Compiler errors in xCode 6 final release - xcode

I was working with a game in swift StriteKit using the beta Xcode. But now with the final release I had a lot of error all of them I been able to fix except this one.
THIS WAS MY ORIGINAL CODE WITH NO ERRORS USING BETA XCODE:
bird.zRotation = self.clamp(-1, max: 0.5, value: bird.physicsBody.velocity.dy * (bird.physicsBody?.velocity.dy < 0 ?0.003 : 0.001 ))
But xCode final release indicates a compiler error on physicsBody stating that: 'SKphysicsBody?' does not have a member called velocity.
I fix this by adding '?' the optional type to physicsBody.
bird.zRotation = self.clamp(-1, max: 0.5, value: bird.physicsBody?.velocity.dy * (bird.physicsBody?.velocity.dy < 0 ?0.003 : 0.001 ))
But stil a new error appears this time on dy stating GGFloat not unwrapped I tried using '!' after dy or '?' still the compiler after doing this suggest deleting it, stating Postfix '?' should have optional type ; type is CGFloat.
I have tried to look for information on what is going on exactly but I can't fix this error. Please help.

It's possible that Xcode is getting confused trying to perform arithmetic on potentially nil values. I'd try moving the optionals out of the self.clamp call. Also, watch out for the spacing around the ternary check: it might be trying to unwrap the value adjacent to the ? operator.
Try
if let dy = bird.physicsBody?.velocity.dy {
self.clamp(-1, max: 0.5, value: dy * ((dy < 0) ? 0.003 : 0.001)
}

Related

How to get same cci values from Trading view in Golang?

I'm trying to replicate values from pine script cci() function in golang. I've found this lib https://github.com/markcheno/go-talib/blob/master/talib.go#L1821
but it gives totally different values than cci function does
pseudo code how do I use the lib
cci := talib.Cci(latest14CandlesHighArray, latest14CandlesLowArray, latest14CandlesCloseArray, 14)
The lib gives me the following data
Timestamp: 2021-05-22 18:59:27.675, Symbol: BTCUSDT, Interval: 5m, Open: 38193.78000000, Close: 38122.16000000, High: 38283.55000000, Low: 38067.92000000, StartTime: 2021-05-22 18:55:00.000, EndTime: 2021-05-22 18:59:59.999, Sma: 38091.41020000, Cci0: -16.63898084, Cci1: -53.92565811,
While current cci values on TradingView are: cci0 - -136, cci1 - -49
could anyone guide what do I miss?
Thank you
P.S. cci0 - current candle cci, cci1 - previous candle cci
PineScript has really great reference when looking for functions, usually even supplying the pine code to recreate it.
https://www.tradingview.com/pine-script-reference/v4/#fun_cci
The code wasn't provided for cci, but a step-by-step explanation was.
Here is how I managed to recreate the cci function using Pine, following the steps in the reference:
// This source code is subject to the terms of the Mozilla Public License 2.0 at https://mozilla.org/MPL/2.0/
// © bajaco
//#version=4
study("CCI Breakdown", overlay=false, precision=16)
cci_breakdown(src, p) =>
// The CCI (commodity channel index) is calculated as the
// 1. difference between the typical price of a commodity and its simple moving average,
// divided by the
// 2. mean absolute deviation of the typical price.
// 3. The index is scaled by an inverse factor of 0.015
// to provide more readable numbers
// 1. diff
ma = sma(src,p)
diff = src - ma
// 2. mad
s = 0.0
for i = 0 to p - 1
s := s + abs(src[i] - ma)
mad = s / p
// 3. Scaling
mcci = diff/mad / 0.015
mcci
plot(cci(close, 100))
plot(cci_breakdown(close,100))
I didn't know what mean absolute deviation meant, but at least in their implementation it appears to be taking the difference from the mean for each value in the range, but NOT changing the mean value as you go back.
I don't know Go but that's the logic.

Pinescript duplicate alerts

I have created a very basic script in pinescript.
study(title='Renko Strat w/ Alerts', shorttitle='S_EURUSD_5_[MakisMooz]', overlay=true)
rc = close
buy_entry = rc[0] > rc[2]
sell_entry = rc[0] < rc[2]
alertcondition(buy_entry, title='BUY')
alertcondition(sell_entry, title='SELL')
plot(buy_entry/10)
The problem is that I get a lot of duplicate alerts. I want to edit this script so that I only get a 'Buy' alert when the previous alert was a 'Sell' alert and visa versa. It seems like such a simple problem, but I have a hard time finding good sources to learn pinescript. So, any help would be appreciated. :)
One way to solve duplicate alters within the candle is by using "Once Per Bar Close" alert. But for alternative alerts (Buy - Sell) you have to code it with different logic.
I Suggest to use Version 3 (version shown above the study line) than version 1 and 2 and you can accomplish the result by using this logic:
buy_entry = 0.0
sell_entry = 0.0
buy_entry := rc[0] > rc[2] and sell_entry[1] == 0? 2.0 : sell_entry[1] > 0 ? 0.0 : buy_entry[1]
sell_entry := rc[0] < rc[2] and buy_entry[1] == 0 ? 2.0 : buy_entry[1] > 0 ? 0.0 : sell_entry[1]
alertcondition(crossover(buy_entry ,1) , title='BUY' )
alertcondition(crossover(sell_entry ,1), title='SELL')
You'll have to do it this way
if("Your buy condition here")
strategy.entry("Buy Alert",true,1)
if("Your sell condition here")
strategy.entry("Sell Alert",false,1)
This is a very basic form of it but it works.
You were getting duplicate alerts because the conditions were fulfulling more often. But with strategy.entry(), this won't happen
When the sell is triggered, as per paper trading, the quantity sold will be double (one to cut the long position and one to create a short position)
PS :You will have to add code to create alerts and enter this not in study() but strategy()
The simplest solution to this problem is to use the built-in crossover and crossunder functions.
They consider the entire series of in-this-case close values, only returning true the moment they cross rather than every single time a close is lower than the close two candles ago.
//#version=5
indicator(title='Renko Strat w/ Alerts', shorttitle='S_EURUSD_5_[MakisMooz]', overlay=true)
c = close
bool buy_entry = false
bool sell_entry = false
if ta.crossover(c[1], c[3])
buy_entry := true
alert('BUY')
if ta.crossunder(c[1], c[3])
sell_entry := true
alert('SELL')
plotchar(buy_entry, title='BUY', char='B', location=location.belowbar, color=color.green, offset=-1)
plotchar(sell_entry, title='SELL', char='S', location=location.abovebar, color=color.red, offset=-1)
It's important to note why I have changed to the indices to 1 and 3 with an offset of -1 in the plotchar function. This will give the exact same signals as 0 and 2 with no offset.
The difference is that you will only see the character print on the chart when the candle actually closes rather than watch it flicker on and off the chart as the close price of the incomplete candle moves.

Error in setting max features parameter in Isolation Forest algorithm using sklearn

I'm trying to train a dataset with 357 features using Isolation Forest sklearn implementation. I can successfully train and get results when the max features variable is set to 1.0 (the default value).
However when max features is set to 2, it gives the following error:
ValueError: Number of features of the model must match the input.
Model n_features is 2 and input n_features is 357
It also gives the same error when the feature count is 1 (int) and not 1.0 (float).
How I understood was that when the feature count is 2 (int), two features should be considered in creating each tree. Is this wrong? How can I change the max features parameter?
The code is as follows:
from sklearn.ensemble.iforest import IsolationForest
def isolation_forest_imp(dataset):
estimators = 10
samples = 100
features = 2
contamination = 0.1
bootstrap = False
random_state = None
verbosity = 0
estimator = IsolationForest(n_estimators=estimators, max_samples=samples, contamination=contamination,
max_features=features,
bootstrap=boostrap, random_state=random_state, verbose=verbosity)
model = estimator.fit(dataset)
In the documentation it states:
max_features : int or float, optional (default=1.0)
The number of features to draw from X to train each base estimator.
- If int, then draw `max_features` features.
- If float, then draw `max_features * X.shape[1]` features.
So, 2 should mean take two features and 1.0 should mean take all of the features, 0.5 take half and so on, from what I understand.
I think this could be a bug, since, taking a look in IsolationForest's fit:
# Isolation Forest inherits from BaseBagging
# and when _fit is called, BaseBagging takes care of the features correctly
super(IsolationForest, self)._fit(X, y, max_samples,
max_depth=max_depth,
sample_weight=sample_weight)
# however, when after _fit the decision_function is called using X - the whole sample - not taking into account the max_features
self.threshold_ = -sp.stats.scoreatpercentile(
-self.decision_function(X), 100. * (1. - self.contamination))
then:
# when the decision function _validate_X_predict is called, with X unmodified,
# it calls the base estimator's (dt) _validate_X_predict with the whole X
X = self.estimators_[0]._validate_X_predict(X, check_input=True)
...
# from tree.py:
def _validate_X_predict(self, X, check_input):
"""Validate X whenever one tries to predict, apply, predict_proba"""
if self.tree_ is None:
raise NotFittedError("Estimator not fitted, "
"call `fit` before exploiting the model.")
if check_input:
X = check_array(X, dtype=DTYPE, accept_sparse="csr")
if issparse(X) and (X.indices.dtype != np.intc or
X.indptr.dtype != np.intc):
raise ValueError("No support for np.int64 index based "
"sparse matrices")
# so, this check fails because X is the original X, not with the max_features applied
n_features = X.shape[1]
if self.n_features_ != n_features:
raise ValueError("Number of features of the model must "
"match the input. Model n_features is %s and "
"input n_features is %s "
% (self.n_features_, n_features))
return X
So, I am not sure on how you can handle this. Maybe figure out the percentage that leads to just the two features you need - even though I am not sure it'll work as expected.
Note: I am using scikit-learn v.0.18
Edit: as #Vivek Kumar commented this is an issue and upgrading to 0.20 should do the trick.

volemont/insights:chart.EquityCurve.R: a bug in graphing peaks of cumulative return?

I came cross a function of graphing cumulative return of a strategy and the peaks of the return in a great example of combining shiny and quantstrat, thanks to Simon Otziger. The source code is here. The code works fine most of time, but for some data it won't graph the peaks properly.
The code is simplified but the key logic is not changed. I ran the code with three set of data (cumPNL1, cumPNL2, cumPNL3) copied from three example strategies, in which the first data will cause the code to fail to graph peaks properly.
I ran the following codes with cumPNL1, cumPNL2, cumPNL3 separately. with both cumPNL2 and cumPNL3 the code can produce cumulative return line and peak points successfully. however, with cumPNL1 the code can only produce line, but peaks are not at the right positions.
I noticed that both peakIndex based on cumPNL2 and cumPNL3 have their first value being TRUE, so when I change the code by adding a line peakIndex[1] <- TRUE, cumPNL1 will work fine with the modified code.
Though now it works with modified code, I have no idea why it is behaving like this. Could anyone have a look? Thanks
cumPNL1 <- c(-193,-345,-406,-472,-562,-543,-450,-460,-544,-659,-581,-342,-384,276,-858,-257.99)
cumPNL2 <- c(35.64,4.95,-2.97,-6.93,11.88,-19.8,-26.73,-39.6,-49.5,-50.49,-51.48,-48.51,-50.49,-55.44,143.55,770.22,745.47,691.02,847.44,1141.47,1007.82,1392.93,1855.26,1863.18,2536.38,2778.93,2811.6,2859.12,2417.58)
cumPNL3 <- c(35.64,4.95,-2.97,-6.93,11.88,-19.8,-26.73,-39.6,-49.5,-50.49,-51.48,-48.51,-50.49,-55.44,143.55,770.22,745.47,691.02,847.44,1141.47,1007.82,1392.93,1855.26,1863.18,2536.38,2778.93,2811.6,2859.12,2417.58)
peakIndex <- c(cumPNL3[1] > 0, diff(cummax(cumPNL3)) > 0)
# peakIndex[1] <- TRUE
dev.new()
plot(cumPNL3, type='n', xlab="index of trades", ylab="returns in cash", main="cumulative returns and peaks")
grid()
lines(cumPNL3)
points(cbind(1 : length(cumPNL3), cumPNL3)[peakIndex, ],
pch=19, col='green', cex=0.6)
legend(
x='bottomright', inset=0.1,
legend=c('Net Profit','Peaks'),
lty=c(1, NA), pch=c(NA, 19),
col=c('black','green')
)
cumPNL1 has a single peak and R reduces the dimension from a numerical matrix to a numerical vector of length 2. The points function plots the two numerical vector values on the y-axis using the x-axis index 1 and 2:
peakIndex1 <- c(cumPNL1[1] > 0, diff(cummax(cumPNL1)) > 0)
peakIndex3 <- c(cumPNL3[1] > 0, diff(cummax(cumPNL3)) > 0)
str(cbind(1 : length(cumPNL1), cumPNL1)[peakIndex1,])
str(cbind(1 : length(cumPNL3), cumPNL3)[peakIndex3,])
Output:
> str(cbind(1 : length(cumPNL1), cumPNL1)[peakIndex1,])
num [1:12, 1:2] 1 15 16 19 20 22 23 24 25 26 ...
- attr(*, "dimnames")=List of 2
..$ : NULL
..$ : chr [1:2] "" "cumPNL1"
> str(cbind(1 : length(cumPNL3), cumPNL3)[peakIndex3,])
Named num [1:2] 14 276
- attr(*, "names")= chr [1:2] "" "cumPNL3"
Usually setting plot = FALSE preserves the object, e.g., str(cbind(1 : length(cumPNL3), cumPNL3)[peakIndex3, drop = FALSE]), which somehow does not work in this case. However, changing the points line to the following fixes the problem:
points(seq_along(cumPNL3)[peakIndex], cumPNL3[peakIndex], pch = 19,
col = 'green', cex = 0.6)
Thanks for reporting the issue. I will push the fix to GitHub tomorrow.

lldb - How to display float with decimals using "type format add"

I have a variable of type float. Xcode displays it using scientific notation (i.e. 3.37626e+07). I'm trying to get it to display using dot notation (i.e. 33762616.00).
I've tried every format provided by lldb, but none displays the float using decimals. I read other posts and watched the WWDC2012 session 415 (as suggested here), but I must be too close the forest to see the trees. Any help would be greatly appreciated!
Try adding a custom data formatter in your ~/.lldbinit file for type float. e.g.
Process 13204 stopped
* thread #1: tid = 0xb6f8d, 0x0000000100000f33 a.out`main + 35 at a.c:5, stop reason = step over
#0: 0x0000000100000f33 a.out`main + 35 at a.c:5
2 int main ()
3 {
4 float f = 33762616.0;
-> 5 printf ("%f\n", f);
6 }
(lldb) p f
(float) $0 = 3.37626e+07
(lldb) type summ add -v -o "return '%f' % valobj.GetData().GetFloat(lldb.SBError(), 0)" float
(lldb) p f
(float) $1 = 33762616.000000
(lldb)
The default set of formatters provided by lldb can't do this, but dropping into Python allows you a lot of flexibility.

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