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 = ["Yes", "No"]
criteria = get_list("Enter your criteria ...", "Criterion %d: ")

# 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 = {}
print
print "Enter score for each option on each criterion"
print
for o in options:
        print
        print "Score for a %s decision "% o
        print
        for c in criteria:
                score[o, c] = get_number("Criterion %s: " % c)

# 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:
                
                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 "The results are"
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)


if results[0] > results[1]:
        print "You should decide Yes!!!"

else:
        print "You should decide No!!!"