This is a matlab code that uses guide to run a timer. The timer function counts 10 numbers starting from the number provided in the text field. I would like to know how to enter two numbers successively and make matlab counts 10 numbers for both values in parallel.
let's say I entered 0 then I pressed the start button, then I immediately entered 10 and pressed the start button again. what happens now is that it counts only from 0 to 10. I would appreciate if you can share a way to make my code count from 0 to 10 and from 10 to 20 simultaneously in parallel.
guide code :
function startbutton_Callback(hObject, eventdata, handles)
t=timer;
t.TimerFcn = #counter;
t.Period = 15;
t.ExecutionMode = 'fixedRate';
t.TasksToExecute = 1;
start(t);
stop (t);
delete (t);
timer callback function:
function counter(~,~)
handles = guidata(counterFig);
num = str2double(get(handles.edit1,'String'));
for i = 0:10
disp (num);
num = num+1;
pause (1);
end
you can use parrallel toolbox for real parallel computation.
but if you dont have that , you can create another timer object that count from 10 to 20
and run it
Related
I am trying to simulate n-dimensional game of life for first t=6 time steps. My Nim code is a straightforward port from Python and it works correctly but instead of the expected speedup, for n=4, t=6 it takes 2 seconds to run, which is order of magnitude slower than my CPython version. Why is my code so slow? What can I do to speed it up? I am compiling with -d:release and --opt:speed
I represent each point in space with a single 64bit integer.
That is, I map (x_0, x_1, ..., x_{n-1}) to sum x_i * 32^i. I can do that since I know that after 6 time steps each coordinate -15<=x_i<=15 so I have no overflow.
The rules are:
alive - has 2 or 3 alive neigbours: stays alive
- different number of them: becomes alive
dead - has 3 alive neighbours: becomes alive
- else: stays dead
Below is my code. The critical part is the proc nxt which gets set of active cells and outputs set of active cells next time step. This proc is called 6 times. The only thing I'm interested in is the number of alive cells.
I run the code on the following input:
.##...#.
.#.###..
..##.#.#
##...#.#
#..#...#
#..###..
.##.####
..#####.
Code:
import sets, tables, intsets, times, os, math
const DIM = 4
const ROUNDS = 6
const REG_SIZE = 5
const MAX_VAL = 2^(REG_SIZE-1)
var grid = initIntSet()
# Inits neighbours
var neigbours: seq[int]
proc initNeigbours(base,depth: int) =
if depth == 0:
if base != 0:
neigbours.add(base)
else:
initNeigbours(base*2*MAX_VAL-1, depth-1)
initNeigbours(base*2*MAX_VAL+0, depth-1)
initNeigbours(base*2*MAX_VAL+1, depth-1)
initNeigbours(0,DIM)
echo neigbours
# Calculates next iteration:
proc nxt(grid: IntSet): IntSet =
var counting: CountTable[int]
for x in grid:
for dx in neigbours:
counting.inc(x+dx)
for x, count in counting.pairs:
if count == 3 or (count == 2 and x in grid):
result.incl(x)
# Loads input
var row = 0
while true:
var line = stdin.readLine
if line == "":
break
for col in 0..<line.len:
if line[col] == '#':
grid.incl((row-MAX_VAL)*2*MAX_VAL + col-MAX_VAL)
inc row
# Run computation
let time = cpuTime()
for i in 1..ROUNDS:
grid = nxt(grid)
echo "Time taken: ", cpuTime() - time
echo "Result: ", grid.len
discard stdin.readLine
Your code runs in my computer in about 0.02:
Time taken: 0.020875947
Result: 2276
Time taken: 0.01853268
Result: 2276
Time taken: 0.021355269
Result: 2276
I changed the part where the input is read to this:
# Loads input
var row = 0
let input = open("input.txt")
for line in input.lines:
for i, col in line:
if col == '#':
grid.incl((row-MAX_VAL)*2*MAX_VAL + i-MAX_VAL)
inc row
input.close()
But it shouldn't impact the performance, it just looks better to my eyes. I compiled with:
nim c -d:danger script.nim
Using Nim 1.4.2. -d:danger is the flag for maximum speed before entering deeper waters.
But even compiling in debug mode:
$ nim c -r script.nim
Time taken: 0.07699487199999999
Result: 2276
Way faster than 2 seconds. There has to be other problem in your end. Sorry for the non-answer.
I have the following:
def process(msg):
print(msg)
for i in range(10): # This is in place of the
print(i, end=' ') # actual task, which takes
time.sleep(1) # around 10 seconds to run
ws = websocket()
ws.start_socket_abc(callback=process)
The websocket pushes info every second and makes a call to process each time.
Say the websocket is returning "one" for this second, "two" for the next second, "three" for the second after, and so on. Running the above code, the ouput would be:
one
0 1 2 3 4 5 6 7 8 9 <- this line takes 10 seconds to finish
two
0 1 2 3 4 5 6 7 8 9 <- this line takes 10 seconds to finish
...
It's clear that process("two") gets queued somewhere and is waited to be called after process("one") finishes. By the time the program prints out two, the real-time message should be twelve.
What I'm trying to achieve is to completely skip the "queue". I want it to take the message, do the task, skip all the calls in between (this is important, I actually need to skip them), take the next real-time message and do the task. In short, I want it to print out whatever real-time message is at the moment process("one") finishes. i.e.
one
0 1 2 3 4 5 6 7 8 9 <- this line takes 10 seconds to finish
twelve
...
This is what I have in mind.
message = ''
def process(msg):
message = msg
ws = websocket()
ws.start_socket_abc(callback=process)
def task(msg=message):
print(msg)
for i in range(10): # This is in place of the
print(i, end=' ') # actual task, which takes
time.sleep(1) # around 10 seconds to run
start_thread(target=task)
This performs the task with the message variable, which changes in real-time with the messages received from the websocket so after task("one") finishes, message is at "twelve" and task("twelve") will be called, skipping all the messages in between.
I strongly doubt that this is the proper way of doing it. Someone please enlighten me, thanks in advance!
So I'm making a Quiz for a school project and I want to start a timer at the start of the Quiz which will be stopped at the end and printed the time took for the player to complete the Quiz, I put the Timer that I created at the start but my program gets stuck at the 'While loop' and doesn't continue with the program.
Heres the beginning of my code:
import time
score = 0
lives = 3
o = 1
timer = 0
while o == 1:
time.sleep(1)
timer += 1
I add 1 to 'o' just before the Quiz ends so I can print out the time taken but I can get past the start
Your While loop isnt comparing the value of 'o'. Use == not = and check.
I have this algorithm that two threads run at the same time
n = 0
int tmp
do 10 times
tmp = n
n = tmp + 1
I know this can get 20, and 10 by each thread executing all the way to the end in sequence, and by one thread loading n into tmp, then the other finishing will result in 10.
I'm just not sure if this can get 2.
I would say that n will be >= 10 and <= 20, and that I really don't see how you could get anything < 10. But the whole thing is so wrong, I don't even know why I am answering :-)
tl;dr: My code "works", in that it gives me the answer I need. I just can't get it to stop running when it reaches that answer. I'm stuck with scrolling back through the output.
I'm a complete novice at programming/Python. In order to hone my skills, I decided to see if I could program my own "solver" for Implied Equity Risk Premium from Prof. Damodaran's Valuation class. Essentially, the code takes some inputs and "guesses and tests" a series of interest rates until it gets a "close" value to the input.
Right now my code spits out an output list, and I can scroll back through it to find the answer. It's correct. However, I cannot for the life of me get the code to "stop" at the correct value with the while function.
I have the following code:
per = int(input("Enter the # of periods forecast ->"))
divbb = float(input("Enter the initial dividend + buyback value ->"))
divgr = float(input("Enter the div + buyback growth rate ->"))
tbondr = float(input("Enter the T-Bond rate ->"))+0.000001
sp = int(input("Enter the S&P value->"))
total=0
pv=0
for i in range(1,10000):
erp = float(i/10000)
a = divbb
b = divgr
pv = 0
temppv = 0
print (sp-total, erp)
for i in range(0, per):
a=a * (1+b)
temppv = a / pow((1+erp),i)
pv=pv+temppv
lastterm=(a*1+tbondr)/((erp-tbondr)*pow(1+erp,per))
total=(pv+lastterm)
From his example, with the inputs:
per = 5
divbb = 69.46
divgr = 0.0527
tbondr = 0.0176
sp = 1430
By scrolling back through the output, I can see my code produces the correct minimum at epr=0.0755.
My question is: where do I stick the while to stop this code at that minimum? I've tried a lot of variations, but can't get it. What I'm looking for is, basically:
while (sp-total) > |1|, keep running the code.
per = 5
divbb = 69.46
divgr = 0.0527
tbondr = 0.0176
sp = 1430
total=0
pv=0
i = 1
while(abs(sp-total)) > 1:
erp = i/10000.
a = divbb
b = divgr
pv = 0
temppv = 0
print (sp-total, erp)
for j in range(0, per):
a=a * (1+b)
temppv = a / pow((1+erp),j)
pv=pv+temppv
lastterm=(a*1+tbondr)/((erp-tbondr)*pow(1+erp,per))
total=(pv+lastterm)
i += 1
should work. Obviously, there are a million ways to do this. But the general gist here is that the while loop will stop as soon as it meets the condition. You could also test every time in the for loop and include a break statement, but because you don't know when it will stop, I think a while loop is better in this case.
Let me give you a quick rundown of two different ways you could solve a problem like this:
Using a while loop:
iterator = start value
while condition(iterator):
do some stuff
increment iterator
Using a for loop:
for i in xrange(startvalue, maxvalue):
do some stuff
if condition:
break
Two more thing: if you're doing large ranges, use the generator xrange. Also, it's probably a bad idea to reuse i inside your for loop.
I recommend CS101 from Udacity.com for learning Python. Also, if you're interested in algorithms, work through the problems at projecteuler.com