Decision_Analysis is a mini-expert system or AI program to help a user make decisions. It includes a general program to help in making any decisions, big or small, as well as four sub-programs that illustrate the use of the program by applying it to four specific kinds of decisions.
Python, 473 lines
Download
Copy to clipboard1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 | 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"
|
This program uses a weighted average algorithm. The user chooses any number of alternative options she could choose. Then, the program looks at various criteria that are important in making the decision. Then, the criteria are ranked or weighted based on importance to the user. Then, the various alternative options are scored on each criteria. The computer then calculates and recommends a particular decision.
Many people have trouble making decisions. This program helps folks to sort out what is really important and why, and encourages them to make assumptions explicit in order to search for the best solution.
