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Ignoring Changed Index Check (Python)

I have made a script:
our_word = "Success"
def duplicate_encode(word):
char_list = []
final_str = ""
changed_index = []
base_wrd = word.lower()
for k in base_wrd:
char_list.append(k)
for i in range(0, len(char_list)):
count = 0
for j in range(i + 1, len(char_list)):
if j not in changed_index:
if char_list[j] == char_list[i]:
char_list[j] = ")"
changed_index.append(j)
count += 1
else:
continue
if count > 0:
char_list[i] = ")"
else:
char_list[i] = "("
print(changed_index)
print(char_list)
final_str = "".join(char_list)
return final_str
print(duplicate_encode(our_word))
essentialy the purpose of this script is to convert a string to a new string where each character in the new string is "(", if that character appears only once in the original string, or ")", if that character appears more than once in the original string. I have made a rather layered up script (I am relatively new to the python language so didn't want to use any helpful in-built functions) that attempts to do this. My issue is that where I check if the current index has been previously edited (in order to prevent it from changing), it seems to ignore it. So instead of the intended )())()) I get )()((((. I'd really appreciate an insightful answer to why I am getting this issue and ways to work around this, since I'm trying to gather an intuitive knowledge surrounding python. Thanks!

word = "Success"
print(''.join([')' if word.lower().count(c) > 1 else '(' for c in word.lower()]))

The issue here has nothing to do with your understanding of Python. It's purely algorithmic. If you retain this 'layered' algorithm, it is essential that you add one more check in the "i" loop.
our_word = "Success"
def duplicate_encode(word):
char_list = list(word.lower())
changed_index = []
for i in range(len(word)):
count = 0
for j in range(i + 1, len(word)):
if j not in changed_index:
if char_list[j] == char_list[i]:
char_list[j] = ")"
changed_index.append(j)
count += 1
if i not in changed_index: # the new inportant check to avoid reversal of already assigned ')' to '('
char_list[i] = ")" if count > 0 else "("
return "".join(char_list)
print(duplicate_encode(our_word))

Your algorithm can be greatly simplified if you avoid using char_list as both the input and output. Instead, you can create an output list of the same length filled with ( by default, and then only change an element when a duplicate is found. The loops will simply walk along the entire input list once for each character looking for any matches (other than self-matches). If one is found, the output list can be updated and the inner loop will break and move on to the next character.
The final code should look like this:
def duplicate_encode(word):
char_list = list(word.lower())
output = list('(' * len(word))
for i in range(len(char_list)):
for j in range(len(char_list)):
if i != j and char_list[i] == char_list[j]:
output[i] = ')'
break
return ''.join(output)
for our_word in (
'Success',
'ChJsTk(u cIUzI htBp#qX)OTIHpVtHHhQ',
):
result = duplicate_encode(our_word)
print(our_word)
print(result)
Output:
Success
)())())
ChJsTk(u cIUzI htBp#qX)OTIHpVtHHhQ
))(()(()))))())))()()((())))()))))

Recursive Decompression of Strings

I'm trying to decompress strings using recursion. For example, the input:
3[b3[a]]
should output:
baaabaaabaaa
but I get:
baaaabaaaabaaaabbaaaabaaaabaaaaa
I have the following code but it is clearly off. The first find_end function works as intended. I am absolutely new to using recursion and any help understanding / tracking where the extra letters come from or any general tips to help me understand this really cool methodology would be greatly appreciated.
def find_end(original, start, level):
if original[start] != "[":
message = "ERROR in find_error, must start with [:", original[start:]
raise ValueError(message)
indent = level * " "
index = start + 1
count = 1
while count != 0 and index < len(original):
if original[index] == "[":
count += 1
elif original[index] == "]":
count -= 1
index += 1
if count != 0:
message = "ERROR in find_error, mismatched brackets:", original[start:]
raise ValueError(message)
return index - 1
def decompress(original, level):
# set the result to an empty string
result = ""
# for any character in the string we have not looked at yet
for i in range(len(original)):
# if the character at the current index is a digit
if original[i].isnumeric():
# the character of the current index is the number of repetitions needed
repititions = int(original[i])
# start = the next index containing the '[' character
x = 0
while x < (len(original)):
if original[x].isnumeric():
start = x + 1
x = len(original)
else:
x += 1
# last = the index of the matching ']'
last = find_end(original, start, level)
# calculate a substring using `original[start + 1:last]
sub_original = original[start + 1 : last]
# RECURSIVELY call decompress with the substring
# sub = decompress(original, level + 1)
# concatenate the result of the recursive call times the number of repetitions needed to the result
result += decompress(sub_original, level + 1) * repititions
# set the current index to the index of the matching ']'
i = last
# else
else:
# concatenate the letter at the current index to the result
if original[i] != "[" and original[i] != "]":
result += original[i]
# return the result
return result
def main():
passed = True
ORIGINAL = 0
EXPECTED = 1
# The test cases
provided = [
("3[b]", "bbb"),
("3[b3[a]]", "baaabaaabaaa"),
("3[b2[ca]]", "bcacabcacabcaca"),
("5[a3[b]1[ab]]", "abbbababbbababbbababbbababbbab"),
]
# Run the provided tests cases
for t in provided:
actual = decompress(t[ORIGINAL], 0)
if actual != t[EXPECTED]:
print("Error decompressing:", t[ORIGINAL])
print(" Expected:", t[EXPECTED])
print(" Actual: ", actual)
print()
passed = False
# print that all the tests passed
if passed:
print("All tests passed")
if __name__ == '__main__':
main()

From what I gathered from your code, it probably gives the wrong result because of the approach you've taken to find the last matching closing brace at a given level (I'm not 100% sure, the code was a lot). However, I can suggest a cleaner approach using stacks (almost similar to DFS, without the complications):
def decomp(s):
stack = []
for i in s:
if i.isalnum():
stack.append(i)
elif i == "]":
temp = stack.pop()
count = stack.pop()
if count.isnumeric():
stack.append(int(count)*temp)
else:
stack.append(count+temp)
for i in range(len(stack)-2, -1, -1):
if stack[i].isnumeric():
stack[i] = int(stack[i])*stack[i+1]
else:
stack[i] += stack[i+1]
return stack[0]
print(decomp("3[b]")) # bbb
print(decomp("3[b3[a]]")) # baaabaaabaaa
print(decomp("3[b2[ca]]")) # bcacabcacabcaca
print(decomp("5[a3[b]1[ab]]")) # abbbababbbababbbababbbababbbab
This works on a simple observation: rather tha evaluating a substring after on reading a [, evaluate the substring after encountering a ]. That would allow you to build the result AFTER the pieces have been evaluated individually as well. (This is similar to the prefix/postfix evaluation using programming).
(You can add error checking to this as well, if you wish. It would be easier to check if the string is semantically correct in one pass and evaluate it in another pass, rather than doing both in one go)

Here is the solution with the similar idea from above:
we go through string putting everything on stack until we find ']', then we go back until '[' taking everything off, find the number, multiply and put it back on stack
It's much less consuming as we don't add strings, but work with lists
Note: multiply number can't be more than 9 as we parse it as one element string
def decompress(string):
stack = []
letters = []
for i in string:
if i != ']':
stack.append(i)
elif i == ']':
letter = stack.pop()
while letter != '[':
letters.append(letter)
letter = stack.pop()
word = ''.join(letters[::-1])
letters = []
stack.append(''.join([word for j in range(int(stack.pop()))]))
return ''.join(stack)

How to rearrange a string's characters such that none of it's adjacent characters are the same, using Python

In my attempt to solve the above question, I've written the following code:
Logic: Create a frequency dict for each character in the string (key= character, value= frequency of the character). If any character's frequency is greater than ceil(n/2), there is no solution. Else, print the most frequent character followed by reducing its frequency in the dict/
import math, operator
def rearrangeString(s):
# Fill this in.
n = len(s)
freqDict = {}
for i in s:
if i not in freqDict.keys():
freqDict[i] = 1
else:
freqDict[i] += 1
for j in list(freqDict.values()):
if j > math.ceil(n / 2):
return None
return maxArrange(freqDict)[:-4]
temp = ""
def maxArrange(inp):
global temp
n = len(inp)
if list(inp.values()) != [0] * n:
resCh = max(inp.items(), key=operator.itemgetter(1))[0]
if resCh is not None and resCh != temp:
inp[resCh] -= 1
# Terminates with None
temp = resCh
return resCh + str(maxArrange(inp))
# Driver
print(rearrangeString("abbccc"))
# cbcabc
print(rearrangeString("abbcccc"))
In the first try, with input abbccc, it gives the right answer, i.e. cbcabc, but fails for the input abbcccc, returning ccbcabc, without handling it using the temp variable, else returning cbcabc and skipping c altogether when handled using temp
How should I modify the logic, or is there a better approach?

How to process character by character in a line

I have a file that has sequence on line 2 and variable called tokenizer, which give me an old position value. I am trying to find the new position.. For example tokenizer for this line give me position 12, which is E by counting letters only until 12. So i need to figure out the new position by counting dashes...
---------------LL---NE--HVKTHTEEK---PF-ICTVCR-KS----------
This is what i have so far it still doesn't work.
with open(filename) as f:
countletter = 0
countdash = 0
for line, line2 in itertools.izip_longest(f, f, fillvalue=''):
tokenizer=line.split()[4]
print tokenizer
for i,character in enumerate(line2):
for countletter <= tokenizer:
if character != '-':
countletter += 1
if character == '-':
countdash +=1
my new position should be 32 for this example

First answer, edited by Chad D to make it 1-indexed (but incorrect):
def get_new_index(string, char_index):
chars = 0
for i, char in enumerate(string):
if char != '-':
chars += 1
if char_index == chars:
return i+1
Rewritten version:
import re
def get(st, char_index):
chars = -1
for i, char in enumerate(st):
if char != '-':
chars += 1
if char_index == chars:
return i
def test():
st = '---------------LL---NE--HVKTHTEEK---PF-ICTVCR-KS----------'
initial = re.sub('-', '', st)
for i, char in enumerate(initial):
print i, char, st[get_1_indexed(st, i)]
def get_1_indexed(st, char_index):
return 1 + get(st, char_index - 1)
def test_1_indexed():
st = '---------------LL---NE--HVKTHTEEK---PF-ICTVCR-KS----------'
initial = re.sub('-', '', st)
for i, char in enumerate(initial):
print i+1, char, st[get_1_indexed(st, i + 1) - 1]

my original text looks like this and the position i was interested in was 12 which is 'E'
Actually, it's K, assuming you're using zero indexed strings. Python uses zero indexing so unless you're jumping through hoops to 1-index things (and you're not) it will give you K. If you were running into issues, try addressing this.
Here's some code for you that does what you need it to (albeit with 0-indexing, not 1-indexing). This can be found online here:
def get_new_index(oldindex, str):
newindex = 0
for c in str:
if c != '-':
if oldindex == 0:
return newindex
oldindex -= 1
newindex += 1
return 1 / 0 # throw a shitfit if we don't find the index

This is a silly way to get the second line, it would be clearer to use an islice, or next(f)
for line, line2 in itertools.izip_longest(f, f, fillvalue=''):
Here count_letter seems to be an int while tokenizer is a str. Probably not what you expect.
for countletter <= tokenizer:
It's also a syntax error, so I think this isn't the code you are running
Perhaps you should have
tokenizer = int(line.split()[4])
to make tokenizer into an int
print tokenizer can be misleading because int and str look identical, so you see what you expect to see. Try print repr(tokenizer) instead when you are debugging.
once you make sure tokenizer is an int, you can change this line
for i,character in enumerate(line2[:tokenizer]):

How do you reverse the words in a string using python (manually)? [duplicate]

This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Reverse the ordering of words in a string
I know there are methods that python already provides for this, but I'm trying to understand the basics of how those methods work when you only have the list data structure to work with. If I have a string hello world and I want to make a new string world hello, how would I think about this?
And then, if I can do it with a new list, how would I avoid making a new list and do it in place?

Split the string, make a reverse iterator then join the parts back.
' '.join(reversed(my_string.split()))
If you are concerned with multiple spaces, change split() to split(' ')
As requested, I'm posting an implementation of split (by GvR himself from the oldest downloadable version of CPython's source code: Link)
def split(s,whitespace=' \n\t'):
res = []
i, n = 0, len(s)
while i < n:
while i < n and s[i] in whitespace:
i = i+1
if i == n:
break
j = i
while j < n and s[j] not in whitespace:
j = j+1
res.append(s[i:j])
i = j
return res
I think now there are more pythonic ways of doing that (maybe groupby) and the original source had a bug (if i = n:, corrrected to ==)

Original Answer
from array import array
def reverse_array(letters, first=0, last=None):
"reverses the letters in an array in-place"
if last is None:
last = len(letters)
last -= 1
while first < last:
letters[first], letters[last] = letters[last], letters[first]
first += 1
last -= 1
def reverse_words(string):
"reverses the words in a string using an array"
words = array('c', string)
reverse_array(words, first=0, last=len(words))
first = last = 0
while first < len(words) and last < len(words):
if words[last] != ' ':
last += 1
continue
reverse_array(words, first, last)
last += 1
first = last
if first < last:
reverse_array(words, first, last=len(words))
return words.tostring()
Answer using list to match updated question
def reverse_list(letters, first=0, last=None):
"reverses the elements of a list in-place"
if last is None:
last = len(letters)
last -= 1
while first < last:
letters[first], letters[last] = letters[last], letters[first]
first += 1
last -= 1
def reverse_words(string):
"""reverses the words in a string using a list, with each character
as a list element"""
characters = list(string)
reverse_list(characters)
first = last = 0
while first < len(characters) and last < len(characters):
if characters[last] != ' ':
last += 1
continue
reverse_list(characters, first, last)
last += 1
first = last
if first < last:
reverse_list(characters, first, last=len(characters))
return ''.join(characters)
Besides renaming, the only change of interest is the last line.

You have a string:
str = "A long string to test this algorithm"
Split the string (at word boundary -- no arguments to split):
splitted = str.split()
Reverse the array obtained -- either using ranges or a function
reversed = splitted[::-1]
Concatenate all words with spaces in between -- also known as joining.
result = " ".join(reversed)
Now, you don't need so many temps, combining them into one line gives:
result = " ".join(str.split()[::-1])

str = "hello world"
" ".join(str.split()[::-1])