Python Question 10 POINTS?

Question

Hi I really need help....

I am stuck, if you can help i would really appreciate it.

So i have been given a task to make a funciton that return a string as hidden
for example:

>>> get_view ('apple'):
'^^^^^'
>>> get_view ('apple-banana')
'^^^^^-^^^^^^'

So we are only trying to hide lower-case letters and nothing else...

My func so far is:

def get_view (puzzle):
''' (str) -> str
>>>get_view ('apple')
'^^^^^'
>>>get_view ('apple-banana')
'^^^^^-^^^^^^'
'''
return len(puzzle) * HIDDEN <-- But it hides EVERYTHING...

Also, in the next function i have 3 strings...

def update_view (puzzle, view, guessed_letter):
''' (str, str, str) -> str

Return the view of the puzzle with each occurrence of the letter in
the puzzle revealed.

>>>update_view ('apple', '^^^^^', 'a')
'a^^^^'
>>>update_view ('apple', '^^^^^', 'p')
'^pp^^'

'''
view == puzzle
for i in puzzle[0:]:
if (guessed_letter in puzzle):
return guessed_letter + view[0+1:] <-- But it only reveals the first letter...i want a condition that would match the placement...

THANKS FOR ALL YOUR HELP:)

Answer 1

We've discussed this in email, but for the record:



def hide_unguessed(word, guessed):

... """hide any letters in word that aren't in the guessed list"""

... result = ""

... for ch in word:

... ... if ch.islower() and ch not in guessed:

... ... ... result += "^"

... ... else:

... ... ... result += ch

... return result



Edit: For the benefit of Python fanatics who may be reading, here's a way to do the whole thing in one line:



return "".join( [(c if c in guessed or not c.islower() else '^') for c in word] )



That depends on three things that you probably haven't learned yet. The expression in [] brackets is a "list comprehension", which is a way of building a list using any sequence, plus a rule. The "for c in word" trailing part defines the sequence as every character in the word string. The front part is a "conditional expression" that returns c if the condition (c in guessed or not c.islower()) is true, else returns '^'. That creates a list of characters that belong in the output. Finally, the "".join(list) expression concatenates a list of strings into a single string.



I don't suggest that you go out of your way to write code like that. However, "".join() is about the fastest way to concatenate a large number of strings together. Sooner or later, it should be in your Python toolbox.

Answer 2

Sounds like you want something like hangman.

The basic idea for the hide function is straightforward. Make an empty string (we'll call it RET). Next, scan through each character of the string to hide using a for loop. For each scanned character, if it's a lowercase letter, add a "hidden character" to RET. Otherwise just add whatever character shows up in the original string.



To check if a character is lower case, convert the character to an integer with the python's built-in ord() function. The lowercase letters of the alphabet correspond to integers 97-122 for a-z.

(Remember to take out the arrows if you want to run this)



def HideMe(str):

>>ret = ""

>>for i in str:

>>>>if ord(i)>=97 and ord(i)<=122:

>>>>>>ret += "^"

>>>>else:

>>>>>>ret += i

>>return ret



For the next function, it's mostly the same. But, when you go to add hidden characters, don't just check that they are lower case, also check that they are not equal to the "show" variable (as I've called it in my example), and that the "prev" variable (as I've called it) has a "hidden character" at the same location.

That way, when you do something like: func('apple','a^^^^','p') it will return 'app^^'.



def HideMeMostly(str,prev,show):

>>HideChar = "^"

>>if(len(str)!=len(prev)) return false; //prev must be a valid hidden version of str

>>ret = ""

>>for i in range(0,len(str)):

>>>>if str[i]!=show and prev[i]==HideChar and ord(i)>=97 and ord(i)<=122:

>>>>>>ret += "^"

>>>>else:

>>>>>>ret += i

>>return ret



BTW, you'll still need to add a way of checking whether the user revealed a character or not, if you are making a hangman game.

See the sources for info on ord() and range().



Good luck! :-)

Answer 3

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