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.
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
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