta} (sum{j in Miasta} ( Dystans[i,j]*x[i,j, z] ) ))/(Godziny_Pracy*Srednia_Predkosc) * Cena_Noclegu;
s.t. OGR_KOSZTY_WYNAGRODZENIA{z in numery_Zlecen}:
Koszty_Wynagrodzenia[z] =
((sum{i in Miasta} (sum{j in Miasta} ( Dystans[i,j]*x[i,j, z] ) ))/(Srednia_Predkosc)) * Wynagrodzenie_za_Godzine;
s.t. OGR_Y_JEDEN{z in numery_Zlecen}:
sum{k in Rodzaj_Transportu}(y[k,z]) = 1;
solve;
How is it possible to get rid of this error? Any hints how to solve this kind of problem are welcome.
First I think the parentheses are incorrect (note that y[k,z] depends on z). The expression
sum{k in Rodzaj_Transportu}(sum{z in numery_Zlecen}Koszty_Suma[k,z])*y[k,z];
is not mathematically correct. So, I assume what you meant is:
sum{k in Rodzaj_Transportu}(sum{z in numery_Zlecen}Koszty_Suma[k,z]*y[k,z]);
Let me restate the problem a little bit. I assume we can write this as:
sum((i,j), x[i,j]*y[i,j])
with y a binary variable and x a continuous variable. I also assume 0 <= x[i,j] <= U[i,j]. (U is an upper bound).
Here is a way to linearize this quadratic term. We can introduce a variable z[i,j]=x[i,j]*y[i,j] using the following inequalities:
z[i,j] <= U[i,j]*y[i,j]
z[i,j] <= x[i,j]
z[i,j] >= x[i,j]-U[i,j]*(1-y[i,j])
0 <= z[i,j] <= U[i,j]
Now you just can minimize sum((i,j),z[i,j]). For a similar linearization see link.
I have a sql Statement with multiple cases now I have to write the statement in Dax
I tried using switch but the output is not same
the sql statement :
CASE WHEN #STRIKER!='' OR #NONSTRIKER!='' THEN (RUNS + (CASE WHEN BYES = 0 AND LEGBYES = 0 THEN OVERTHROW ELSE 0 END))
WHEN #BOWLER!='' THEN (RUNS + WIDE+ (CASE WHEN BALL.NOBALL > 0 AND (BALL.BYES > 0 OR BALL.LEGBYES > 0) THEN BALL.BYES + BALL.NOBALL + BALL.LEGBYES
WHEN BALL.NOBALL > 0 AND BALL.BYES = 0 AND BALL.LEGBYES = 0 THEN BALL.NOBALL
ELSE 0 END) + (CASE WHEN BYES = 0 AND LEGBYES = 0 THEN OVERTHROW ELSE 0 END))
ELSE GRANDTOTAL END ) AS RUNS
First, I would simplify the case statement,
I assume you have the case statements because NULLS can exist in your result? you could simply try
RUNS +
WIDE +
ISNULL(BALL.NOBALL,0) +
ISNULL(BALL.BYES,0) +
ISNULL(BALL.LEGBYES,0) +
ISNULL(OVERTHROW,0)
But if you've already got the model in PowerBI with those columns listed, you could simply add a calculated column that would sum all the columns which would ignore the nulls...
Total = RUNS + WIDE + NOBALL + BYES + LEGBYES + OVERTHROW
? but without the schema, its really hard to know what the solution might be.
good luck!
I'm trying to figure out how I would go about formatting a large number to the shorter version by appending 'k' or 'm' using Lua. Example:
17478 => 17.5k
2832 => 2.8k
1548034 => 1.55m
I would like to have the rounding in there as well as per the example. I'm not very good at Regex, so I'm not sure where I would begin. Any help would be appreciated. Thanks.
Pattern matching doesn't seem like the right direction for this problem.
Assuming 2 digits after decimal point are kept in the shorter version, try:
function foo(n)
if n >= 10^6 then
return string.format("%.2fm", n / 10^6)
elseif n >= 10^3 then
return string.format("%.2fk", n / 10^3)
else
return tostring(n)
end
end
Test:
print(foo(17478))
print(foo(2832))
print(foo(1548034))
Output:
17.48k
2.83k
1.55m
Here a longer form, which uses the hint from Tom Blodget.
Maybe its not the perfect form, but its a little more specific.
For Lua 5.0, replace #steps with table.getn(steps).
function shortnumberstring(number)
local steps = {
{1,""},
{1e3,"k"},
{1e6,"m"},
{1e9,"g"},
{1e12,"t"},
}
for _,b in ipairs(steps) do
if b[1] <= number+1 then
steps.use = _
end
end
local result = string.format("%.1f", number / steps[steps.use][1])
if tonumber(result) >= 1e3 and steps.use < #steps then
steps.use = steps.use + 1
result = string.format("%.1f", tonumber(result) / 1e3)
end
--result = string.sub(result,0,string.sub(result,-1) == "0" and -3 or -1) -- Remove .0 (just if it is zero!)
return result .. steps[steps.use][2]
end
print(shortnumberstring(100))
print(shortnumberstring(200))
print(shortnumberstring(999))
print(shortnumberstring(1234567))
print(shortnumberstring(999999))
print(shortnumberstring(9999999))
print(shortnumberstring(1345123))
Result:
> dofile"test.lua"
100.0
200.0
1.0k
1.2m
1.0m
10.0m
1.3m
>
And if you want to get rid of the "XX.0", uncomment the line before the return.
Then our result is:
> dofile"test.lua"
100
200
1k
1.2m
1m
10m
1.3m
>