Python: random options

[masque7]

I've written some Python code. 3 possible answers to simple sums are presented. Out of the 3 possible answers, sometimes 2 of them are the same. Sometimes even all 3 are the same. Example:

Question: 2+2
Answers:
a. 1
b. 4
c. 4


import random
from pygame import * # import all pygame functions

count = 0 # Sentry variables
score = 0

file1 = open('words.txt', 'r') # Open text files
file2 = open('images.txt', 'r')

f1content = file1.readlines() # Get content
f2content = file2.readlines()

init() # Start PyGame
screen = display.set_mode((640, 480)) # Produces window
display.set_caption('Numeracy') # Sets title

font = font.Font(None, 48) # Font object 'None' = system default, size 48

while count < 10: # Main loop, runs 10 times
screen.fill(0) # Fills screen with black

# Random number for image shown
# limited to number of entries in images.txt
wordnum = random.randint(0, len(f2content)-1)

# Randnum indicates the line of images.txt
# corresponding to an image file
mainpic = image.load(f2content[wordnum].rstrip('\n')) # Strip newline char

# Draw images at top-center of screen
screen.blit(mainpic, (255, 50))

# Choose 3 random options
options = [random.randint(0, len(f1content)-1),
random.randint(0, len(f1content)-1),
random.randint(0, len(f1content)-1)]

# One of them the real answer
options[random.randint(0, 2)] = wordnum

# Draw the 3 possible answer words from f1content[] array
#' True' for the second = anti-alias.
# colours (red, green, blue) aka white
text1 = font.render('(PRESS 1) = ' + f1content[options[0]].rstrip('\n'),
True, (255, 255, 255))
text2 = font.render('(PRESS 2) = ' + f1content[options[1]].rstrip('\n'),
True, (255, 255, 255))
text3 = font.render('(PRESS 3) = ' + f1content[options[2]].rstrip('\n'),
True, (255, 255, 255))

# Draw those words to the screen (horiz, vertical)
screen.blit(text1, (30, 200))
screen.blit(text2, (30, 300))
screen.blit(text3, (30, 400))

# Update everything
display.update()

done = False # Sentry var for next loop

# Wait for keyboard event. If key KEYUP, check if 1 key (K_1),
# or 2, etc. When answer key, set to 'done' = True to exit loop
while not done:
for e in event.get():
if e.type == KEYUP:
if e.key == K_1:
answer = 1
done = True
if e.key == K_2:
answer = 2
done = True
if e.key == K_3:
answer = 3
done = True

# Prints correct in green, increments to score
if options[answer-1] == wordnum:
resulttext = font.render('Correct!', True, (50, 255, 50))
score += 1
# Wrong - red text
else:
resulttext = font.render('Wrong!', True, (255, 50, 50))

# Show the text
screen.blit(resulttext, (450, 300))
display.update()

# Keypress before the next question
while event.wait().type != KEYDOWN: pass

count += 1

# Print the score when loop finished
print '\nYour score is:', score


How can I fix it so that each of the 3 answers are individual yet randomised options? I suppose this would contain a lot of comparing but the issue is that they aren't 3 separate varibles.

[andrew1992]

A simple way would be to store the random items in a list if they are unique, and use a while loop until the list is of length 3 in your case.



my_list = []
while (len(my_list) < 3):
number = get_random_number()
if number not in my_list:
my_list.append(number)

[DaithiF]

alas, if only the python random module contained a function to return a random sample of a certain size from a population, that would be pretty useful, huh? But probably not worth checking the docs, because it hardly seems likely...

but wait ... whats this? http://docs.python.org/library/random.html#random.sample


options = random.sample(xrange(len(f1content)), 3)

[andrew1992]

Haha nice. I too often default to re-inventing the wheel :)