print "A General Decision Analysis Program."
print
print "Have You Ever Had to Make Up Your Mind?"
print
def decisionanalysis():
# This code is placed in the public domain
print
print "This is a general decision program, and you can define your choices and criteria."
print
print "When prompted, please enter the options or choices that you need to decide amongst."
print
print "Then, when prompted, enter the criteria for making this decision."
def get_list(heading, prompt):
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):
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 = get_list("Enter your options:", "Option %d: ")
criteria = get_list("Now, 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.
# number of the criterion)
print "A program to help you make decisions."
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 "Scores for option %s" % 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:
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)
def ProgramLanguageFinal():
print "This is a program to help give you an idea which programming languages you should consider learning."
print "While there are any number of languages you might consider, this program considers only 11 of the most popluar ones."
print
print "The program will ask you to input a ranking or weighting for a number of criteria that may be of importance"
print "in choosing your next programming language."
def get_list(heading, prompt):
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
options = ["Python", "Perl", "Ruby", "Tcl", "JavaScript", "Visual Basic", "Java", "C++", "C", "Lisp", "Delphi"]
criteria = ["ease of learning", "ease of use", "speed of program execution", "quality of available tools", "popularity", "power & expressiveness", "cross platform?", "cost"]
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"):70, ("Python", "popularity"):50, ("Python", "power & expressiveness"):100, ("Python", "cross platform?"):100, ("Python", "cost"):100,
("Perl", "ease of learning"):50, ("Perl", "ease of use"):60, ("Perl", "speed of program execution"):20, ("Perl", "quality of available tools"):50, ("Perl", "popularity"):85, ("Perl", "power & expressiveness"):70, ("Perl", "cross platform?"):100, ("Perl", "cost"):100,
("Ruby", "ease of learning"):50, ("Ruby", "ease of use"):100, ("Ruby", "speed of program execution"):20, ("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"):10, ("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,
("Java", "ease of learning"):15, ("Java", "ease of use"):50, ("Java", "speed of program execution"):50, ("Java", "quality of available tools"):100, ("Java", "popularity"):90, ("Java", "power & expressiveness"):100, ("Java", "cross platform?"):100, ("Java", "cost"):100,
("C++", "ease of learning"):10, ("C++", "ease of use"):25, ("C++", "speed of program execution"):90, ("C++", "quality of available tools"):90, ("C++", "popularity"):80, ("C++", "power & expressiveness"):100, ("C++", "cross platform?"):90, ("C++", "cost"):100,
("C", "ease of learning"):15, ("C", "ease of use"):20, ("C", "speed of program execution"):100, ("C", "quality of available tools"):80, ("C", "popularity"):80, ("C", "power & expressiveness"):80, ("C", "cross platform?"):110, ("C", "cost"):100,
("Lisp", "ease of learning"):40, ("Lisp", "ease of use"):50, ("Lisp", "speed of program execution"):80, ("Lisp", "quality of available tools"):60, ("Lisp", "popularity"):25, ("Lisp", "power & expressiveness"):110, ("Lisp", "cross platform?"):80, ("Lisp", "cost"):90,
("Delphi", "ease of learning"):50, ("Delphi", "ease of use"):110, ("Delphi", "speed of program execution"):85, ("Delphi", "quality of available tools"):100, ("Delphi", "popularity"):30, ("Delphi", "power & expressiveness"):100, ("Delphi", "cross platform?"):80, ("Delphi", "cost"):10}
# Calculate the resulting score for each option.
# 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 "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)
def ProgramLanguageScript():
print
print "This is a program to help you choose a scripting language."
print
print "You will be asked to rank some important criteria as to their relative importance to you."
print "These criteria are 'ease of learning', 'ease of use', 'speed of program execution'"
"'quality of available tools', 'popularity', and 'power & expressiveness'"
print
print "Please rank each of the criteria with a number from 1 to 100 when prompted."
print
print "100 means of highest relative importance, 1 means of least importance."
def get_list(heading, prompt):
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):
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)
def Basketball():
print "This is a program to help you decide which team will win a basketball game"
print
print "When prompted, enter a number ranking each team on the prompted team skill"
print "on a scale from 1 to 100, with 1 being terrible and 100 being the best imaginable"
print
team_one = raw_input ("What is the name of team one:")
team_two = raw_input ("What is the name of team two:")
criteria = {"speed":100, "size":66, "jumping_ability":50, "defense":60, "shooting":75, "ballhandling":50, "rebounding":50}
scoreonespeed = input ("rank the team speed of %s on a scale of 1 to 100:" % team_one)
scoretwospeed = input ("rank the team speed of %s on a scale of 1 to 100:" % team_two)
scoreonesize= input ("rank the team size of %s on scale of 1 to 100" % team_one)
scoretwosize= input ("rank the team size of %s on scale of 1 to 100" % team_two)
scoreonejumping_ability= input("rank the jumping ability of %s" % team_one)
scoretwojumping_ability= input("rank the jumping ability of %s" % team_two)
scoreonedefense = input ("ramk the defense of %s" % team_one)
scoretwodefense = input ("ramk the defense of %s" % team_two)
scoreoneshooting = input ("rank the shooting ability of %s" % team_one)
scoretwoshooting = input ("rank the shooting ability of %s" % team_two)
scoreoneballhandling= input("rank the ballhandling ability of %s:" % team_one)
scoretwoballhandling= input("rank the ballhandling ability of %s:" % team_two)
scoreonerebounding = input ("rank the rebounding ability of %s" % team_one)
scoretworebounding = input ("rank the rebounding ability of %s" % team_two)
scoreteamone = (criteria["speed"])*(scoreonespeed) + (criteria["size"])*(scoreonesize) +(criteria["jumping_ability"])*(scoreonejumping_ability) +(criteria["defense"])*(scoreonedefense) + (criteria["shooting"])*(scoreoneshooting) +(criteria["ballhandling"])*(scoreoneballhandling) +(criteria["rebounding"])*(scoreonerebounding)
scoreteamtwo = (criteria["speed"])*(scoretwospeed) + (criteria["size"])*(scoretwosize) +(criteria["jumping_ability"])*(scoretwojumping_ability) +(criteria["defense"])*(scoretwodefense) + (criteria["shooting"])*(scoretwoshooting) +(criteria["ballhandling"])*(scoretwoballhandling) +(criteria["rebounding"])*(scoretworebounding)
print "%s has a power ranking of %d" % (team_one, scoreteamone)
print
print "%s has a power ranking of %d" % (team_two, scoreteamtwo)
if scoreteamone > scoreteamtwo:
print "%s wins!!!" % team_one
elif scoreteamone == scoreteamtwo:
print "the two teams are a toss-up!!!"
else:
print "%s wins!!!" % team_two
def YesNo():
def get_list(heading, prompt):
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):
res = None
while res is None:
try:
res = float(raw_input(prompt))
except ValueError: pass
return res
options = ["Yes", "No"]
criteria = get_list("Enter your criteria ...", "Criterion %d: ")
rankings = {}
print
print "Enter relative importance of criteria (higher is more important)"
print
for c in criteria:
rankings[c] = get_number("Criterion %s: " % c)
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
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")
for option, result in results:
print "%5s %s" % (result, option)
if results[0] > results[1]:
print "You should decide Yes!!!"
else:
print
print "You should decide No!!!"
print
while 1: # loop forever
print "Please enter the number for the type of decision you wish to analayze:"
print "1. General Decision Analysis, you choose the options, criteria, etc."
print "2. Help in Choosing Programming Language amongst 11 popular languages"
print "3. Help in choosing scripting programming language amongst 6 scripting languages"
print "4. Which Basketball Team will win the Game???"
print "5. Questions with Yes or No type answers"
choice = input("Please type in the number of the type of decision-program you wish to run from above and hit enter:")
if choice ==1:
decisionanalysis()
elif choice ==2:
ProgramLanguageFinal()
elif choice ==3:
ProgramLanguageScript()
elif choice ==4:
Basketball()
elif choice ==5:
YesNo()
elif choice =="quit":
break # exit from infinite loop
else:
print "Invalid operation"
