I = np.argmax(np.maximum.accumulate(xs) - xs) # end of the period Related code (from behzad.nouri) below: n = 1000 Recent Corona Virus crisis, drop 33.9%, 1,1148.75 pointsīy applying this method to period after 2000, you'll see Corona Virus Crisis rather than 2007-08 Financial Crisis.
For example, if you would apply this to time series that is ascending over the long run (for example stock market index S&P 500), the most recent drop in value (higher nominal value drops) will be prioritized over the older decrease in value as long as the drop in nominal value/points is higher.
What you end up having is the maximum drop in the nominal value rather than a relative drop in value (percentage drop). S = S0*np.exp(X) # geometric brownian motion #īehzad.nouri solution is very clean, but it's not a maximum drawdow (couldn't comment as I just opened my account and I don't have enough reputation atm). W = np.cumsum(W)*np.sqrt(dt) # standard brownian motion # # create random walk which I want to calculate maximum drawdown for:
#Excel maximum drawdown how to#
If anyone knows how to identify the places where the drawdown begins and ends, I'd really appreciate it! import pandas as pd So far I've got code to generate a random time series, and I've got code to calculate the max drawdown. I want to mark the beginning and end of the drawdown on a plot of the timeseries like this: Given a time series, I want to calculate the maximum drawdown, and I also want to locate the beginning and end points of the maximum drawdown so I can calculate the duration.
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