Dispatching functions from a dictionary

[tkpmep@gmail.com]

To keep a simulation tidy, I created a dispatcher that generates
random variables drawn from various distributions as follows:

import random

RVType = 1 #Type of random variable - pulled from RVDict

RVDict= {'1': random.betavari ate(1,1), '2': random.expovari ate(1),
'3': random.gammavar iate(1,1), '4': random.gauss(0, 1),
'5': random.lognormv ariate(1,1), '6':
random.paretova riate(1),
'7': random.uniform( -1,1), '8':
random.weibullv ariate(1,2) }

x = []
y=[]

rv = RVDict[str(RVType)]
for i in range(N):
x.append(rv)
y.append(rv)


Oddly, x and y get filled with a single value repeated N times. I
expected to get a different random number appear each time I called
rv ,but this does not happen. Instead, my first call to rv generates a
random number from the appropriate distribution, while subsequent
calls simply repeat the random number generated in the first call.
Where am I going wrong?

Thanks in advance for your help.

Sincerely


Thomas Philips

[Paul Rubin]

Re: Dispatching functions from a dictionary

tkpmep@gmail.co m writes:

RVDict= {'1': random.betavari ate(1,1), '2': random.expovari ate(1), ...}

This actually calls the functions random.betavari ate, etc. when
initializing RVDict. If you print out the contents of RVDict you'll see
that each value in it is just a floating point number, not a callable.

You want something like:

RVDict = {'1': lambda: random.betavari ate(1,1),
'2': lambda: random.expovari ate(1), etc.

The "lambda" keyword creates a function that when called evaluates the
expression that you gave it. For example, lambda x: x*x is a function
that squares its argument, so saying

y = (lambda x: x*x) (3)

is similar to saying:

def square(x): return x*x
y = square(3)

Both of them set y to 9. In the case of lambda: random.expovari ate(1)
you have made a function with no args, so you'd call it like this:

rvfunc = RVDict[str(RVType)]
for i in range(N):
x.append(rvfunc ())
y.append(rvfunc ())

rvfunc (the retrieved dictionary item) is now a callable function
instead of just a number. It takes no args, so you call it by saying
rvfunc().

[George Sakkis]

Re: Dispatching functions from a dictionary

On Mar 30, 5:06 pm, Paul Rubin <http://phr...@NOSPAM.i nvalidwrote:

tkp...@gmail.co m writes:

RVDict= {'1': random.betavari ate(1,1), '2': random.expovari ate(1), ...}

>
This actually calls the functions random.betavari ate, etc. when
initializing RVDict. If you print out the contents of RVDict you'll see
that each value in it is just a floating point number, not a callable.
>
You want something like:
>
RVDict = {'1': lambda: random.betavari ate(1,1),
'2': lambda: random.expovari ate(1), etc.

In Python 2.5, you can also write this as:

from functools import partial

RVDict = {'1': partial(random. betavariate,1,1 ),
'2': partial(random. expovariate,1),
etc.


George

[tkpmep@gmail.com]

Re: Dispatching functions from a dictionary

Paul, George,

Thanks a mill - the help is greatly appreciated.

Thomas Philips