How to escape parentheses python - python

I am the newbie in programming. The goal of this exercise 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. Ignore capitalization when determining if a character is a duplicate.
But when the code meet ) -- closing parentheses it generate wrong output. As I figured out, the problem is with regular expressions, but I didn't get how to fix code.
from collections import Counter
def duplicate_encode(word):
counter = Counter(word.lower())
counter2 = dict.copy(counter)
print(counter2)
for k,v in counter2.items():
if counter2[k]==1:
counter2[k]='('
else:
counter2[k]=')'
for key in counter2.keys():
word = str(word.lower()).replace(key, str(counter2[key]))
return word
For example:
duplicate_encode('yy! R)!QvdG') should return )))((()(((( but I got (((((((((((.

from collections import Counter
def duplicate_encode(word):
res = list(word.lower())
counter = Counter(word.lower())
counter2 = dict.copy(counter)
print(counter2)
for k, value in enumerate(res):
if counter2[value] == 1:
res[k] = '('
else:
res[k] = ')'
# for key in counter2.keys():
# word = str(word.lower()).replace(key, str(counter2[key]))
return "".join(res)
res = duplicate_encode('yy! R)!QvdG')
print("res", res)

The problem arises when your input string contains a brace, like ( or ).
When that happens, like it does in your example, the wrong characters get replaced, which you can verify by adding print() statements in your code, every time you change word.
I've replaced that part in your code.
from collections import Counter
def duplicate_encode(word):
counter = Counter(word.lower())
counter2 = dict.copy(counter)
print(counter2)
for k,v in counter2.items():
if counter2[k]==1:
counter2[k]='('
else:
counter2[k]=')'
print(counter2)
new_word = ''
for character in word:
new_word += counter2[character.lower()]
return new_word
Note, however, that the code has quite a few redundancies and unnecessary usage of memory. It can be simplified to the following:
from collections import Counter
def duplicate_encode(word):
lower_word = word.lower()
new_word = ''
counter = Counter(lower_word)
for char in lower_word:
if counter[char] > 1:
new_word += ')'
else:
new_word += '('
return new_word

Another solution. Start with a set of (unique) characters, loop over characters in the string, removing the character from that set, and if it is already removed then it must be a duplicate. Use this to build up a set of duplicated characters.
def duplicate_encode(word):
word = word.upper()
s = set(word)
dups = set()
for char in word:
if char in s:
s.remove(char)
else:
dups.add(char)
return "".join(")" if char in dups else "("
for char in word)
print(duplicate_encode("'yy! R)!QvdG'"))
gives:
))))((()(((()

Related

How to use a list of numbers as index inputs

So I have a list of numbers (answer_index) which correlate to the index locations (indicies) of a characters (char) in a word (word). I would like to use the numbers in the list as index inputs later (indexes) on in code to replace every character except my chosen character(char) with "*" so that the final print (new_word) in this instance would be (****ee) instead of (coffee). it is important that (word) maintains it's original value while (new_word) becomes the modified version. Does anyone have a solution for turning a list into valid index inputs? I will also except easier ways to meet my goal. (Note: I am extremely new to python so I'm sure my code looks horrendous) Code below:
word = 'coffee'
print(word)
def find(string, char):
for i, c in enumerate(string):
if c == char:
yield i
string = word
char = "e"
indices = (list(find(string, char)))
answer_index = (list(indices))
print(answer_index)
for t in range(0, len(answer_index)):
answer_index[t] = int(answer_index[t])
indexes = [(answer_index)]
new_character = '*'
result = ''
for i in indexes:
new_word = word[:i] + new_character + word[i+1:]
print(new_word)
You hardly ever need to work with indices directly:
string = "coffee"
char_to_reveal = "e"
censored_string = "".join(char if char == char_to_reveal else "*" for char in string)
print(censored_string)
Output:
****ee
If you're trying to implement a game of hangman, you might be better off using a dictionary which maps characters to other characters:
string = "coffee"
map_to = "*" * len(string)
mapping = str.maketrans(string, map_to)
translated_string = string.translate(mapping)
print(f"All letters are currently hidden: {translated_string}")
char_to_reveal = "e"
del mapping[ord(char_to_reveal)]
translated_string = string.translate(mapping)
print(f"'{char_to_reveal}' has been revealed: {translated_string}")
Output:
All letters are currently hidden: ******
'e' has been revealed: ****ee
The easiest and fastest way to replace all characters except some is to use regular expression substitution. In this case, it would look something like:
import re
re.sub('[^e]', '*', 'coffee') # returns '****ee'
Here, [^...] is a pattern for negative character match. '[^e]' will match (and then replace) anything except "e".
Other options include decomposing the string into an iterable of characters (#PaulM's answer) or working with bytearray instead
In Python, it's often not idiomatic to use indexes, unless you really want to do something with them. I'd avoid them for this problem and instead just iterate over the word, read each character and and create a new word:
word = "coffee"
char_to_keep = "e"
new_word = ""
for char in word:
if char == char_to_keep:
new_word += char_to_keep
else:
new_word += "*"
print(new_word)
# prints: ****ee

Split a string, loop through it character by character, and replace specific ones?

I'm working on an assignment and have gotten stuck on a particular task. I need to write two functions that do similar things. The first needs to correct capitalization at the beginning of a sentence, and count when this is done. I've tried the below code:
def fix_capitalization(usrStr):
count = 0
fixStr = usrStr.split('.')
for sentence in fixStr:
if sentence[0].islower():
sentence[0].upper()
count += 1
print('Number of letters capitalized: %d' % count)
print('Edited text: %s' % fixStr)
Bu receive an out of range error. I'm getting an "Index out of range error" and am not sure why. Should't sentence[0] simply reference the first character in that particular string in the list?
I also need to replace certain characters with others, as shown below:
def replace_punctuation(usrStr):
s = list(usrStr)
exclamationCount = 0
semicolonCount = 0
for sentence in s:
for i in sentence:
if i == '!':
sentence[i] = '.'
exclamationCount += 1
if i == ';':
sentence[i] = ','
semicolonCount += 1
newStr = ''.join(s)
print(newStr)
print(semicolonCount)
print(exclamationCount)
But I'm struggling to figure out how to actually do the replacing once the character is found. Where am I going wrong here?
Thank you in advance for any help!
I would use str.capitalize over str.upper on one character. It also works correctly on empty strings. The other major improvement would be to use enumerate to also track the index as you iterate over the list:
def fix_capitalization(s):
sentences = [sentence.strip() for sentence in s.split('.')]
count = 0
for index, sentence in enumerate(sentences):
capitalized = sentence.capitalize()
if capitalized != sentence:
count += 1
sentences[index] = capitalized
result = '. '.join(sentences)
return result, count
You can take a similar approach to replacing punctuation:
replacements = {'!': '.', ';': ','}
def replace_punctuation(s):
l = list(s)
counts = dict.fromkeys(replacements, 0)
for index, item in enumerate(l):
if item in replacements:
l[index] = replacements[item]
counts[item] += 1
print("Replacement counts:")
for k, v in counts.items():
print("{} {:>5}".format(k, v))
return ''.join(l)
There are better ways to do these things but I'll try to change your code minimally so you will learn something.
The first function's issue is that when you split the sentence like "Hello." there will be two sentences in your fixStr list that the last one is an empty string; so the first index of an empty string is out of range. fix it by doing this.
def fix_capitalization(usrStr):
count = 0
fixStr = usrStr.split('.')
for sentence in fixStr:
# changed line
if sentence != "":
sentence[0].upper()
count += 1
print('Number of letters capitalized: %d' % count)
print('Edited text: %s' % fixStr)
In second snippet you are trying to write, when you pass a string to list() you get a list of characters of that string. So all you need to do is to iterate over the elements of the list and replace them and after that get string from the list.
def replace_punctuation(usrStr):
newStr = ""
s = list(usrStr)
exclamationCount = 0
semicolonCount = 0
for c in s:
if c == '!':
c = '.'
exclamationCount += 1
if c == ';':
c = ','
semicolonCount += 1
newStr = newStr + c
print(newStr)
print(semicolonCount)
print(exclamationCount)
Hope I helped!
Python has a nice build in function for this
for str in list:
new_str = str.replace('!', '.').replace(';', ',')
You can write a oneliner to get a new list
new_list = [str.replace('!', '.').replace(';', ',') for str in list]
You also could go for the split/join method
new_str = '.'.join(str.split('!'))
new_str = ','.join(str.split(';'))
To count capitalized letters you could do
result = len([cap for cap in str if str(cap).isupper()])
And to capitalize them words just use the
str.capitalize()
Hope this works out for you

Replace Duplicate String Characters

I need to convert a string word where each character that appears only once should be appear as '(' in the new string. Any duplicate characters in the original string should be replaced with ')'.
My code below...
def duplicate_encode(word):
new_word = ''
for char in word:
if len(char) > 1:
new_word += ')'
else:
new_word += '('
return new_word
The test I'm not passing is as follows:
'((((((' should equal '()()()'
This would suggest that, if for example, the input is "recede," the output should read ()()().
Your Code is Good just need some alteration it will be great.
def duplicate_encode(word):
"""
To replace the duplicate letter with ")" in a string.
if given letter is unique it replaced with "("
"""
word_dict = {} # initialize a dictionary
new_word = ""
for i in set(word): # this loop is used to count duplicate words
word_count = word.count(i)
word_dict[i] = word_count # add letter and count of the letter to dictionary
for i in word:
if word_dict[i] > 1:
new_word += ")"
else:
new_word += "("
print new_word
duplicate_encode("recede")
I think you got the answer :)
Just because (it's late and) it's possible:
def duplicate_encode(word):
return (lambda w: ''.join(('(', ')')[c in w[:i] + w[i+1:]] for i, c in enumerate(w)))(word.lower())
print(duplicate_encode("rEcede"))
OUTPUT
> python3 test.py
()()()
>
Seems like your result is based on the number of occurrences of a character in the word, you can use Counter to keep track of that:
def duplicate_encode(word):
from collections import Counter
word = word.lower() # to disregard case
counter = Counter(word)
new_word = ''
for char in word:
if counter[char] > 1: # if the character appears more than once in the word
# translate it to )
new_word += ')'
else:
new_word += '('
return new_word
duplicate_encode('recede')
# '()()()'

Swapping uppercase and lowercase in a string [duplicate]

This question already has answers here:
How can I invert (swap) the case of each letter in a string?
(8 answers)
How can I use `return` to get back multiple values from a loop? Can I put them in a list?
(2 answers)
Closed 6 months ago.
I would like to change the chars of a string from lowercase to uppercase.
My code is below, the output I get with my code is a; could you please tell me where I am wrong and explain why?
Thanks in advance
test = "AltERNating"
def to_alternating_case(string):
words = list(string)
for word in words:
if word.isupper() == True:
return word.lower()
else:
return word.upper()
print to_alternating_case(test)
If you want to invert the case of that string, try this:
>>> 'AltERNating'.swapcase()
'aLTernATING'
There are two answers to this: an easy one and a hard one.
The easy one
Python has a built in function to do that, i dont exactly remember what it is, but something along the lines of
string.swapcase()
The hard one
You define your own function. The way you made your function is wrong, because
iterating over a string will return it letter by letter, and you just return the first letter instead of continuing the iteration.
def to_alternating_case(string):
temp = ""
for character in string:
if character.isupper() == True:
temp += character.lower()
else:
temp += word.upper()
return temp
Your loop iterates over the characters in the input string. It then returns from the very first iteration. Thus, you always get a 1-char return value.
test = "AltERNating"
def to_alternating_case(string):
words = list(string)
rval = ''
for c in words:
if word.isupper():
rval += c.lower()
else:
rval += c.upper()
return rval
print to_alternating_case(test)
That's because your function returns the first character only. I mean return keyword breaks your for loop.
Also, note that is unnecessary to convert the string into a list by running words = list(string) because you can iterate over a string just as you did with the list.
If you're looking for an algorithmic solution instead of the swapcase() then modify your method this way instead:
test = "AltERNating"
def to_alternating_case(string):
res = ""
for word in string:
if word.isupper() == True:
res = res + word.lower()
else:
res = res + word.upper()
return res
print to_alternating_case(test)
You are returning the first alphabet after looping over the word alternating which is not what you are expecting. There are some suggestions to directly loop over the string rather than converting it to a list, and expression if <variable-name> == True can be directly simplified to if <variable-name>. Answer with modifications as follows:
test = "AltERNating"
def to_alternating_case(string):
result = ''
for word in string:
if word.isupper():
result += word.lower()
else:
result += word.upper()
return result
print to_alternating_case(test)
OR using list comprehension :
def to_alternating_case(string):
result =[word.lower() if word.isupper() else word.upper() for word in string]
return ''.join(result)
OR using map, lambda:
def to_alternating_case(string):
result = map(lambda word:word.lower() if word.isupper() else word.upper(), string)
return ''.join(result)
You should do that like this:
test = "AltERNating"
def to_alternating_case(string):
words = list(string)
newstring = ""
if word.isupper():
newstring += word.lower()
else:
newstring += word.upper()
return alternative
print to_alternating_case(test)
def myfunc(string):
i=0
newstring=''
for x in string:
if i%2==0:
newstring=newstring+x.lower()
else:
newstring=newstring+x.upper()
i+=1
return newstring
contents='abcdefgasdfadfasdf'
temp=''
ss=list(contents)
for item in range(len(ss)):
if item%2==0:
temp+=ss[item].lower()
else:
temp+=ss[item].upper()
print(temp)
you can add this code inside a function also and in place of print use the return key
string=input("enter string:")
temp=''
ss=list(string)
for item in range(len(ss)):
if item%2==0:
temp+=ss[item].lower()
else:
temp+=ss[item].upper()
print(temp)
Here is a short form of the hard way:
alt_case = lambda s : ''.join([c.upper() if c.islower() else c.lower() for c in s])
print(alt_case('AltERNating'))
As I was looking for a solution making a all upper or all lower string alternating case, here is a solution to this problem:
alt_case = lambda s : ''.join([c.upper() if i%2 == 0 else c.lower() for i, c in enumerate(s)])
print(alt_case('alternating'))
You could use swapcase() method
string_name.swapcase()
or you could be a little bit fancy and use list comprehension
string = "thE big BROWN FoX JuMPeD oVEr thE LAZY Dog"
y = "".join([val.upper() if val.islower() else val.lower() for val in string])
print(y)
>>> 'THe BIG brown fOx jUmpEd OveR THe lazy dOG'
This doesn't use any 'pythonic' methods and gives the answer in a basic logical format using ASCII :
sentence = 'aWESOME is cODING'
words = sentence.split(' ')
sentence = ' '.join(reversed(words))
ans =''
for s in sentence:
if ord(s) >= 97 and ord(s) <= 122:
ans = ans + chr(ord(s) - 32)
elif ord(s) >= 65 and ord(s) <= 90 :
ans = ans + chr(ord(s) + 32)
else :
ans += ' '
print(ans)
So, the output will be : Coding IS Awesome

Python: Best Way to remove duplicate character from string

How can I remove duplicate characters from a string using Python? For example, let's say I have a string:
foo = "SSYYNNOOPPSSIISS"
How can I make the string:
foo = SYNOPSIS
I'm new to python and What I have tired and it's working. I knew there is smart and best way to do this.. and only experience can show this..
def RemoveDupliChar(Word):
NewWord = " "
index = 0
for char in Word:
if char != NewWord[index]:
NewWord += char
index += 1
print(NewWord.strip())
NOTE: Order is important and this question is not similar to this one.
Using itertools.groupby:
>>> foo = "SSYYNNOOPPSSIISS"
>>> import itertools
>>> ''.join(ch for ch, _ in itertools.groupby(foo))
'SYNOPSIS'
This is a solution without importing itertools:
foo = "SSYYNNOOPPSSIISS"
''.join([foo[i] for i in range(len(foo)-1) if foo[i+1]!= foo[i]]+[foo[-1]])
Out[1]: 'SYNOPSIS'
But it is slower than the others method!
How about this:
oldstring = 'SSSYYYNNNOOOOOPPPSSSIIISSS'
newstring = oldstring[0]
for char in oldstring[1:]:
if char != newstring[-1]:
newstring += char
def remove_duplicates(astring):
if isinstance(astring,str) :
#the first approach will be to use set so we will convert string to set and then convert back set to string and compare the lenght of the 2
newstring = astring[0]
for char in astring[1:]:
if char not in newstring:
newstring += char
return newstring,len(astring)-len(newstring)
else:
raise TypeError("only deal with alpha strings")
I've discover that solution with itertools and with list comprehesion even the solution when we compare the char to the last element of the list doesn't works
def removeDuplicate(s):
if (len(s)) < 2:
return s
result = []
for i in s:
if i not in result:
result.append(i)
return ''.join(result)
How about
foo = "SSYYNNOOPPSSIISS"
def rm_dup(input_str):
newstring = foo[0]
for i in xrange(len(input_str)):
if newstring[(len(newstring) - 1 )] != input_str[i]:
newstring += input_str[i]
else:
pass
return newstring
print rm_dup(foo)
You can try this:
string1 = "example1122334455"
string2 = "hello there"
def duplicate(string):
temp = ''
for i in string:
if i not in temp:
temp += i
return temp;
print(duplicate(string1))
print(duplicate(string2))

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