Implementation of Maximum Drawdown in python working directly with returns

Question

I have a strategy on a stock (such as Buy and Hold) on which I have to calculate the maximum drawdown. The problem is that I'm working on returns expressed in percentages, so I do not have the time series of prices but the one of returns obtained at each step. So I wrote this code:

def MDD(returns):
 rend_cum=returns.cumsum()
 rend_max=pd.Series(rend_cum).cummax()
 drawdown=rend_cum-rend_max
 MDD=max(abs(drawdown))

 return(MDD)

Is it correct?

Answer 1

You are missing a few things. The function below assumes that returns is either a pandas series or a column of a pandas dataframe. Try this:

def MDD(returns):
 
    cum_rets = (1 + returns).cumprod() - 1
    nav = ((1 + cum_rets) * 100).fillna(100)
    hwm = nav.cummax()
    dd = nav / hwm - 1

    return min(dd)

Answer 2

The empyrical package has an efficient function for max drawdown. Here's an example:

import empyrical as ep

returns = np.array([-0.02089651,
                    -0.023142165,
                    0.016320209,
                    0.009323824,
                    -0.048883758,
                    -0.003912041,
                    0.005281875,
                    0.029637668,
                    -0.012299053,
                    0.040005685])

max_dd = ep.max_drawdown(returns)

The function that @amdopt provided in an answer to your question will sometimes produce a different answer than the empyrical function because empyrical uses a starting value of 100. This modified version would provide the equivalent results:

def MDD(returns):
    cum_rets = (1 + returns).cumprod() - 1
    nav = ((1 + cum_rets) * 100).fillna(100)
    nav = pd.Series([100]).append(nav) # start at 100
    hwm = nav.cummax()
    dd = nav / hwm - 1
    return min(dd)