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| author | Alan McIntyre <alan.mcintyre@local> | 2008-07-05 14:26:16 +0000 |
|---|---|---|
| committer | Alan McIntyre <alan.mcintyre@local> | 2008-07-05 14:26:16 +0000 |
| commit | 36e02207c1a82fe669531dd24ec799eca2989c80 (patch) | |
| tree | 104f800d6800c4a01a0aecac323a8a70517aa94b /numpy/lib/financial.py | |
| parent | f07e385b69ee59ef6abe05f164138dc6a7279291 (diff) | |
| download | numpy-36e02207c1a82fe669531dd24ec799eca2989c80.tar.gz | |
Use the implicit "import numpy as np" made available to all doctests instead
of explicit imports or dependency on the local scope where the doctest is
defined..
Diffstat (limited to 'numpy/lib/financial.py')
| -rw-r--r-- | numpy/lib/financial.py | 8 |
1 files changed, 4 insertions, 4 deletions
diff --git a/numpy/lib/financial.py b/numpy/lib/financial.py index a3552ebc0..9c5d2753a 100644 --- a/numpy/lib/financial.py +++ b/numpy/lib/financial.py @@ -69,7 +69,7 @@ What is the future value after 10 years of saving $100 now, with an additional monthly savings of $100. Assume the interest rate is 5% (annually) compounded monthly? ->>> fv(0.05/12, 10*12, -100, -100) +>>> np.fv(0.05/12, 10*12, -100, -100) 15692.928894335748 By convention, the negative sign represents cash flow out (i.e. money not @@ -94,7 +94,7 @@ Examples What would the monthly payment need to be to pay off a $200,000 loan in 15 years at an annual interest rate of 7.5%? ->>> pmt(0.075/12, 12*15, 200000) +>>> np.pmt(0.075/12, 12*15, 200000) -1854.0247200054619 In order to pay-off (i.e. have a future-value of 0) the $200,000 obtained @@ -122,7 +122,7 @@ Examples If you only had $150 to spend as payment, how long would it take to pay-off a loan of $8,000 at 7% annual interest? ->>> nper(0.07/12, -150, 8000) +>>> np.nper(0.07/12, -150, 8000) 64.073348770661852 So, over 64 months would be required to pay off the loan. @@ -130,7 +130,7 @@ So, over 64 months would be required to pay off the loan. The same analysis could be done with several different interest rates and/or payments and/or total amounts to produce an entire table. ->>> nper(*(ogrid[0.06/12:0.071/12:0.01/12, -200:-99:100, 6000:7001:1000])) +>>> np.nper(*(np.ogrid[0.06/12:0.071/12:0.01/12, -200:-99:100, 6000:7001:1000])) array([[[ 32.58497782, 38.57048452], [ 71.51317802, 86.37179563]], |