I've also toyed with dictionaries, but couldn't get them to work either. Any help would be greatly appreciated.
When "it doesn't work" in python, most of the time, there is an exception traceback (python complains). Here is what I get in my terminal when I try to run your program
[t215809] python cypher.py
Traceback (most recent call last):
File "cypher.py", line 31, in <module>
cypher = string.maketrans(base, FIVEHIGH)
NameError: name 'string' is not defined
Python tells us that when it reaches the line cypher = string.maketrans(base, FIVEHIGH) , it fails because you didn't say what the variable "string" is. When it doesn't work, you must study the traceback and understand what it means. Now your problem is to tell python what "string" is. After this problem, there are other problems waiting for you, with other tracebacks.
So, go one step after the other, and when you have a traceback, put it in your posts because it's very useful for helpers.
Gribouillis
Posting Maven
2,786 posts since Jul 2008
Reputation Points: 1,044
Solved Threads: 691
"".stringmethod() works if you need an empty string to use the string methods to.
willygstyle
Junior Poster in Training
60 posts since Aug 2009
Reputation Points: 17
Solved Threads: 9
word = raw_input("What word should we convert? ").upper()
word.translate(cypher)
If the first statement, you define 'word' as a string. In the second statement, it appears to be a class with the method "translate". You can not use the same variable name for two different types of memory.
woooee
Nearly a Posting Maven
2,454 posts since Dec 2006
Reputation Points: 777
Solved Threads: 714
Look at the documentation of maketrans . It can only replace a character by a single character. This means that you can't use maketrans and translate for your purpose. Here is how you could iterate over the characters
if word == "":
print "No word entered"
else:
print "Your word is",word
for c in word:
print(c)
Now, instead of printing the character, you must find the triple quoted string which corresponds to this character (if any). You should then put all these strings in a list and finally concatenate all the strings of the list.
Gribouillis
Posting Maven
2,786 posts since Jul 2008
Reputation Points: 1,044
Solved Threads: 691
You can create a dictionary to associate a string to each letter
translation = dict()
for i in range(len(base)):
translation[base[i]] = FIVEHIGH[i]
and then use translation[c] to get the string corresponding to c. Post your new code when it's written.
Gribouillis
Posting Maven
2,786 posts since Jul 2008
Reputation Points: 1,044
Solved Threads: 691
You must not add the triple quoted strings in FIVEHIGH.
Gribouillis
Posting Maven
2,786 posts since Jul 2008
Reputation Points: 1,044
Solved Threads: 691
First, make a list with the strings which correspond to CAB, then think about how to produce your output with this list.
Gribouillis
Posting Maven
2,786 posts since Jul 2008
Reputation Points: 1,044
Solved Threads: 691
Here is how we could do in pseudo-algorithmic language. Let's call the elements of FIVEHIGH "ideograms"
create an empty list L
for each character c in word:
find the ideogram x corresponding to c
append x to our list L
# now the problem is to display the result
# we display the 5 lines one after the other
for each i in 1, 2, 3, 4, 5:
initialize the i-th line S to the empty string
for each ideogram x in L:
find the i-th line y of x # for example 2nd line of the first ideogram is 0 X 0 X
concatenate y to S
print S
Now, you write the python code for this strategy !
Gribouillis
Posting Maven
2,786 posts since Jul 2008
Reputation Points: 1,044
Solved Threads: 691
I for one am not sure what you are trying to do. It is something like this? If not, please post some input and desired output examples. Also, if joining, remember that each triple quoted string is one string and not a list.
base_dict = {}
base_dict["A"] = """X 0 X X
0 X 0 X
0 0 0 X
0 X 0 X
0 X 0 X"""
base_dict["B"] = """0 0 X X
0 X 0 X
0 0 X X
0 X 0 X
0 0 X X"""
base_dict["C"] = """X 0 X X
0 X 0 X
0 X X X
0 X 0 X
X 0 X X
"""
for key in ("A", "B", "C"):
print key, "-" *60
print base_dict[key]
woooee
Nearly a Posting Maven
2,454 posts since Dec 2006
Reputation Points: 777
Solved Threads: 714
Ok, here is a solution
base = "ABC"
FIVEHIGH = (
"""
X 0 X X
0 X 0 X
0 0 0 X
0 X 0 X
0 X 0 X
""",
"""
0 0 X X
0 X 0 X
0 0 X X
0 X 0 X
0 0 X X
""",
"""
X 0 X X
0 X 0 X
0 X X X
0 X 0 X
X 0 X X
""")
class Ideogram(tuple):
def __new__(cls, string):
return tuple.__new__(cls, (y for y in (x.strip() for x in string.split("\n")) if y))
def __init__(self, *args):
assert(len(self) == 5)
base_dict = dict(zip(tuple(base), (Ideogram(x) for x in FIVEHIGH)))
def translate(word):
lines = list(list() for i in range(5))
for letter in (c for c in word.upper() if c in base_dict): # skip characters that dont belong to base
for i, x in enumerate(base_dict[letter]):
lines[i].append(x)
return "\n".join(" ".join(L) for L in lines)
if __name__ == "__main__":
word = raw_input("Enter word:")
print translate(word)
Gribouillis
Posting Maven
2,786 posts since Jul 2008
Reputation Points: 1,044
Solved Threads: 691
My attempt. Once again, if not correct, post back with more info. The trick (I think) is to split each string on the "\n" to create individual rows, and is similiar to Gribouillis' solution but doesn't use list comprehension (which is an amazing tool). It instead uses a dictionary with row number as the key, in this case 0 through 4, and appends each row to the corresponding key. Finally, the joined rows are spaced and printed. Hopefully, using a dictionary with the row number is easy to understand.
def join_letters( letter_list, base_dict ):
combined_rows_dict = {}
##--- base length to test for equality
len_letter = len(base_dict[letter_list[0]])
for j in range(0, len(letter_list)):
this_letter = letter_list[j]
##--- all letters must be in the dictionary
if this_letter not in base_dict:
print "letter %s not in base_dict" % (this_letter)
return "***"
##--- test for all translations are equal in length
if len(base_dict[this_letter]) != len_letter:
print "unequal letter lengths", len_letter, len(base_dict[this_letter])
return "***"
##--- convert triple quote string into rows in a list
string_list = base_dict[this_letter].split("\n")
print "string_list", string_list
##--- append each row to the corresponding dictionary key
for row in range(0, len(string_list)):
if row not in combined_rows_dict:
combined_rows_dict[row] = []
combined_rows_dict[row].append(string_list[row])
print
for row in range(0, len(combined_rows_dict)):
this_row = combined_rows_dict[row]
print " ".join(this_row)
return "Success"
base_dict = {}
base_dict["A"] = """X 0 X X
0 X 0 X
0 0 0 X
0 X 0 X
0 X 0 X"""
base_dict["B"] = """0 0 X X
0 X 0 X
0 0 X X
0 X 0 X
0 0 X X"""
base_dict["C"] = """X 0 X X
0 X 0 X
0 X X X
0 X 0 X
X 0 X X"""
##for key in ("A", "B", "C"):
## print key, "-" *60
## print base_dict[key]
print join_letters(["A", "C", "B"], base_dict)
woooee
Nearly a Posting Maven
2,454 posts since Dec 2006
Reputation Points: 777
Solved Threads: 714
