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()
    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
            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


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()
  • $\begingroup$ 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? $\endgroup$
    – DickyBrown
    Jun 22 '20 at 21:58
  • $\begingroup$ @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. $\endgroup$ Jun 22 '20 at 21:59
  • $\begingroup$ @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% $\endgroup$
    – amdopt
    Jun 22 '20 at 22:06
  • $\begingroup$ @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. $\endgroup$ 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. $$MDD=\frac{Trough\ Value−Peak\ Value}{Peak\ Value}$$


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