I have a function
makeMarks: (first, nextIncrement, classifier) ->
results = new Array()
t = first(#minT)
while t<=#maxT
mark =
t: t
x: this.tToX(t)
class: classifier(t)
results.push(mark)
t = nextIncrement(t)
results
this function works great with the following two functions as parameters
# parameters for hour tickmarks
#firstHour = (t) ->
msPerHour = 1000*60*60
Math.floor(t / msPerHour) * msPerHour
#nextHour = (currentHour) ->
msPerHour = 1000*60*60
currentHour + msPerHour
when called as such
marks = markMaker.makeMarks( #firstMonth, #nextMonth, #classifier)
Now to the problem:
# parameters for month tickmarks
#firstMonth = (minT) ->
msPerDay = 1000*60*60*24
t = Math.floor(minT/msPerDay) * msPerDay
d = new Date(t)
while(d.getDate() isnt 0)
t += msPerDay
d.setTime(t)
t
#nextMonth = (currentMonth) ->
msPerDay = 1000*60*60*24
t = currentMonth + msPerDay
d = new Date(t)
while(d.getDate() isnt 0)
t += msPerDay
d.setTime(t)
t
The hour code works fine, but the month code doesn't seem to terminate.
The getDate function never returns 0. Its minimum value is 1 and it's maximum value is 31. If you're looking for anything outside that range, that's a long wait for a train that don't come.
Related
Below is my implementation of a2c using PyTorch. Upon learning about backpropagation in PyTorch, I have known to zero_grad() the optimizer after each update iteration. However, there is still a RunTime error on second-time backpropagation.
def torchworker(number, model):
worker_env = gym.make("Taxi-v3").env
max_steps_per_episode = 2000
worker_opt = optim.Adam(lr=5e-4, params=model.parameters())
p_history = []
val_history = []
r_history = []
running_reward = 0
episode_count = 0
under = 0
start = time.time()
for i in range(2):
state = worker_env.reset()
episode_reward = 0
penalties = 0
drop = 0
print("Episode {} begins ({})".format(episode_count, number))
worker_env.render()
criterion = nn.SmoothL1Loss()
time_solve = 0
for _ in range(1, max_steps_per_episode):
#worker_env.render()
state = torch.tensor(state, dtype=torch.long)
action_probs = model.forward(state)[0]
critic_value = model.forward(state)[1]
val_history.append((state, critic_value[0]))
# Choose action
action = np.random.choice(6, p=action_probs.detach().numpy())
p_history.append(torch.log(action_probs[action]))
# Apply chosen action
state, reward, done, _ = worker_env.step(action)
r_history.append(reward)
episode_reward += reward
time_solve += 1
if reward == -10:
penalties += 1
elif reward == 20:
drop += 1
if done:
break
# Update running reward to check condition for solving
running_reward = (running_reward * (episode_count) + episode_reward) / (episode_count + 1)
# Calculate discounted returns
returns = deque(maxlen=3500)
discounted_sum = 0
for r in r_history[::-1]:
discounted_sum = r + gamma * discounted_sum
returns.appendleft(discounted_sum)
# Calculate actor losses and critic losses
loss_actor_value = 0
loss_critic_value = 0
history = zip(p_history, val_history, returns)
for log_prob, value, ret in history:
diff = ret - value[1]
loss_actor_value += -log_prob * diff
ret_tensor = torch.tensor(ret, dtype=torch.float32)
loss_critic_value += criterion(value[1], ret_tensor)
loss = loss_actor_value + 0.1 * loss_critic_value
print(loss)
# Update params
loss.backward()
worker_opt.step()
worker_opt.zero_grad()
# Log details
end = time.time()
episode_count += 1
if episode_count % 1 == 0:
worker_env.render()
if running_reward > -50: # Condition to consider the task solved
under += 1
if under > 5:
print("Solved at episode {} !".format(episode_count))
break
I believe there may be something to do with the architecture of my AC model, so I also include it here for reference.
class ActorCriticNetwork(nn.Module):
def __init__(self, num_inputs, num_hidden, num_actions):
super(ActorCriticNetwork, self).__init__()
self.embed = nn.Embedding(500, 10)
self.fc1 = nn.Linear(10, num_hidden * 2)
self.fc2 = nn.Linear(num_hidden * 2, num_hidden)
self.c = nn.Linear(num_hidden, 1)
self.fc3 = nn.Linear(num_hidden, num_hidden)
self.a = nn.Linear(num_hidden, num_actions)
def forward(self, x):
out = F.relu(self.embed(x))
out = F.relu(self.fc1(out))
out = F.relu(self.fc2(out))
critic = self.c(out)
out = F.relu(self.fc3(out.detach()))
actor = F.softmax(self.a(out), dim=-1)
return actor, critic
Would you please tell me what the mistake here is? Thank you in advance.
SOLVED: I forgot to clear the history of probabilities, action-values and rewards after iterations. It is clear why that would cause the issue, as the older elements would cause propagating through old dcgs.
I’m dealing with BigDecimal in Java and I need to make 2 check against BigDecimal fields in my DTO:
Number of digits of full part (before point) < 15
Total number of
digits < 32 including scale (zeros after point)
What is the best way to implement it? I extremely don’t want toBigInteger().toString() and .toString()
I think this will work.
BigDecimal d = new BigDecimal("921229392299229.2922929292920000");
int fractionCount = d.scale();
System.out.println(fractionCount);
int wholeCount = (int) (Math.ceil(Math.log10(d.longValue())));
System.out.println(wholeCount);
I did some testing of the above method vs using indexOf and subtracting lengths of strings. The above seems to be signficantly faster if my testing methodology is reasonable. Here is how I tested it.
Random r = new Random(29);
int nRuns = 1_000_000;
// create a list of 1 million BigDecimals
List<BigDecimal> testData = new ArrayList<>();
for (int j = 0; j < nRuns; j++) {
String wholePart = r.ints(r.nextInt(15) + 1, 0, 10).mapToObj(
String::valueOf).collect(Collectors.joining());
String fractionalPart = r.ints(r.nextInt(31) + 1, 0, 10).mapToObj(
String::valueOf).collect(Collectors.joining());
BigDecimal d = new BigDecimal(wholePart + "." + fractionalPart);
testData.add(d);
}
long start = System.nanoTime();
// Using math
for (BigDecimal d : testData) {
int fractionCount = d.scale();
int wholeCount = (int) (Math.ceil(Math.log10(d.longValue())));
}
long time = System.nanoTime() - start;
System.out.println(time / 1_000_000.);
start = System.nanoTime();
//Using strings
for (BigDecimal d : testData) {
String sd = d.toPlainString();
int n = sd.indexOf(".");
int m = sd.length() - n - 1;
}
time = System.nanoTime() - start;
System.out.println(time / 1_000_000.);
}
i have problems regarding passing value with formatting issues.
The description i have put in the codes as i am having problem passing my PreEditedCheque in my code for the if ValidateMMonCheque <> MM part. The output for if length(RawChequenumber) = 15 will be in 1 digit instead of 00001 ( example)
MM = HostGetFLD('','MM')
YY = HostGetFLD('','YY')
PreEditedCheque = substr(RawChequenumber,11,5)
ValidateMMonCheque = substr(RawChequenumber,7,2)
if ValidateMMonCheque <> MM Then *From this statement*
Do
PreEditedCheque = substr('00000',1,5) *This part where those 0 can't be properly shown if pass to the next statement*
EditedCheque = '00'||'2'||'0'||YY||MM||'00'||PreEditedCheque
rc = message(2,2,EditedCheque)
End
if length(RawChequenumber) = 15 Then
EditedCheque = '00'||'2'||'0'||YY||MM||'00'||PreEditedCheque + 1 *Second statement if <>MM ran, this part, the PreEditedCheque will be not in 00001, it will be 1.
rc = PanSetCtlData('PREVIEW',EditedCheque)
What you're asking for is to have the cheque number padded to the left with zeroes in a 5-character field. The Right() function is your friend:
Right(PreEditedCheque, 5, '0') /* "1" -> "00001" */
I have a Google Sheet with three columns:
- Date and time (timestamp)
- Duration
- Description
I have an script that when I write something in 'Description', inserts in 'Date' the date and time at this moment, and the 'Duration':
function onEdit(e) {
if(e.source.getActiveSheet().getName() == "Sheet2" ) {
var col = e.source.getActiveCell().getColumn();
if(col == 3 ) {
// I'm in column three
var cellTimeStamp = e.range.offset(0,-2); // First column of the same row
var cellTimeDiff = e.range.offset(0,-1); // Second column of the same row
var cellTimePrev = e.range.offset(-1,-2); // First column of the previous row
var timeTimeStamp = new Date();
var iniTime = cellTimePrev.getValue().getTime();
var finTime = timeTimeStamp.getTime() ;
var timeDiff = String(finTime - iniTime) ;
cellTimeStamp.setValue(timeTimeStamp);
cellTimeDiff.setValue(timeDiff); // [***]
}
}
}
When this executes (as an event) in the column of 'Duration' there is NOT something in the format of 'HH:mm:ss'.
But if I remove the last line in this script and adds this formulae in the sheet:
=A3-A2 (in row 3)
=A4-A3 (in row 4)
...
then it works ok.
I'd like to know how to meet the same result but with a script.
Thanks in advance.
timeDiff is the result of finTime - iniTime which are both native date object values, which means we have milliseconds .
converting that in hh:mm:ss is simple math... : 60 seconds in a minute and 60 minutes in an hour...
A simple code could be like this :
function msToTime(s) {
var ms = s % 1000;
s = (s - ms) / 1000;
var secs = s % 60;
s = (s - secs) / 60;
var mins = s % 60;
var hrs = (s - mins) / 60;
return hrs + ':' + mins + ':' + secs; // milliSecs are not shown but you can use ms if needed
}
If you prefer formating your string more conventionally (2 digits for each value) don't forget you can use Utilities.formatString() to do so.
example below :
return Utilities.formatString("%02d",hrs) + ':' + Utilities.formatString("%02d",mins) + ':' + Utilities.formatString("%02d",secs);
EDIT
Following your comment :
Spreadsheets are smarter than you think, you can try the code below and you will see that the result is actually a time value.(check by double clicking on it)
function test() {
var sh = SpreadsheetApp.getActiveSheet();
var t1 = sh.getRange('a1').getValue().getTime();
var t2 = sh.getRange('b1').getValue().getTime();
sh.getRange('c1').setValue(msToTime(t1-t2)).setNumberFormat('hh:mm:ss');
}
function msToTime(s) {
var ms = s % 1000;
s = (s - ms) / 1000;
var secs = s % 60;
s = (s - secs) / 60;
var mins = s % 60;
var hrs = (s - mins) / 60;
return hrs + ':' + mins + ':' + secs; // milliSecs are not shown but you can use ms if needed
}
note that setNumberFormat('hh:mm:ss') is optional, it's only there to force the spreadsheet to display hour:min:sec format but automatic mode works as well.
I am trying to calculate the time difference between 2 date time strings.
I have 2 inputs where the input string is something like this "1:00 PM" and the second one "3:15 PM". I want to know the time difference. So for the above example I want to display 3.15
What I have done:
Converted the time to a 24 hours format. So "1:00 PM" becomes "13:00:00"
Appended the new time to a date like so: new Date("1970-1-1 13:00:00")
Calculated the difference like so:
Code:
var total = Math.round(((new Date("1970-1-1 " + end_time) -
new Date("1970-1-1 " + start_time) ) / 1000 / 3600) , 2 )
But the total is always returning integers and not decimals, so the difference between "1:00 PM" and "3:15 PM" is 2 not 2.15.
I have also tried this (using jQuery, but that is irrelevant):
$('#to_ad,#from_ad').change(function(){
$('#total_ad').val( getDiffTime() );
});
function fixTimeString(time){
var hours = Number(time.match(/^(\d+)/)[1]);
var minutes = Number(time.match(/:(\d+)/)[1]);
var AMPM = time.match(/\s(.*)$/)[1];
if(AMPM == "PM" && hours<12) hours = hours+12;
if(AMPM == "AM" && hours==12) hours = hours-12;
var sHours = hours.toString();
var sMinutes = minutes.toString();
if(hours<10) sHours = "0" + sHours;
if(minutes<10) sMinutes = "0" + sMinutes;
return sHours + ':' + sMinutes + ':00';
}
function getDiffTime(){
var start_time = fixTimeString($('#from_ad').val());
var end_time = fixTimeString($('#to_ad').val());
var start = new Date("1970-1-1 " + end_time).getTime(),
end = new Date("1970-1-1 " + start_time).getTime();
return parseInt(((start - end) / 1000 / 3600, 10)*100) / 100;
}
But the total_ad input is displaying only integer values.
How can I fix this problem?
Math.round rounds to the nearest integer, multiply and divide instead
var start = new Date("1970-1-1 " + start_time).getTime(),
end = new Date("1970-1-1 " + end_time).getTime();
var total = (parseInt(((start-end) / 1000 / 3600)*100, 10)) / 100;
FIDDLE
When you take the time 15:15:00 and subtract 13:00:00, you're left with 2.15 hours, not 3.15, and this example would return 2.15 even without making sure there is only two decimals, but for other times that might not be the case.
You could also use toFixed(2), but that would leave you with 3.00 and not 3 etc.
This is how I calculate it:
calculateDiff();
function calculateDiff(){
_start = "7:00 AM";
_end = "1:00 PM";
_start_time = parseAMDate(_start);
_end_time = parseAMDate(_end);
if (_end_time < _start_time){
_end_time = parseAMDate(_end,1);
}
var difference= _end_time - _start_time;
var hours = Math.floor(difference / 36e5),
minutes = Math.floor(difference % 36e5 / 60000);
if (parseInt(hours) >= 0 ){
if (minutes == 0){
minutes = "00";
}
alert(hours+":"+minutes);
}
}
function parseAMDate(input, next_day) {
var dateReg = /(\d{1,2}):(\d{2})\s*(AM|PM)/;
var hour, minute, result = dateReg.exec(input);
if (result) {
hour = +result[1];
minute = +result[2];
if (result[3] === 'PM' && hour !== 12) {
hour += 12;
}
}
if (!next_day) {
return new Date(1970, 01, 01, hour, minute).getTime();
}else{
return new Date(1970, 01, 02, hour, minute).getTime();
}
}