By using this site, you agree to our updated Privacy Policy and our Terms of Use. Manage your Cookies Settings.
443,908 Members | 1,860 Online
Bytes IT Community
+ Ask a Question
Need help? Post your question and get tips & solutions from a community of 443,908 IT Pros & Developers. It's quick & easy.

Dispatching functions from a dictionary

P: n/a
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?

Thanks in advance for your help.

Sincerely
Thomas Philips

Mar 30 '08 #1
Share this Question
Share on Google+
3 Replies


P: n/a
tk****@gmail.com 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().
Mar 30 '08 #2

P: n/a
On Mar 30, 5:06 pm, Paul Rubin <http://phr...@NOSPAM.invalidwrote:
tkp...@gmail.com 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
Mar 30 '08 #3

P: n/a
Paul, George,

Thanks a mill - the help is greatly appreciated.

Thomas Philips

Mar 31 '08 #4

This discussion thread is closed

Replies have been disabled for this discussion.