# This code is placed in the public domain

def get_list(heading, prompt):
        """ get_list(heading, prompt) -> list

This function prompts for a list of things.  The heading is printed on
a line by itself, and the prompt must have a %d substitution for the
number of the item within the list.
"""
        print heading
        print
        print "(enter a blank line to end the list)"
        ret = []
        i = 1
        while 1:
                line = raw_input(prompt % i)
                if not line:
                        break
                ret.append(line)
                i=i+1
        print
        return ret

def get_number(prompt):
        """ get_number(prompt) -> float

This function prompts for a number.  If the user enters bad input, such as
"cat" or "3l", it will prompt again.
"""
        res = None
        while res is None:
                try:
                        res = float(raw_input(prompt))
                except ValueError: pass
        return res

# First, ask the user to enter the lists
options = ["Python", "Perl", "Ruby", "Tcl", "JavaScript", "Visual Basic"]
criteria = ["ease of learning", "ease of use", "speed of program execution", "quality of available tools", "popularity", "power & expressiveness", "cross platform?", "cost"]

# Next, get the user to rank his criteria.  I use a system where higher
# is better, so that an undesirable characteristic can be given a negative
# weight.
#
# {} is a dictionary, it can be indexed by (nearly) any expression,
# and we will index it with the names of the criteria.
# (For a more traditional program, we could use a list and index by the
# number of the criterion)

rankings = {}
print
print "Enter relative importance of criteria (higher is more important)"
print
for c in criteria:
                rankings[c] = get_number("Criterion %s: " % c)

# Next, get the user to score each option on all the criteria.
# Here, we index the dictionary on the pair (option, criterion). 
# This is similar to a two-dimensional array in other languages

score = {("Python", "ease of learning"):100, ("Python", "ease of use"):100, ("Python", "speed of program execution"):10, ("Python", "quality of available tools"):50, ("Python", "popularity"):50, ("Python", "power & expressiveness"):100, ("Python", "cross platform?"):100, ("Python", "cost"):100,
("Perl", "ease of learning"):50, ("Perl", "ease of use"):90, ("Perl", "speed of program execution"):30, ("Perl", "quality of available tools"):50, ("Perl", "popularity"):75, ("Perl", "power & expressiveness"):100, ("Perl", "cross platform?"):100, ("Perl", "cost"):100,
("Ruby", "ease of learning"):50, ("Ruby", "ease of use"):100, ("Ruby", "speed of program execution"):10, ("Ruby", "quality of available tools"):20, ("Ruby", "popularity"):10, ("Ruby", "power & expressiveness"):100, ("Ruby", "cross platform?"):80, ("Ruby", "cost"):100, 
("Tcl", "ease of learning"):100, ("Tcl", "ease of use"):100, ("Tcl", "speed of program execution"):5, ("Tcl", "quality of available tools"):50, ("Tcl", "popularity"):40, ("Tcl", "power & expressiveness"):10, ("Tcl", "cross platform?"):100, ("Tcl", "cost"):100,
("JavaScript", "ease of learning"):70, ("JavaScript", "ease of use"):75, ("JavaScript", "speed of program execution"):10, ("JavaScript", "quality of available tools"):50, ("JavaScript", "popularity"):100, ("JavaScript", "power & expressiveness"):40, ("JavaScript", "cross platform?"):50, ("JavaScript", "cost"):100, 
("Visual Basic", "ease of learning"):50, ("Visual Basic", "ease of use"):100, ("Visual Basic", "speed of program execution"):20, ("Visual Basic", "quality of available tools"):100, ("Visual Basic", "popularity"):100, ("Visual Basic", "power & expressiveness"):50, ("Visual Basic", "cross platform?"):1, ("Visual Basic", "cost"):1}
# Calculate the resulting score for each option.  This equation
# is different from Rod Stephen's original program, because I
# make more important criteria have higher rankings, and even let
# bad criteria have negative rankings.

# The "result" dictionary is indexed with the names of the options.
result = {}
for o in options:
        value = 0
        for c in criteria:
                print o, c, rankings[c], score[o, c]
                value = value + rankings[c] * score[o, c]
        result[o] = value

# Now, I want to take the dictionary result, and turn it into a ranked list

results = result.items()        # A list of tuples (key, value)
results.sort(lambda x, y: -cmp(x[1], y[1]))
                                # Sort the list using the reverse of the
                                # "value" of the entry, so that higher
                                # values come first

print
print "Results, in order from highest to lowest score"
print
print "%5s %s" % ("Score", "Option")

# Take the pairs out of results in order, and print them out
for option, result in results:
        print "%5s %s" % (result, option)