Calculate max draw down with a vectorized solution in python

Question

Maximum Drawdown is a common risk metric used in quantitative finance to assess the largest negative return that has been experienced.

Recently, I became impatient with the time to calculate max drawdown using my looped approach.

def max_dd_loop(returns):
    """returns is assumed to be a pandas series"""
    max_so_far = None
    start, end = None, None
    r = returns.add(1).cumprod()
    for r_start in r.index:
        for r_end in r.index:
            if r_start < r_end:
                current = r.ix[r_end] / r.ix[r_start] - 1
                if (max_so_far is None) or (current < max_so_far):
                    max_so_far = current
                    start, end = r_start, r_end
    return max_so_far, start, end

I'm familiar with the common perception that a vectorized solution would be better.

The questions are:

• can I vectorize this problem?
• What does this solution look like?
• How beneficial is it?

---

Edit

I modified Alexander's answer into the following function:

def max_dd(returns):
    """Assumes returns is a pandas Series"""
    r = returns.add(1).cumprod()
    dd = r.div(r.cummax()).sub(1)
    mdd = dd.min()
    end = dd.argmin()
    start = r.loc[:end].argmax()
    return mdd, start, end

Answer 1

df_returns is assumed to be a dataframe of returns, where each column is a seperate strategy/manager/security, and each row is a new date (e.g. monthly or daily).

cum_returns = (1 + df_returns).cumprod()
drawdown =  1 - cum_returns.div(cum_returns.cummax())

Answer 2

I had first suggested using .expanding() window but that's obviously not necessary with the .cumprod() and .cummax() built ins to calculate max drawdown up to any given point:

df = pd.DataFrame(data={'returns': np.random.normal(0.001, 0.05, 1000)}, index=pd.date_range(start=date(2016,1,1), periods=1000, freq='D'))

df = pd.DataFrame(data={'returns': np.random.normal(0.001, 0.05, 1000)},
                  index=pd.date_range(start=date(2016, 1, 1), periods=1000, freq='D'))
df['cumulative_return'] = df.returns.add(1).cumprod().subtract(1)
df['max_drawdown'] = df.cumulative_return.add(1).div(df.cumulative_return.cummax().add(1)).subtract(1)

            returns  cumulative_return  max_drawdown
2016-01-01 -0.014522          -0.014522      0.000000
2016-01-02 -0.022769          -0.036960     -0.022769
2016-01-03  0.026735          -0.011214      0.000000
2016-01-04  0.054129           0.042308      0.000000
2016-01-05 -0.017562           0.024004     -0.017562
2016-01-06  0.055254           0.080584      0.000000
2016-01-07  0.023135           0.105583      0.000000
2016-01-08 -0.072624           0.025291     -0.072624
2016-01-09 -0.055799          -0.031919     -0.124371
2016-01-10  0.129059           0.093020     -0.011363
2016-01-11  0.056123           0.154364      0.000000
2016-01-12  0.028213           0.186932      0.000000
2016-01-13  0.026914           0.218878      0.000000
2016-01-14 -0.009160           0.207713     -0.009160
2016-01-15 -0.017245           0.186886     -0.026247
2016-01-16  0.003357           0.190869     -0.022979
2016-01-17 -0.009284           0.179813     -0.032050
2016-01-18 -0.027361           0.147533     -0.058533
2016-01-19 -0.058118           0.080841     -0.113250
2016-01-20 -0.049893           0.026914     -0.157492
2016-01-21 -0.013382           0.013173     -0.168766
2016-01-22 -0.020350          -0.007445     -0.185681
2016-01-23 -0.085842          -0.092648     -0.255584
2016-01-24  0.022406          -0.072318     -0.238905
2016-01-25  0.044079          -0.031426     -0.205356
2016-01-26  0.045782           0.012917     -0.168976
2016-01-27 -0.018443          -0.005764     -0.184302
2016-01-28  0.021461           0.015573     -0.166797
2016-01-29 -0.062436          -0.047836     -0.218819
2016-01-30 -0.013274          -0.060475     -0.229189
...              ...                ...           ...
2018-08-28  0.002124           0.559122     -0.478738
2018-08-29 -0.080303           0.433921     -0.520597
2018-08-30 -0.009798           0.419871     -0.525294
2018-08-31 -0.050365           0.348359     -0.549203
2018-09-01  0.080299           0.456631     -0.513004
2018-09-02  0.013601           0.476443     -0.506381
2018-09-03 -0.009678           0.462153     -0.511158
2018-09-04 -0.026805           0.422960     -0.524262
2018-09-05  0.040832           0.481062     -0.504836
2018-09-06 -0.035492           0.428496     -0.522411
2018-09-07 -0.011206           0.412489     -0.527762
2018-09-08  0.069765           0.511031     -0.494817
2018-09-09  0.049546           0.585896     -0.469787
2018-09-10 -0.060201           0.490423     -0.501707
2018-09-11 -0.018913           0.462235     -0.511131
2018-09-12 -0.094803           0.323611     -0.557477
2018-09-13  0.025736           0.357675     -0.546088
2018-09-14 -0.049468           0.290514     -0.568542
2018-09-15  0.018146           0.313932     -0.560713
2018-09-16 -0.034118           0.269104     -0.575700
2018-09-17  0.012191           0.284576     -0.570527
2018-09-18 -0.014888           0.265451     -0.576921
2018-09-19  0.041180           0.317562     -0.559499
2018-09-20  0.001988           0.320182     -0.558623
2018-09-21 -0.092268           0.198372     -0.599348
2018-09-22 -0.015386           0.179933     -0.605513
2018-09-23 -0.021231           0.154883     -0.613888
2018-09-24 -0.023536           0.127701     -0.622976
2018-09-25  0.030160           0.161712     -0.611605
2018-09-26  0.025528           0.191368     -0.601690