CODEWITHSUNDEEP
pythonpython-3.xpython-3.4
F. Borges
F. Borges

Reputation: 23

Did I get this randrange() range correct?

I'm trying to get this homework done I've completed almost all of it besides formatting to make it look readable and this little problem I've been having, I can't seem to be getting this randrange() to work properly. It works most of the time, but every so often I get the error at the bottom.

dict_of_strings = {
    1:"PoopyPants",     #my go-to programming word... i donno what to tell you...
    2:"SleighBells",    #i asked my girlfriend or the next two
    3:"CookieMonster",
    4:"bomb",           #i have noidea how this one came to me
    5:"Blellow",        #when you mix blue and yellow, according to reece from Malcom in the Middle
    6:"RandomWordOne",  #because, random
    7:"Flaccid",        #theres a pattern to this word
    8:"Solid",          #maybe this is the pattern
    9:"VaultOneHundredEleven",  #FALLOUT4 HYPETRAIN
    10:"SirDoctorMcFlippertonTheTwentySecondOfTheHouseOfLordProfessorSteveWilkerton"    #this was a fun, convoluted phrase to come up with
}

def choice_is_rand(dict_of_strings):
    rand_word = str(dict_of_strings[random.randrange(0,10)]) #did i get this range correctly? i've always had trouble. I thought I had it, but i guess not...

    rand_backward = rand_word[::-1]

    print(rand_word)
    print(rand_backward.lower())

I'm trying to get the rand_word randrange to work, but I seem to be getting it wrong. My particular error is:

Traceback (most recent call last):
  File "I:\User\My Documents\Komodo Projects\COP 1000\Module 11\Reverse the Word.py", line 116, in <module>
    main()
  File "I:\User\My Documents\Komodo Projects\COP 1000\Module 11\Reverse the Word.py", line 35, in main
    choice_is_rand(dict_of_strings)
  File "I:\User\My Documents\Komodo Projects\COP 1000\Module 11\Reverse the Word.py", line 73, in choice_is_rand
    rand_word = str(dict_of_strings[random.randrange(0,10)])
KeyError: 0

Upvotes: 0

Views: 181

Answers (2)

Christian Witts
Christian Witts

Reputation: 11585

You can either change your range choice to start at your first value, being 1, or you can directly pick out the word with choice

# Select the word directly
random_word = random.choice(list(dict_of_strings.values()))
# Or select the key, and then access the word
random_word_index = random.choice(list(dict_of_strings.keys()))
random_word = dict_of_strings[random_word_index]

# As Pynchia pointed out, you can simply do the following without the
# need to fetch the keys, and cast them to a list.
# This will work on Python 3.5+, but not Python 3.4< or Python 2.7<
# in such case, you must make your keys start from zero, not one
random_word = random.choice(dict_of_strings)

Upvotes: 2

Mike Covington
Mike Covington

Reputation: 2157

Instead of this:

rand_word = str(dict_of_strings[random.randrange(0,10)])

Use this:

rand_word = str(dict_of_strings[random.randint(1, len(dict_of_strings))])

Two reasons:

  • Your keys start at 1 and randrange() does not include the endpoint, while randint() does, so starting at 0 causes an error about 10% of the time for you.

  • Getting the endpoint with len(dict_of_strings) means you can use different dicts.

However, random.choice(dict_of_strings) is more appropriate.

Upvotes: 0

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