Calculating Variable Cash-flow IRR in Python (pandas)

Viewed 10093

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.

1 Answers
Related