I want to compare if n1-10 one of them is = box
then write nx1 and ny1 and so on ...
what could I do to compare them faster .
the below way will take years from me
box=readString('[[[[[_name]+11c]+670]+394]+2fc]+1c',20)
n1=readString('[_player]+184',20)
n2=readString('[_player]+f94',20)
n3=readString('[_player]+1da4',20)
nx1= readInteger('[_player]+48d8')
nx2= readInteger('[_player]+56e8')
nx3= readInteger('[_player]+64f8')
ny1= readInteger('[_player]+48dc')
ny2= readInteger('[_player]+56ec')
ny3= readInteger('[_player]+64fc')
if box =n1 or n2 or n3 --lets say n1=box
then writeInteger('[_player]+28c',n1x)
writeInteger('[_player]+28c',n1y)
if box =n1 or n2 or n3 --lets say n2=box
then writeInteger('[_player]+28c',n2x)
writeInteger('[_player]+28c',n2y)
You implementation is not good practive here. You'd better store these variable in to an array like
ns[1] = ...
ns[2] = ...
...
nys[1] = ...
nys[2] = ...
then
key = -1
for k,v in na do
if box == v then
key = k
end
end
if key == -1 then
print("no match values")
else
writeInteger('[_player]+28c',nys[key])
writeInteger('[_player]+28c',nys[key])
end
Related
I'm trying to plot a function of two variables with pyplot in Julia. The working starting-point is the following (found here at StackOverflow):
function f(z,t)
return z*t
end
z = linspace(0,5,11)
t = linspace(0,40,4)
for tval in t
plot(z, f(z, tval))
end
show()
This works right for me and is giving me exactly what I wanted:
a field of lines.
My own functions are as follows:
## needed functions ##
const gamma_0 = 6
const Ksch = 1.2
const Kver = 1.5
function Kvc(vc)
if vc <= 0
return 0
elseif vc < 20
return (100/vc)^0.1
elseif vc < 100
return 2.023/(vc^0.153)
elseif vc == 100
return 1
elseif vc > 100
return 1.380/(vc^0.07)
else
return 0
end
end
function Kgamma(gamma_t)
return 1-((gamma_t-gamma_0)/100)
end
function K(gamma_t, vc)
return Kvc(vc)*Kgamma(gamma_t)*Ksch*Kver
end
I've tried to plot them as follows:
i = linspace(0,45,10)
j = linspace(0,200,10)
for i_val in i
plot(i,K(i,j))
end
This gives me the following Error:
isless has no method matching isless(::Int64, ::Array{Float64,1})
while loading In[51], in expression starting on line 3
in Kvc at In[17]:2 in anonymous at no file:4
Obviously, my function cant deal with an array.
Next try:
i = linspace(0,200,11)
j = linspace(0,45,11)
for i_val in i
plot(i_val,map(K,i_val,j))
end
gives me a empty plot only with axes
Can anybody please give me a hint...
EDIT
A simpler example:
using PyPlot
function P(n,M)
return (M*n^3)/9550
end
M = linspace(1,5,5)
n = linspace(0,3000,3001)
for M_val in M
plot(n,P(n,M_val))
end
show()
Solution
OK, with your help I found this solution for the shortened example which works for me as intended:
function P(n,M)
result = Array(Float64, length(n))
for (idx, val) in enumerate(n)
result[idx] = (M*val^3)/9550
end
return result
end
n = linspace(0,3000,3001)
for M_val = 1:5
plot(n,P(n,M_val))
end
show()
This gives me what I wanted for this shortened example. The remainig question is: could it be done in a simpler more elegant way?
I'll try to apply it to the original example and post it when I'll succed.
I don't completely follow all the details of what you're trying to accomplish, but here are examples on how you can modify a couple of your functions so that they accept and return arrays:
function Kvc(vc)
result = Array(Float64, length(vc))
for (idx, val) in enumerate(vc)
if val <= 0
result[idx] = 0
elseif val < 20
result[idx] = (100/val)^0.1
elseif val < 100
result[idx] = 2.023/(val^0.153)
elseif val == 100
result[idx] = 1
elseif val > 100
result[idx] = 1.380/(val^0.07)
else
result[idx] = 0
end
end
return result
end
function Kgamma(gamma_t)
return ones(length(gamma_t))-((gamma_t - gamma_0)/100)
end
Also, for your loop, I think you probably want something like:
for i_val in i
plot(i_val,K(i_val,j))
end
rather than plot(i, K(i,j), as that would just print the same thing over and over.
< is defined for scalars. I think you need to broadcast it for arrays, i.e. use .<. Example:
julia> x = 2
2
julia> x < 3
true
julia> x < [3 4]
ERROR: MethodError: no method matching isless(::Int64, ::Array{Int64,2})
Closest candidates are:
isless(::Real, ::AbstractFloat)
isless(::Real, ::Real)
isless(::Integer, ::Char)
in <(::Int64, ::Array{Int64,2}) at .\operators.jl:54
in eval(::Module, ::Any) at .\boot.jl:234
in macro expansion at .\REPL.jl:92 [inlined]
in (::Base.REPL.##1#2{Base.REPL.REPLBackend})() at .\event.jl:46
julia> x .< [3 4]
1x2 BitArray{2}:
true true
What is the fastest way to iterate over all elements in a 3D NumPy array? If array.shape = (r,c,z), there must be something faster than this:
x = np.asarray(range(12)).reshape((1,4,3))
#function that sums nearest neighbor values
x = np.asarray(range(12)).reshape((1, 4,3))
#e is my element location, d is the distance
def nn(arr, e, d=1):
d = e[0]
r = e[1]
c = e[2]
return sum(arr[d,r-1,c-1:c+2]) + sum(arr[d,r+1, c-1:c+2]) + sum(arr[d,r,c-1]) + sum(arr[d,r,c+1])
Instead of creating a nested for loop like the one below to create my values of e to run the function nn for each pixel :
for dim in range(z):
for row in range(r):
for col in range(c):
e = (dim, row, col)
I'd like to vectorize my nn function in a way that extracts location information for each element (e = (0,1,1) for example) and iterates over ALL elements in my matrix without having to manually input each locational value of e OR creating a messy nested for loop. I'm not sure how to apply np.vectorize to this problem. Thanks!
It is easy to vectorize over the d dimension:
def nn(arr, e):
r,c = e # (e[0],e[1])
return np.sum(arr[:,r-1,c-1:c+2],axis=2) + np.sum(arr[:,r+1,c-1:c+2],axis=2) +
np.sum(arr[:,r,c-1],axis=?) + np.sum(arr[:,r,c+1],axis=?)
now just iterate over the row and col dimensions, returning a vector, that is assigned to the appropriate slot in x.
for row in <correct range>:
for col in <correct range>:
x[:,row,col] = nn(data, (row,col))
The next step is to make
rows = [:,None]
cols =
arr[:,rows-1,cols+2] + arr[:,rows,cols+2] etc.
This kind of problem has come up many times, with various descriptions - convolution, smoothing, filtering etc.
We could do some searches to find the best, or it you prefer, we could guide you through the steps.
Converting a nested loop calculation to Numpy for speedup
is a question similar to yours. There's only 2 levels of looping, and sum expression is different, but I think it has the same issues:
for h in xrange(1, height-1):
for w in xrange(1, width-1):
new_gr[h][w] = gr[h][w] + gr[h][w-1] + gr[h-1][w] +
t * gr[h+1][w-1]-2 * (gr[h][w-1] + t * gr[h-1][w])
Here's what I ended up doing. Since I'm returning the xv vector and slipping it in to the larger 3D array lag, this should speed up the process, right? data is my input dataset.
def nn3d(arr, e):
r,c = e
n = np.copy(arr[:,r-1:r+2,c-1:c+2])
n[:,1,1] = 0
n3d = np.ma.masked_where(n == nodata, n)
xv = np.zeros(arr.shape[0])
for d in range(arr.shape[0]):
if np.ma.count(n3d[d,:,:]) < 2:
element = nodata
else:
element = np.sum(n3d[d,:,:])/(np.ma.count(n3d[d,:,:])-1)
xv[d] = element
return xv
lag = np.zeros(shape = data.shape)
for r in range(1,data.shape[1]-1): #boundary effects
for c in range(1,data.shape[2]-1):
lag[:,r,c] = nn3d(data,(r,c))
What you are looking for is probably array.nditer:
a = np.arange(6).reshape(2,3)
for x in np.nditer(a):
print(x, end=' ')
which prints
0 1 2 3 4 5
I have two tables in lua (In production, a has 18 elements and b has 8):
local a = {1,2,3,4,5,6}
local b = {3,5,7,8,9}
I need to return 'a' omitting any common elements from 'b' -- {1,2,4,6} similar to the ruby command a-b (if a and b were arrays).
The best lua logic I have been able to come up with is:
local function find(a, tbl)
for _,a_ in ipairs(tbl) do if a_==a then return true end end
end
function difference(a, b)
local ret = {}
for _,a_ in ipairs(a) do
if not find(a_,b) then table.insert(ret, a_) end
end
return ret
end
local a = {1,2,3,4,5,6}
local b = {3,5,7,8,9}
local temp = {}
temp = difference(a,b)
print(temp[1],temp[2],temp[3],temp[4])
I need to loop through these table comparison pretty quick (min 10K times a second in production). Is there a cleaner way to do this?
======
This is part of a redis server side script and I have to protect my Redis CPU. Outside of a clean Lua process I have two other options:
1.Create two redis temp keys then run sinter for a big(O) of 42
18 sadd(a)
8 sadd(b)
16 sinter(a,b)
2.Return both a and b to ruby do the array comparison and send back the result.
Network cost of the back and fourth of a few thousand connections a second will be a drain on resources.
Try this:
function difference(a, b)
local aa = {}
for k,v in pairs(a) do aa[v]=true end
for k,v in pairs(b) do aa[v]=nil end
local ret = {}
local n = 0
for k,v in pairs(a) do
if aa[v] then n=n+1 ret[n]=v end
end
return ret
end
You could do this (not tested):
function difference(a, b)
local ai = {}
local r = {}
for k,v in pairs(a) do r[k] = v; ai[v]=true end
for k,v in pairs(b) do
if ai[v]~=nil then r[k] = nil end
end
return r
end
If you can modify a then it is even shorter:
function remove(a, b)
local ai = {}
for k,v in pairs(a) do ai[v]=true end
for k,v in pairs(b) do
if ai[v]~=nil then a[k] = nil end
return r
end
If your tables are sorted you could also do a parallel sweep of the two tables together, something like the following pseudo code:
function diff(a, b)
Item = front of a
Diff = front of b
While Item and Diff
If Item < Diff then
Item is next of a
Else if Item == Diff then
remove Item from a
Item = next of a
Diff = next of b
Else # else Item > Diff
Diff = next of b
This doesn't use any extra tables. Even if you want new table instead of in-place diff, only one new table. I wonder how it would compare to the hash table method (remove).
Note that it doesn't matter how many times you loop, if a and b are small then there will be no major difference between these and your alg. You'll need at least 100 items in a and in b, maybe even 1000.
Maybe something like that:
function get_diff (t1, t2)
local diff = {}
local bool = false
for i, v in pairs (t1) do
if t2 and type (v) == "table" then
local deep_diff = get_diff (t1[i], t2[i])
if deep_diff then
diff[i] = deep_diff
bool = true
end
elseif t2 then
if not (t1[i] == t2[i]) then
diff[i] = t1[i] .. ' -- not [' .. t2[i] .. ']'
bool = true
end
else
diff[i] = t1[i]
bool = true
end
end
if bool then
return diff
end
end
local t1 = {1, 2, 3, {1, 2, 3}}
local t2 = {1, 2, 3, {1, 2, 4}}
local diff = get_diff (t1, t2)
Result:
diff = {
nil,
nil,
nil,
{
[3] = "3 -- not [4]"
}
}
I may be asking a silly question but I am self teaching myself VBA and I am just stumped and I am not even sure what terms I can use to look up a solution.
I am writing a code that will compare three variables to three other variables then I want to display which variables have changed.
So if x = a but y <> b and z <> c then the output should be b/c
I have worked out a code that works fine
Dim Str As String
If X <> A Then
If Y <> B Then
If Z <> C Then
Str = "a/b/c"
Else
Str = "a/b"
End If
ElseIf Z <> C Then
Str = "a/c"
Else
Str = "a"
End If
ElseIf Y <> B Then
If Z <> C Then
Str = "b/c"
Else
Str = "b"
End If
Else
Str = "c"
End If
But as I increase the number of variables this becomes extremely complex very quickly.
If anyone can help direct me to a simpler method without the exponential complexity I would be very grateful.
Thank you all so much!
You need to test each variable pair independently from each other -- not link them together in one giant If construct tree.
Example:
str = "" 'Start with blank string. Append as required.
If x <> a Then str = str & "a/"
If y <> b Then str = str & "b/"
If z <> c Then str = str & "c/"
'Remove the extra / at the end
If Right(str, 1) = "/" Then str = Left(str, Len(str - 1))
You could put the 2 strings in 2 arrays, and then use a FOR...NEXT construct to loop both arrays. You can use UBound(arValues) to dynamically find out the number of items in the array.
Good luck
I have an assignment in which I need to swap two integers without using third variable.
I'm not sure how to do this. How would I code this?
Yes, it's possible:
Dim var1 = 1
Dim var2 = 2
var1 = var1 + var2
var2 = var1 - var2
var1 = var1 - var2
But why do you need it? The code becomes abstruse.
Lets assume
a = 10;
b = 20;
a = a + b; // a = 30
b = a - b; // b = 10
a = a - b; // a = 20
Values swapped.
Read up on the "xor swap algorithm."
You can find an answer here:
http://www.java2s.com/Tutorial/VB/0040__Data-Type/Swaptwointegerswithoutusingathird.htm
firstValue = firstValue Xor secondValue
secondValue = firstValue Xor secondValue
firstValue = firstValue Xor secondValue
Dim a As Integer
Dim b As Integer
a= 1
b= 2
a = a Xor b
b = a Xor b
a = a Xor b
To swap two numeric variables do like this
a = a + b;
b = a - b;
a = a - b;
OR
a = a xor b;
b = a xor b;
a = a xor b;
where a and b are variables to be swapped
theoretically 3 ways
a = 4 , b = 5
1. Using XOR
a = a XOR b = 4 XOR 5 = 9
b = a XOR b = 9 XOR 5 = 4
a = a XOR b = 9 XOR 4 = 5
2. Using +,-
a = a+b = 4+5 = 9 // should not overflow
b = a-b = 9-5 = 4
a = a-b = 9-4 = 5
3. Using *,/
a = a*b = 4*5 = 20 // should not overflow
b = a/b = 20/5 = 4 // should not overflow and should not be irrational number
a = a/b = 20/4 = 5 // should not overflow and should not be irrational number
The Xor or a+b algorithms above work and are the best way to do this, but just an example of a weird way to do it. Still not sure why you would want to do this. Just build a function that you supply two values ByRef and have it do the standard swap method.
Dim newList as New List(Of Integer)
newList.Add firstvalue
newList.Add secondValue
newList.Reverse
secondValue = newList.Item(0)
firstValue = newList.Item(1)
Take two text boxes and a command box.In command box type this code.
text1.text=val(text1.text) + val(text2.text)
text2.text=val(text1.text) - val(text2.text)
text1.text=val(text1.text) - val(text2.text)
Check link written for you
Approach#1.
Addition and Subtraction Method
Integer a, b
read a and b
a= a+b;
b=a-b;
a=a-b;
Problem:
Incorrect result when sum of numbers will exceed the Integer range.
Approach#2.
Multiplication and Division Method
Integer a, b
read a and b
a=a*b;
b=a/b;
a=a/b;
Problems:
If the value of a*b exceeds the range of integer.
If the value of a or b is zero then it will give wrong results.
Approach#3.
XOR Method
Integer a , b
read a and b
a=a^b;
b=a^b;
a=a^b;
Best approach to solve this problem without any pitfalls.