Calculating Variable Cash-flow IRR in Python (pandas)

Question

I have a DataFrame of unpredictable cashflows and unpredictable period lengths, and I need to generate a backward-looking IRR.

Doing it in Excel is pretty straightforward using the solver, wondering if there's a good way to pull it off in Python. (I think I could leverage openpyxl to get solver to work in excel from python, but that feels unnecessarily cumbersome).

The problem is pretty straightforward:

NPV of Cash Flow = ((cash_flow)/(1+IRR)^years_ago)

GOAL: Find IRR where SUM(NPV) = 0

My dataframe looks something like this:

cash_flow    |years_ago
-----------------------
-3.60837e+06 |4.09167    
31462        |4.09167    
1.05956e+06  |3.63333    
-1.32718e+06 |3.28056    
-4.46554e+06 |3.03889    

It seems as though other IRR calculators (such as numpy.irr) assume strict period cutoffs (every 3 months, 1 year, etc), which won't work. The other option seems to be the iterative route, where I continually guess, check, and iterate, but that feels like the wrong way to tackle this. Ideally, I'm looking for something that would do this:

irr = calc_irr((cash_flow1,years_ago1),(cash_flow2,years_ago2),etc)

EDIT: Here is the code I'm running the problem from. I have a list of transactions, and I've chosen to create temporary tables by id.

for id in df_tran.id.unique():
   temp_df = df_tran[df_tran.id == id] 

   cash_flow = temp_df.cash_flows.values
   years = temp_df.years.values

   print(id, cash_flow)
   print(years)
#irr_calc = irr(cfs=cash_flow, yrs=years,x0=0.100000)
#print(sid, irr_calc)

where df_tran (which temp_df is based on) looks like:

    cash_flow       |years     |id
0   -3.60837e+06     4.09167    978237
1   31462            4.09167    978237
4   1.05956e+06      3.63333    978237
6   -1.32718e+06     3.28056    978237
8   -4.46554e+06     3.03889    978237
10  -3.16163e+06     2.81944    978237
12  -5.07288e+06     2.58889    978237
14  268833           2.46667    978237
17  -4.74703e+06     1.79167    978237
20  -964987          1.40556    978237
22  -142920          1.12222    978237
24  163894           0.947222   978237
26  -2.2064e+06      0.655556   978237
27  1.23804e+06      0.566667   978237
29  180655           0.430556   978237
30  -85297           0.336111   978237
34  -2.3529e+07      0.758333   1329483
36  21935            0.636111   1329483
38  -3.55067e+06     0.366667   1329483
41  -4e+06           4.14167    1365051

temp_df looks identical to df_tran, except it only holds transactions for a single id.

Answer 1

You can use scipy.optimize.fsolve:

Return the roots of the (non-linear) equations defined by func(x) = 0 given a starting estimate.

First define the function that will be the func parameter to fsolve. This is NPV as a result of your IRR, cash flows, and years. (Vectorize with NumPy.)

import numpy as np
def npv(irr, cfs, yrs):  
    return np.sum(cfs / (1. + irr) ** yrs)

An example:

cash_flow = np.array([-2., .5, .75, 1.35])
years = np.arange(4)

# A guess
print(npv(irr=0.10, cfs=cash_flow, yrs=years))
0.0886551465064

Now to use fsolve:

from scipy.optimize import fsolve
def irr(cfs, yrs, x0):
    return np.asscalar(fsolve(npv, x0=x0, args=(cfs, yrs)))

Your IRR is:

print(irr(cfs=cash_flow, yrs=years, x0=0.10))
0.12129650313214262

And you can confirm that this gets you to a 0 NPV:

res = irr(cfs=cash_flow, yrs=years, x0=0.10)
print(np.allclose(npv(res, cash_flow, years), 0.))
True

All code together:

import numpy as np
from scipy.optimize import fsolve

def npv(irr, cfs, yrs):  
    return np.sum(cfs / (1. + irr) ** yrs)

def irr(cfs, yrs, x0, **kwargs):
    return np.asscalar(fsolve(npv, x0=x0, args=(cfs, yrs), **kwargs))

To make this compatible with your pandas example, just use

cash_flow = df.cash_flow.values
years = df.years_ago.values

Update: the values in your question seem a bit nonsensical (your IRR is going to be some astronomical number if it even exists) but here is how you'd run:

cash_flow = np.array([-3.60837e+06, 31462, 1.05956e+06, -1.32718e+06, -4.46554e+06])    
years_ago = np.array([4.09167, 4.09167, 3.63333, 3.28056, 3.03889])

print(irr(cash_flow, years_ago, x0=0.10, maxfev=10000))
1.3977721900669127e+82

Second update: there are a couple minor typos in your code, and your actual flows of $ and timing work out to nonsensical IRRs, but here's what you're looking to do, below. For instance, notice you have one id with one single negative transaction, a negatively infinite IRR.

for i, df in df_tran.groupby('id'):
   cash_flow = df.cash_flow.values
   years = df.years.values
   print('id:', i, 'irr:', irr(cash_flow, years, x0=0.))

id: 978237 irr: 347.8254979851405
id: 1329483 irr: 3.2921314448062817e+114
id: 1365051 irr: 1.0444951674872467e+25