# Dispatching functions from a dictionary

Discussion in 'Python' started by tkpmep@gmail.com, Mar 30, 2008.

1. ### Guest

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.betavariate(1,1), '2': random.expovariate(1),
'3': random.gammavariate(1,1), '4': random.gauss(0,1),
'5': random.lognormvariate(1,1), '6':
random.paretovariate(1),
'7': random.uniform( -1,1), '8':
random.weibullvariate(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?

Sincerely

Thomas Philips

, Mar 30, 2008

2. ### Paul RubinGuest

writes:
> RVDict= {'1': random.betavariate(1,1), '2': random.expovariate(1), ...}

This actually calls the functions random.betavariate, 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.betavariate(1,1),
'2': lambda: random.expovariate(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.expovariate(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().

Paul Rubin, Mar 30, 2008

3. ### George SakkisGuest

On Mar 30, 5:06 pm, Paul Rubin <http://> wrote:
> writes:
> > RVDict= {'1': random.betavariate(1,1), '2': random.expovariate(1), ...}

>
> This actually calls the functions random.betavariate, 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.betavariate(1,1),
> '2': lambda: random.expovariate(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

George Sakkis, Mar 30, 2008
4. ### Guest

Paul, George,

Thanks a mill - the help is greatly appreciated.

Thomas Philips

, Mar 31, 2008