def calc_MDD(networth):ĭf = pd.Series(networth, name="nw").to_frame() My vectorized implementation is also based on Investopedia. Also, in my case, I was supposed to take the MDD of each strategy alone and thus wasn't required to apply the cumprod. I recently had a similar issue, but instead of a global MDD, I was required to find the MDD for the interval after each peak. I modified his code into the following function: def max_dd(returns): Same test using modified code 10: 0.000 seconds The same test for the looped solution is below: 10: 0.153 secondsĪlexander's answer provides superior results. The time it took is below: 10: 0.032 seconds Return z.min(), z.argmin(), z.argmin()įor the vectorized solution I ran 10 iterations over the time series of lengths. # my choice is to return the periods and the actual max # let's make sure we only calculate a return when start X.columns.name, x.index.name = 'start', 'end' # I copy r.T to ensure r's index is not the same R = pd.DataFrame(returns).add(1).cumprod() # multiplication and end up with an nxn matrix # matrix such that I can perform an nx1 by 1xn matrix # make into a DataFrame so that it is a 2-dimensional The fourth trick is to ensure that I'm constraining my denominator to represent periods prior to those being represented by the numerator.īelow is my vectorized function. Just subtract 1 and I've actually got returns. The resulting product contains every combination of ri_j / ri_k. The third trick is to take the matrix product of r * (1 / r).Transpose. If r is my series of return indices then 1 / r is my series of inverses. The second trick is to produce a second series of inverses of return indices. Given a series of return indices, I can calculate the return over any sub-period with the return index at the beginning ri_0 and at the end ri_1. The first trick is to convert a time series of returns into a series of return indices. Given a time series of returns, we need to evaluate the aggregate return for every combination of starting point to ending point. cummax() built ins to calculate max drawdown up to any given point: df = pd.DataFrame(data=, expanding() window but that's obviously not necessary with the. I modified Alexander's answer into the following function: def max_dd(returns): I'm familiar with the common perception that a vectorized solution would be better. If (max_so_far is None) or (current < max_so_far): """returns is assumed to be a pandas series"""Ĭurrent = r.ix / r.ix - 1 Recently, I became impatient with the time to calculate max drawdown using my looped approach. I thought I had it figured out, but I'm getting the following error: return self._engine.get_value(s, k, tz=getattr(series.dtype, "tz", None))įile "pandas/_libs/index.pyx", line 80, in pandas._圎ngine.get_valueįile "pandas/_libs/index.pyx", line 88, in pandas._圎ngine.get_valueįile "pandas/_libs/index.pyx", line 131, in pandas._圎ngine.get_locįile "pandas/_libs/hashtable_class_helper.pxi", line 992, in pandas._itemįile "pandas/_libs/hashtable_class_helper.pxi", line 998, in pandas._ Drawdown is a common risk metric used in quantitative finance to assess the largest negative return that has been experienced. Max_drawdown, max_drawdown_time = drawdownCalculator(cumulative_returns) #cumulative_returns is a Pandas series Return drawdown.max(), drawdownduration.max() Highwatermark = max(highwatermark, data)ĭrawdown = (1 + highwatermark)/(1 + data) - 1ĭrawdownduration = drawdownduration + 1 He codes it in MATLAB, but I wanted to try my hand at the same code in Python. Chan's book, I'm attempting to calculate the maximum drawdown and the longest drawdown duration from cumulative portfolio returns.
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