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StarKill riddle in Python

Riddle:
Return a version of the given string, where for every star (*) in the string the star and the chars immediately to its left and right are gone. So "ab*cd" yields "ad" and "ab**cd" also yields "ad".
I'm wondering if there's a pythonish way to improve this algorithm:
def starKill(string):
result = ''
for idx in range(len(string)):
if(idx == 0 and string[idx] != '*'):
result += string[idx]
elif (idx > 0 and string[idx] != '*' and (string[idx-1]) != '*'):
result += string[idx]
elif (idx > 0 and string[idx] == '*' and (string[idx-1]) != '*'):
result = result[0:len(result) - 1]
return result
starKill("wacy*xko") yields wacko

Here's a numpy solution just for fun:
def star_kill(string, target='*'):
arr = np.array(list(string))
mask = arr != '*'
mask[1:] &= mask[:-1]
mask[:-1] &= mask[1:]
arr = arr[mask]
return arr[mask].view(dtype=f'U{arr.size}').item()

Regular expression?
>>> import re
>>> for s in "ab*cd", "ab**cd", "wacy*xko", "*Mad*Physicist*":
print(re.sub(r'\w?\*\w?', '', s))
ad
ad
wacko
ahysicis

You can do this by iterating over the string three times in parallel. Each iteration will be shifted relative to the next by one character. The middle one is the one that will provide the valid letters, the other two let us check if adjacent characters are stars. The two flanking iterators require dummy values to represent "before the start" and "after the end" of the string. There are a variety of ways to set that up, I'm using itertools.chain (and .islice) to fill in None for the dummy values. But you could use plain string and iterator manipulation if you prefer (i.e. iter('x' + string) and iter(string[1:] + 'x')):
import itertools
def star_kill(string):
main_iterator = iter(string)
look_behind = itertools.chain([None], string)
look_ahead = itertools.chain(itertools.islice(string, 1, None), [None])
return "".join(a for a, b, c in zip(main_iterator, look_behind, look_ahead)
if a != '*' and b != '*' and c != '*')

Not sure whether or not it's "Pythonic," but the problem can be solved with regular expressions.
import re
def starkill(s):
s = re.sub(".{0,1}\\*{1,}.{0,1}", "", s)
return s
For those not familiar with regex, I'll break that long string down:
Prefix
".{0,1}"
This specifies we want the replaced section to begin with either 0 or 1 of any character. If there is a character before the star, we want to replace it; otherwise, we still want the expression to hit if the star is at the very beginning of the input string.
Star
"\\*{1,}"
This specifies that the middle of the expression must contain an asterisk character, but it can also contain more than one. For instance, "a****b" will still hit, even though there are four stars. We need a backslash before the asterisk because regex has asterisk as a reserved character, and we need a second backslash before that because Python strings reserve the backslash character.
Suffix
.{0,1}
Same as the prefix. The expression can either end with one or zero of any character.
Hope that helps!

Taking long time to execute Python code for the definition

This is the problem definition:
Given a string of lowercase letters, determine the index of the
character whose removal will make a palindrome. If is already a
palindrome or no such character exists, then print -1. There will always
be a valid solution, and any correct answer is acceptable. For
example, if "bcbc", we can either remove 'b' at index or 'c' at index.
I tried this code:
# !/bin/python
import sys
def palindromeIndex(s):
# Complete this function
length = len(s)
index = 0
while index != length:
string = list(s)
del string[index]
if string == list(reversed(string)):
return index
index += 1
return -1
q = int(raw_input().strip())
for a0 in xrange(q):
s = raw_input().strip()
result = palindromeIndex(s)
print(result)
This code works for the smaller values. But taken hell lot of time for the larger inputs.
Here is the sample: Link to sample
the above one is the bigger sample which is to be decoded. But at the solution must run for the following input:
Input (stdin)
3
aaab
baa
aaa
Expected Output
3
0
-1
How to optimize the solution?

Here is a code that is optimized for the very task
def palindrome_index(s):
# Complete this function
rev = s[::-1]
if rev == s:
return -1
for i, (a, b) in enumerate(zip(s, rev)):
if a != b:
candidate = s[:i] + s[i + 1:]
if candidate == candidate[::-1]:
return i
else:
return len(s) - i - 1
First we calculate the reverse of the string. If rev equals the original, it was a palindrome to begin with. Then we iterate the characters at the both ends, keeping tab on the index as well:
for i, (a, b) in enumerate(zip(s, rev)):
a will hold the current character from the beginning of the string and b from the end. i will hold the index from the beginning of the string. If at any point a != b then it means that either a or b must be removed. Since there is always a solution, and it is always one character, we test if the removal of a results in a palindrome. If it does, we return the index of a, which is i. If it doesn't, then by necessity, the removal of b must result in a palindrome, therefore we return its index, counting from the end.

There is no need to convert the string to a list, as you can compare strings. This will remove a computation that is called a lot thus speeding up the process. To reverse a string, all you need to do is used slicing:
>>> s = "abcdef"
>>> s[::-1]
'fedcba'
So using this, you can re-write your function to:
def palindromeIndex(s):
if s == s[::-1]:
return -1
for i in range(len(s)):
c = s[:i] + s[i+1:]
if c == c[::-1]:
return i
return -1
and the tests from your question:
>>> palindromeIndex("aaab")
3
>>> palindromeIndex("baa")
0
>>> palindromeIndex("aaa")
-1
and for the first one in the link that you gave, the result was:
16722
which computed in about 900ms compared to your original function which took 17000ms but still gave the same result. So it is clear that this function is a drastic improvement. :)

How to append even and odd chars python

I want to convert all the even letters using one function and all the odd numbers using another function. So, each letter represents 0-25 correspsonding with a-z, so a,c,e,g,i,k,m,o,q,s,u,w,y are even characters.
However, only my even letters are converting correctly.
def encrypt(plain):
charCount = 0
answer=[]
for ch in plain:
if charCount%2==0:
answer.append(pycipher.Affine(7,6).encipher(ch))
else:
answer.append(pycipher.Affine(3,0).encipher(ch))
return ''.join(answer)

You never change charCount in your loop -- So it starts at 0 and stays at 0 which means that each ch will be treated as "even".
Based on your update, you actually want to check if the character is odd or even based on it's "index" in the english alphabet. Having some sort of mapping of characters to numbers is helpful here. You could build it yourself:
alphabet = 'abcde...' # string.ascii_lowercase?
mapping = {k: i for i, k in enumerate(alphabet)}
OR we can use the builtin ord noticing that ord('a') produces an odd result, ord('b') is even, etc.
def encrypt(plain):
answer=[]
for ch in plain:
if ord(ch) % 2 == 1: # 'a', 'c', 'e', ...
answer.append(pycipher.Affine(7,6).encipher(ch))
else: # 'b', 'd', 'f', ...
answer.append(pycipher.Affine(3,0).encipher(ch))
return ''.join(answer)

Your basic approach is to re-encrypt a letter each time you see it. With only 26 possible characters to encrypt, it is probably worth pre-encrypting them, then just performing a lookup for each character in the plain text. While doing that, you don't need to compute the position of each character, because you know you are alternating between even and odd the entire time.
import string
def encrypt(plain):
# True == 1, False == 0
fs = [pycipher.Affine(3,0).encipher,
pycipher.Affine(7,6).encipher]
is_even = True # assuming "a" is even; otherwise, just set this to False
d = dict()
for ch in string.ascii_lowercase:
f = fs[is_even]
d[ch] = f(ch)
is_even = not is_even
return ''.join([d[ch] for ch in plain])
You can also use itertools.cycle to simplify the alternation for you.
def encrypt(plain):
# again, assuming a is even. If not, reverse this list
fs = itertools.cycle([pycipher.Affine(3,0).encipher,
pycipher.Affine(7,6).encipher])
d = dict((ch, f(ch)) for f, ch in zip(fs, string.ascii_lowercase))
return ''.join([d[ch] for ch in plain])

This are my two cents on that. What #mgilson is proposing also works of course but not in the way you specified (in the comments). Try to debug your code in your head after writing it.. Go through the for loop and perform 1-2 iterations to see whether the variables take the values you intended them to. charCount is never reassigned a value. It is always 0. And, yes charCount += 1 would make it change but not in the way you want it to..
def encrypt(plain):
alphabet = 'abcdefghijklmnopqrwstuvwxyz'
answer = ''
for letter in plain:
try:
if alphabet.index(letter.lower()) % 2 == 0:
answer += pycipher.Affine(7, 6).encipher(letter)
else:
answer += pycipher.Affine(3, 0).encipher(letter)
except:
answer += letter
return answer
my_text = 'Your question was not very clear OP'
encripted_text = encrypt(my_text)
Also, i would not use ord(ch) because ord('a') = 97 and not 0 therefore odd instead of even.

Since your notion of even letter is based on the position of a character in the alphabet, you could use ord(), like this:
if ord(ch)%2==0:
Note that ord('a') and ord('A') are both odd, so that would make a go in the else part. If you want the opposite, then just negate the condition:
if ord(ch)%2!=0:

Finding common letters between 2 strings in Python

For a homework assignment, I have to take 2 user inputted strings, and figure out how many letters are common (in the same position of both strings), as well as find common letters.. For example for the two strings 'cat' and 'rat', there are 2 common letter positions (which are positions 2 and 3 in this case), and the common letters are also 2 because 'a' is found one and 't' is found once too..
So I made a program and it worked fine, but then my teacher updated the homework with more examples, specifically examples with repetitive letters, and my program isn't working for that.. For example, with strings 'ahahaha' and 'huhu' - there are 0 common letters in same positions, but there's 3 common letters between them (because 'h' in string 2 appears in string 1, three times..)
My whole issue is that I can't figure out how to count if "h" appears multiple times in the first string, as well as I don't know how to NOT check the SECOND 'h' in huhu because it should only count unique letters, so the overall common letter count should be 2..
This is my current code:
S1 = input("Enter a string: ")
S2 = input("Enter a string: ")
i = 0
big_string = 0
short_string = 0
same_letter = 0
common_letters = 0
if len(S1) > len(S2):
big_string = len(S1)
short_string = len(S2)
elif len(S1) < len(S2):
big_string = len(S2)
short_string = len(S1)
elif len(S1) == len(S2):
big_string = short_string = len(S1)
while i < short_string:
if (S1[i] == S2[i]) and (S1[i] in S2):
same_letter += 1
common_letters += 1
elif (S1[i] == S2[i]):
same_letter += 1
elif (S1[i] in S2):
common_letters += 1
i += 1
print("Number of positions with the same letter: ", same_letter)
print("Number of letters from S1 that are also in S2: ", common_letters)
So this code worked for strings without common letters, but when I try to use it with "ahahaha" and "huhu" I get 0 common positions (which makes sense) and 2 common letters (when it should be 3).. I figured it might work if I tried to add the following:
while x < short_string:
if S1[i] in S2[x]:
common_letters += 1
else:
pass
x += 1
However this doesn't work either...
I am not asking for a direct answer or piece of code to do this, because I want to do it on my own, but I just need a couple of hints or ideas how to do this..
Note: I can't use any functions we haven't taken in class, and in class we've only done basic loops and strings..

You need a data structure like multidict. To my knowledge, the most similar data structure in standard library is Counter from collections.
For simple frequency counting:
>>> from collections import Counter
>>> strings = ['cat', 'rat']
>>> counters = [Counter(s) for s in strings]
>>> sum((counters[0] & counters[1]).values())
2
With index counting:
>>> counters = [Counter(zip(s, range(len(s)))) for s in strings]
>>> sum(counters[0] & counters[1].values())
2
For your examples ahahaha and huhu, you should get 2 and 0, respectively since we get two h but in wrong positions.
Since you can't use advanced constructs, you just need to simulate counter with arrays.
Create 26 elements arrays
Loop over strings and update relevant index for each letter
Loop again over arrays simultaneously and sum the minimums of respective indexes.

A shorter version is this:
def gen1(listItem):
returnValue = []
for character in listItem:
if character not in returnValue and character != " ":
returnValue.append(character)
return returnValue
st = "first string"
r1 = gen1(st)
st2 = "second string"
r2 = gen1(st2)
if len(st)> len(st2):
print list(set(r1).intersection(r2))
else:
print list(set(r2).intersection(r1))
Note:
This is a pretty old post but since its got new activity,I posted my version.

Since you can't use arrays or lists,
Maybe try to add every common character to a var_string then test
if c not in var_string:
before incrementing your common counter so you are not counting the same character multiple times.

You are only getting '2' because you're only going to look at 4 total characters out of ahahaha (because huhu, the shortest string, is only 4 characters long). Change your while loop to go over big_string instead, and then add (len(S2) > i) and to your two conditional tests; the last test performs an in, so it won't cause a problem with index length.
NB: All of the above implicitly assumes that len(S1) >= len(S2); that should be easy enough to ensure, using a conditional and an assignment, and it would simplify other parts of your code to do so. You can replace the first block entirely with something like:
if (len(S2) > len(S1)): (S2, S1) = (S1, S2)
big_string = len(S1)
short_string = len(S2)

We can solve this by using one for loop inside of another as follows
int y=0;
for(i=0;i<big_string ;i++)
{
for(j=0;j<d;j++)
{
if(s1[i]==s2[j])
{y++;}
}
If you enter 'ahahaha' and 'huhu' this code take first character of big
string 'a' when it goes into first foor loop. when it enters into second for loop
it takes first letter of small string 'h' and compares them as they are not
equal y is not incremented. In next step it comes out of second for loop but
stays in first for loop so it consider first character of big string 'a' and
compares it against second letter of small string 'u' as 'j' is incremented even
in this case both of them are not equal and y remains zero. Y is incremented in
the following cases:-
when it compares second letter of big string 'h' and small letter of first string y is incremented for once i,e y=1;
when it compares fourth letter of big string 'h' and small letter of first string y is incremented again i,e y=2;
when it compares sixth letter of big string 'h' and small letter of first string y is incremented again i,e y=3;
Final output is 3. I think that is what we want.

Find symmetric words in a text [duplicate]

This question already has answers here:
how to find words that made up of letter exactly facing each other? (python) [closed]
(4 answers)
Closed 9 years ago.
I have to write a function which takes one arguments text containing a block of text in the form of a str, and returns a sorted list of “symmetric” words. A symmetric word is defined as a word where for all values i, the letter i positions from the start of the word and the letter i positions from the end of the word are equi-distant from the respective ends of the alphabet. For example, bevy is a symmetric word as: b (1 position from the start of the word) is the second letter of the alphabet and y (1 position from the end of the word) is the second-last letter of the alphabet; and e (2 positions from the start of the word) is the fifth letter of the alphabet and v (2 positions from the end of the word) is the fifth-last letter of the alphabet.
For example:
>>> symmetrics("boy bread aloz bray")
['aloz','boy']
>>> symmetrics("There is a car and a book;")
['a']
All I can think about the solution is this but I can't run it since it's wrong:
def symmetrics(text):
func_char= ",.?!:'\/"
for letter in text:
if letter in func_char:
text = text.replace(letter, ' ')
alpha1 = 'abcdefghijklmnopqrstuvwxyz'
alpha2 = 'zyxwvutsrqponmlkjihgfedcba'
sym = []
for word in text.lower().split():
n = range(0,len(word))
if word[n] == word[len(word)-1-n]:
sym.append(word)
return sym
The code above doesn't take into account the position of alpha1 and alpha2 as I don't know how to put it. Is there anyone can help me?

Here is a hint:
In [16]: alpha1.index('b')
Out[16]: 1
In [17]: alpha2.index('y')
Out[17]: 1
An alternative way to approach the problem is by using the str.translate() method:
import string
def is_sym(word):
alpha1 = 'abcdefghijklmnopqrstuvwxyz'
alpha2 = 'zyxwvutsrqponmlkjihgfedcba'
tr = string.maketrans(alpha1, alpha2)
n = len(word) // 2
return word[:n] == word[::-1][:n].translate(tr)
print(is_sym('aloz'))
print(is_sym('boy'))
print(is_sym('bread'))
(The building of the translation table can be easily factored out.)

The for loop could be modified as:
for word in text.lower().split():
for n in range(0,len(word)//2):
if alpha1.index(word[n]) != alpha2.index(word[len(word)-1-n]):
break
else:
sym.append(word)
return sym

According to your symmetric rule, we may verify a symmetric word with the following is_symmetric_word function:
def is_symmetric_word(word):
alpha1 = 'abcdefghijklmnopqrstuvwxyz'
alpha2 = 'zyxwvutsrqponmlkjihgfedcba'
length = len(word)
for i in range(length / 2):
if alpha1.index(word[i]) != alpha2.index(word[length - 1 - i]):
return False
return True
And then the whole function to get all unique symmetric words out of a text can be defined as:
def is_symmetrics(text):
func_char= ",.?!:'\/;"
for letter in text:
if letter in func_char:
text = text.replace(letter, ' ')
sym = []
for word in text.lower().split():
if is_symmetric_word(word) and not (word in sym):
sym.append(word)
return sym
The following are two test cases from you:
is_symmetrics("boy bread aloz bray") #['boy', 'aloz']
is_symmetrics("There is a car and a book;") #['a']

Code first. Discussion below the code.
import string
# get alphabet and reversed alphabet
try:
# Python 2.x
alpha1 = string.lowercase
except AttributeError:
# Python 3.x and newer
alpha1 = string.ascii_lowercase
alpha2 = alpha1[::-1] # use slicing to reverse alpha1
# make a dictionary where the key, value pairs are symmetric
# for example symd['a'] == 'z', symd['b'] == 'y', and so on
_symd = dict(zip(alpha1, alpha2))
def is_symmetric_word(word):
if not word:
return False # zero-length word is not symmetric
i1 = 0
i2 = len(word) - 1
while True:
if i1 >= i2:
return True # we have checked the whole string
# get a pair of chars
c1 = word[i1]
c2 = word[i2]
if _symd[c1] != c2:
return False # the pair wasn't symmetric
i1 += 1
i2 -= 1
# note, added a space to list of chars to filter to a space
_filter_to_space = ",.?!:'\/ "
def _filter_ch(ch):
if ch in _filter_to_space:
return ' ' # return a space
elif ch in alpha1:
return ch # it's an alphabet letter so return it
else:
# It's something we don't want. Return empty string.
return ''
def clean(text):
return ''.join(_filter_ch(ch) for ch in text.lower())
def symmetrics(text):
# filter text: keep only chars in the alphabet or spaces
for word in clean(text).split():
if is_symmetric_word(word):
# use of yield makes this a generator.
yield word
lst = list(symmetrics("The boy...is a yob."))
print(lst) # prints: ['boy', 'a', 'yob']
No need to type the alphabet twice; we can reverse the first one.
We can make a dictionary that pairs each letter with its symmetric letter. This will make it very easy to test whether any given pair of letters is a symmetric pair. The function zip() makes pairs from two sequences; they need to be the same length, but since we are using a string and a reversed copy of the string, they will be the same length.
It's best to write a simple function that does one thing, so we write a function that does nothing but check if a string is symmetric. If you give it a zero-length string it returns False, otherwise it sets i1 to the first character in the string and i2 to the last. It compares characters as long as they continue to be symmetric, and increments i1 while decrementing i2. If the two meet or pass each other, we know we have seen the whole string and it must be symmetric, in which case we return True; if it ever finds any pair of characters that are not symmetric, it returns False. We have to do the check for whether i1 and i2 have met or passed at the top of the loop, so it won't try to check if a character is its own symmetric character. (A character can't be both 'a' and 'z' at the same time, so a character is never its own symmetric character!)
Now we write a wrapper that filters out the junk, splits the string into words, and tests each word. Not only does it convert the chosen punctuation characters to spaces, but it also strips out any unexpected characters (anything not an approved punctuation char, a space, or a letter). That way we know nothing unexpected will get through to the inner function. The wrapper is "lazy"... it is a generator that yields up one word at a time, instead of building the whole list and returning that. It's easy to use list() to force the generator's results into a list. If you want, you can easily modify this function to just build a list and return it.
If you have any questions about this, just ask.
EDIT: The original version of the code didn't do the right thing with the punctuation characters; this version does. Also, as #heltonbiker suggested, why type the alphabet when Python has a copy of it you can use? So I made that change too.
EDIT: #heltonbiker's change introduced a dependency on Python version! I left it in with a suitable try:/except block to handle the problem. It appears that Python 3.x has improved the name of the lowercase ASCII alphabet to string.ascii_lowercase instead of plain string.lowercase.