# Global Maximum Drawdown and Maximum Drawdown Duration Implementation in Python

Following along with E.P. Chan's book, I'm attempting to calculate the maximum drawdown and the longest drawdown duration from cumulative portfolio returns. He codes it in MATLAB, but I wanted to try my hand at the same code in Python.

import pandas as pd

def drawdownCalculator(data):
highwatermark = data.copy()
highwatermark[:] = 0
drawdown = data.copy()
drawdown[:] = 0
drawdownduration = data.copy()
drawdownduration[:]=0
t = 1
while t <= len(data):
highwatermark[t] = max(highwatermark[t-1], data[t])
drawdown[t] = (1 + highwatermark[t])/(1 + data[t]) - 1
if drawdown[t] == 0:
drawdownduration[t] = 0
else:
drawdownduration[t] = drawdownduration[t-1] + 1
t += 1
return drawdown.max(), drawdownduration.max()
max_drawdown, max_drawdown_time = drawdownCalculator(cumulative_returns) #cumulative_returns is a Pandas series


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))
File "pandas/_libs/index.pyx", line 80, in pandas._libs.index.IndexEngine.get_value
File "pandas/_libs/index.pyx", line 88, in pandas._libs.index.IndexEngine.get_value
File "pandas/_libs/index.pyx", line 131, in pandas._libs.index.IndexEngine.get_loc
File "pandas/_libs/hashtable_class_helper.pxi", line 992, in pandas._libs.hashtable.Int64HashTable.get_item
File "pandas/_libs/hashtable_class_helper.pxi", line 998, in pandas._libs.hashtable.Int64HashTable.get_item
KeyError: 0


Thank you in advance

## 2 Answers

I'm guessing your Series is indexed by a timestamp, which would explain why accessing by an integer doesn't work. But I can't tell for sure since you haven't shown us any data.

The good news is that I don't need that anyway. Here is a more idiomatic way to compute what you want:

highwatermarks = cumulative_returns.cummax()

drawdowns = (1 + highwatermarks)/(1 + cumulative_returns) - 1

max_drawdown = max(drawdowns)


There is no simple way to compute duration with array notation. Fortunately, this question shows how to use an accumulator for exactly your scenario:

from itertools import accumulate

drawdown_times = (drawdowns > 0).astype(np.int64)

max_drawdown_time = max(accumulate(drawdown_times, lambda x,y: (x+y)*y))


Alternatively, you can group the consecutive durations together. I don't recommend this approach, but I'll include it for posterity:

max_drawdown_time = drawdown_times.groupby((drawdown_times != drawdown_times.shift()).cumsum()).cumsum().max()

• Thank you for your answer! By the way, I was looking through the proposed edits on this answer, are the formulas I was using incorrect? Jun 22 '20 at 21:58
• @DickyBrown I doubt it matters how you define "drawdown" as long as you are consistent. I've used pnl.cummax() - pnl; ie., the positive difference in P&L. Jun 22 '20 at 21:59
• @chrisaycock Your code does not return a max drawdown. For example, it's returns ~122% as the max draw for SPY in 2008...it should be ~55% Jun 22 '20 at 22:06
• @amdopt You are free to add your own answer or comment to make your case. You are not entitled to fundamentally alter someone else's post. Jun 22 '20 at 22:09

In addition to chrisaycock's answer, we could also normalize the maximum drawdown with the high wartermarks instead of the current cumulative returns.

highwatermarks = cumulative_returns.cummax()
drawdowns = 1 - (1 + cumulative_returns) / (1 + highwatermarks)
max_drawdown = max(drawdowns)


Here we observe the drawdown definition defined here. $$MDD=\frac{Trough\ Value−Peak\ Value}{Peak\ Value}$$