I need some advice about how to logically proceed to limit the use of an app, I mean, as Shazam does: you can use it only for some times during a month, then you have to wait until next month for use it again. I'm doing it with Xcode and objective c.
How can I understand if the month is changed?
From a logical point of view, a basic method can be:
use a structure like:
struct run_time_for_month {
int month;
int minutes_left;
}
and save in some option file (maybe with 0 as default).
when the application start, load the structure, then check the month. If it is 0, then is the first run, so set
month = current_month
minutes_left = 100 (for example)
and write it to the file.
If the month is greater than 0, then you use this code (I write some pseudo code here)
if current_month == saved_month then
if minutes_left <= 0 then
*** Running time for month ended ***
*** Notify the user and exit the app ***
else
saved_month = current_month
minutes_left = 100
and save the file
Now, while the application is running, every x minutes (with x = 5 or 10) and when the application is quitting you use this code (again, pseudo code here)
minutes_left = minutes_left - x
if minutes_left <= 0 then
*** Time for month ended ***
*** Notify the user and exit the app ***
this a stripped down version of what I do in my code when I needed something like that, but again: I am not working with XCode and/or Objective C but with C++ so this can be only an idea.
Related
How do you convert base 12 to base 10 in Python 3 and the other way around too? (from 10 to 12)
I wrote this code to make a list of numbers in base 4, but the code never executes and eventually has a timeout error. Any ideas on how to fix it? It also identifies the mode and counts the frequency of the mode. The program should start at 1 and print the next 21 numbers in base 4 and then do the mode stuff.
while num > 0:
if start in yes:
list4.append(start)
start += 1
num -= 1
if start not in yes:
num += 1
new_4 = ''.join([str(n) for n in list4])
print(statistics.mode(new_4))
count_m = new_4.count(statistics.mode(new_4))
print(count_m)
Comment any other questions you may have!
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.
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
I have a table with 2 columns. In column 1, I have a string information, in column 2, I have a logical index
%% Tables and their use
T={'A2P3';'A2P3';'A2P3';'A2P3 with (extra1)';'A2P3 with (extra1) and (extra 2)';'A2P3 with (extra1)';'B2P3';'B2P3';'B2P3';'B2P3 with (extra 1)';'A2P3'};
a={1 1 0 1 1 0 1 1 0 1 1 }
T(:,2)=num2cell(1);
T(3,2)=num2cell(0);
T(6,2)=num2cell(0);
T(9,2)=num2cell(0);
T=table(T(:,1),T(:,2));
class(T.Var1);
class(T.Var2);
T.Var1=categorical(T.Var1)
T.Var2=cell2mat(T.Var2)
class(T.Var1);
class(T.Var2);
if T.Var1=='A2P3' & T.Var2==1
disp 'go on'
else
disp 'change something'
end
UPDATES:
I will update this section as soon as I know how to copy my workspace into a code format
** still don't know how to do that but here it goes
*** why working with tables is a double edged sword (but still cool): I have to be very aware of the class inside the table to refer to it in an if else construct, here I had to convert two columns to categorical and to double from cell to make it work...
Here is what my data looks like:
I want to have this:
if T.Var1=='A2P3*************************' & T.Var2==1
disp 'go on'
else
disp 'change something'
end
I manage to tell matlab to do as i wish, but the whole point of this post is: how do i tell matlab to ignore what comes after A2P3 in the string, where the string length is variable? because otherwise it would be very tiring to look up every single piece of string information left on A2P3 (and on B2P3 etc) just to say thay.
How do I do that?
Assuming you are working with T (cell array) as listed in your code, you may use this code to detect the successful matches -
%%// Slightly different than yours
T={'A2P3';'NotA2P3';'A2P3';'A2P3 with (extra1)';'A2P3 with (extra1) and (extra 2)';'A2P3 with (extra1)';'B2P3';'B2P3';'NotA2P3';'B2P3 with (extra 1)';'A2P3'};
a={1 1 0 1 1 0 1 1 0 1 1 }
T(:,2)=num2cell(1);
T(3,2)=num2cell(0);
T(6,2)=num2cell(0);
T(9,2)=num2cell(0);
%%// Get the comparison results
col1_comps = ismember(char(T(:,1)),'A2P3') | ismember(char(T(:,1)),'B2P3');
comparisons = ismember(col1_comps(:,1:4),[1 1 1 1],'rows').*cell2mat(T(:,2))
One quick solution would be to make a function that takes 2 strings and checks whether the first one starts with the second one.
Later Edit:
The function will look like this:
for i = 0, i < second string's length, i = i + 1
if the first string's character at index i doesn't equal the second string's character at index i
return false
after the for, return true
This assuming the second character's lenght is always smaller the first's. Otherwise, return the function with the arguments swapped.
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