So, basically I want to check if a string is palindrome or not
I know I can simply use
def palindrome(n):
return n == n[::-1]
It works fine, but if I want a string, for example "A cat, one crab, lol: barcenotaca."
to be true as well, what can I do?
def palindrome(n):
n = n.lower()
n = ''.join(char for char in n if char.isalpha())
return n==n[::-1]
Checking for char.isalpha() lets you ignore everything that's not a letter (when combined with the n = n.lower(), it basically checks for only lowercase letters) That way, you ignore all the punctuation and whitespace
def palindrome(n):
x = ''.join([x.lower() for x in n if x.isalpha()])
return x == x[::-1]
try checking if every letter in the given string is in the alphabet first, goes something like this
import string
def palindrome(str):
alphabet=string.ascii_lowercase
temp=""
for c in str.lower():
if c in alphabet:
temp+=c
return temp==temp[::-1]
def palindrome():
stringOne = alphaCheck("give me a lovely palindrome")
stringTwo = stringOne[::-1]
for i in stringOne:
print(i)
for j in stringTwo:
print(j)
if stringOne == stringTwo:
return True
else:
return False
def alphaCheck(prompt):
stringOne = input(prompt)
stringOne = stringOne.replace(" ", "")
stringOne = stringOne.lower()
while not stringOne.isalpha():
stringOne = input(prompt)
return stringOne
print(palindrome())
this method allows you to input your own palindrome and also checks to ensure that it only contains alphabetic characters. if you want to change that to only digits change while not stringOne.isalpha( ) to is
Related
So I'm a little confused as far as putting this small code together. My teacher gave me this info:
Iterate over the string and remove any triplicated letters (e.g.
"byeee mmmy friiiennd" becomes "bye my friennd"). You may assume any
immediate following same letters are a triplicate.
I've mostly only seen examples for duplicates, so how do I remove triplicates? My code doesn't return anything when I run it.
def removeTriplicateLetters(i):
result = ''
for i in result:
if i not in result:
result.append(i)
return result
def main():
print(removeTriplicateLetters('byeee mmmy friiiennd'))
main()
I have generalized the scenario with "n". In your case, you can pass n=3 as below
def remove_n_plicates(input_string, n):
i=0
final_string = ''
if not input_string:
return final_string
while(True):
final_string += input_string[i]
if input_string[i:i+n] == input_string[i]*n:
i += n
else:
i += 1
if i >= len(input_string):
break
return final_string
input_string = "byeee mmmy friiiennd"
output_string = remove_n_plicates(input_string, 3)
print(output_string)
# bye my friennd
You can use this for any "n" value now (where n > 0 and n < length of input string)
Your code returns an empty string because that's exactly what you coded:
result = ''
for i in result:
...
return result
Since result is an empty string, you don't enter the loop at all.
If you did enter the loop you couldn't return anything:
for i in result:
if i not in result:
The if makes no sense: to get to that statement, i must be in result
Instead, do as #newbie showed you. Iterate through the string, looking at a 3-character slice. If the slice is equal to 3 copies of the first character, then you've identified a triplet.
if input_string[i:i+n] == input_string[i]*n:
Without going in to writing the code to resolve the problem.
When you iterate over the string, add that iteration to a new string.
If the next iteration is the same as the previous iteration then do not add that to the new string.
This will catch both the triple and the double characters in your problem.
Tweaked a previous answer to remove a few lines that were not needed.
def remove_n_plicates(input_string, n):
i=0
result = ''
while(True):
result += input_string[i]
if input_string[i:i+n] == input_string[i]*n:
i += n
else:
i += 1
if i >= len(input_string):
break
return result
input_string = "byeee mmmy friiiennd"
output_string = remove_n_plicates(input_string, 3)
print(output_string)
# bye my friennd
Here's a fun way using itertools.groupby:
def removeTriplicateLetters(s):
return ''.join(k*(l//3+l%3) for k,l in ((k,len(list(g))) for k, g in groupby(s)))
>>> removeTriplicateLetters('byeee mmmy friiiennd')
'bye my friennd'
just modifying #newbie solution and using stack data structure as solution
def remove_n_plicates(input_string, n):
if input_string =='' or n<1:
return None
w = ''
c = 0
if input_string!='':
tmp =[]
for i in range(len(input_string)):
if c==n:
w+=str(tmp[-1])
tmp=[]
c =0
if tmp==[]:
tmp.append(input_string[i])
c = 1
else:
if input_string[i]==tmp[-1]:
tmp.append(input_string[i])
c+=1
elif input_string[i]!=tmp[-1]:
w+=str(''.join(tmp))
tmp=[input_string[i]]
c = 1
w+=''.join(tmp)
return w
input_string = "byeee mmmy friiiennd nnnn"
output_string = remove_n_plicates(input_string, 3)
print(output_string)
output
bye my friennd nn
so this is a bit dirty but it's short and works
def removeTriplicateLetters(i):
result,string = i[:2],i[2:]
for k in string:
if result[-1]==k and result[-2]==k:
result=result[:-1]
else:
result+=k
return result
print(removeTriplicateLetters('byeee mmmy friiiennd'))
bye my friennd
You have already got a working solution. But here, I come with another way to achieve your goal.
def removeTriplicateLetters(sentence):
"""
:param sentence: The sentence to transform.
:param words: The words in the sentence.
:param new_words: The list of the final words of the new sentence.
"""
words = sentence.split(" ") # split the sentence into words
new_words = []
for word in words: # loop through words of the sentence
new_word = []
for char in word: # loop through characters in a word
position = word.index(char)
if word.count(char) >= 3:
new_word = [i for i in word if i != char]
new_word.insert(position, char)
new_words.append(''.join(new_word))
return ' '.join(new_words)
def main():
print(removeTriplicateLetters('byeee mmmy friiiennd'))
main()
Output: bye my friennd
I'm trying to write a function which given a string returns another string that flips its lowercase characters to uppercase and viceversa.
My current aproach is as follows:
def swap_case(s):
word = []
for char in s:
word.append(char)
for char in word:
if char.islower():
char.upper()
else:
char.lower()
str1 = ''.join(word)
However the string doesn't actually change, how should I alter the characters inside the loop to do so?
PS: I know s.swapcase() would easily solve this, but I want to alter the characters inside the string during a for loop.
def swap_case(s):
swapped = []
for char in s:
if char.islower():
swapped.append(char.upper())
elif char.isupper():
swapped.append(char.lower())
else:
swapped.append(char)
return ''.join(swapped)
you can use swapcase.
string.swapcase(s)
Return a copy of s, but with lower case letters converted to upper case and vice versa.
Source : Python docs
>>> name = 'Mr.Ed' or name= ["M","r",".","E","d"]
>>> ''.join(c.lower() if c.isupper() else c.upper() for c in name)
'mR.eD'
Your code is not working because you are not storing the chars after transforming them. You created a list with all chars and then, you access the values without actually saving them. Also, you should make your function return str1 or you will never get a result. Here's a workaround:
def swap_case(s):
word = []
for char in s:
if char.islower():
word.append(char.upper())
else:
word.append(char.lower())
str1 = ''.join(word)
return str1
By the way, I know you want to do it with a for loop, but you should know there are better ways to perform this which would be more pythonic.
You can use this code:
s = "yourString"
res = s.swapcase()
print(res)
Which will output:
YOURsTRING
def swap_case(s):
x = ''
for i in s:
if i.isupper():
i = i.lower()
else:
i = i.upper()
x += ''.join(i)
return x
I am trying to write a python program that checks if a given string is a pangram - contains all letters of the alphabet.
Therefore, "We promptly judged antique ivory buckles for the next prize" should return True while any string that does not contain every letter of the alphabet at least once should return False.
I believe I should be using RegEx for this one, but I'm not sure how. It should look similar to this:
import sys
import re
input_string_array = sys.stdin.readlines()
input_string = input_string_array[0]
if (re.search('string contains every letter of the alphabet',input_string):
return True
else:
return False
This is not something I'd solve with a regular expression, no. Create a set of the lowercased string and check if it is a superset of the letters of the alphabet:
import string
alphabet = set(string.ascii_lowercase)
def ispangram(input_string):
return set(input_string.lower()) >= alphabet
Only if every letter of the alphabet is in the set created from the input text will it be a superset; by using a superset and not equality, you allow for punctuation, digits and whitespace, in addition to the (ASCII) letters.
Demo:
>>> import string
>>> alphabet = set(string.ascii_lowercase)
>>> input_string = 'We promptly judged antique ivory buckles for the next prize'
>>> set(input_string.lower()) >= alphabet
True
>>> set(input_string[:15].lower()) >= alphabet
False
This is my solution in python:
alphabet = "abcdefghijklmnopqrstuvwxyz"
sentence = input()
sentence = sentence.lower()
missing = ''
for letter in alphabet:
if letter not in sentence:
missing = missing+letter
if (len(missing) != 0):
print("missing", missing)
else:
print("pangram")
You dont need regex. What you want can be done in two lines with good space efficiency.
ms = itertools.chain(range(ord("a"),ord("z")),range(ord("A"),ord("Z")))
flag = all(chr(o) in string for o in ms)
That's it. string is the string you want to check. flag will be either True or False depending on if all chars are in string or not.
A pangram is a function that contains at least each letter of the alphabet.
I have tried in this way:
def pangram():
n = str(input('give me a word to check if it is a pangram:\n'))
n = n.lower()
n = n.replace(' ','')
if not isinstance(n, str):
return n, False
elif set(n) >= set('abcdefghijklmnopqrstuvxywz'):
return n, True
else:
return n, False
The function isinstance(n, str) checks if n is a string. The function set() gives us a set. For example set('penny') returns {'y', 'e', 'p', 'n'}... as you see it is a set without the repeated letters.
I was doing the same exercise today, maybe it's not the best aproach, but I think it's easy to understand.
def ispangram(s):
stringy = ''
flag = True
validLetters = "abcdefghijklmnopqrstuvwxyz"
#transform the given string in simple string with no symbols only letters
for char in s.lower():
if(char in validLetters):
stringy += char
#check if all the letters of the alphabet exist on the string
for char in validLetters:
if(char in stringy):
pass
else:
flag = False
break
return flag
if(ispangram("We promptly judged antique ivory buckles for the next prize")):
print("It's PANGRAM!")
else:
print("It's not Pangram :(")
import string
def ispangram(str1, alphabet=string.ascii_lowercase):
return ''.join(sorted(set(str1.lower().replace(" ","")))) == alphabet
First changed all alphabets to lowercase and then removed all spaces using replace. Then Converted into set to have unique chars and then used sorted function to sort alphabetically. As sorted function gives a list, so join func to join it without spaces and then compared it to all lowercase chars.
Here is my solution:
def isaPangrams(s):
alph = list(string.ascii_lowercase)
s = s.lower()
s = list(s)
for letter in alph:
if letter not in s:
print('not pangram')
present='false'
break
if letter in s:
present = 'true'
if present == 'true':
print('pangram')
if __name__ == '__main__':
s = input()
answer = isaPangrams(s)
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
Right now, what I have is a code that I can test a single word whether it is a Palindrome or not. I have to input the word, and it will tell me whether it is a Palindrome (True) or if it is not (False)
I need to create one that Asks for a single word, then provides a True of False based on the word that is typed. This is what i have so far.
I really have no idea how to do this, any help would be greatly appreciated.
def isPalindrome(s):
if len(s) <= 1:
return True
else:
if s[0] != s[len(s)-1]:
return False
else:
return isPalindrome(s[1:len(s)-1])
print(isPalindrome("poop"))
Simply create a reversed string and check if both are equal.
def isPalindrome(s):
return s == s[::-1]
print(isPalindrome('poop'))
Also using a reversed string, but can be used in-line, as well (i.e., doesn't require a function).
def is_palindrome(word):
return word == ''.join(reversed(word))
print is_palindrome('hello') #False
print is_palindrome('racecar') #True
If you are asking how to get Python to take user input, then there are a couple options available to you. One option is to make your script run from the command line and use a command line argument and the other is to use the raw_input function. Both are implemented in the following code.
import sys
def isPalindrome(word):
...
if __name__ == '__main__':
if len(sys.argv) > 1:
inp = sys.argv[1]
else:
inp = raw_input("Type a word: ") # Which you might want to strip for white space
if isPalindrome(inp):
print inp,"is a palindrome"
else:
print inp,"is not a palindrome"
Several other people have suggested alternative implementations for isPalindrome that are probably better, but if you are doing this as an assignment and are supposed to use recursion, then keep using yours. Also, the raw_input function can be called anywhere that is convenient in your script and doesn't have to be used when called from the command line.
You might try this function:
def is_palindrome(text):
return text[:len(text)//2] == text[:(len(text)-1)//2:-1]
Here is example usage for reference:
>>> is_palindrome('')
True
>>> is_palindrome('a')
True
>>> is_palindrome('b')
True
>>> is_palindrome('aa')
True
>>> is_palindrome('ab')
False
>>> is_palindrome('ba')
False
>>> is_palindrome('bb')
True
>>> is_palindrome('aaa')
True
>>> is_palindrome('aab')
False
>>> is_palindrome('aba')
True
here is mine:
def isPalindrome(word):
word=str(word)
a=word[::-1]
if a!=word:
return False
elif word=='':
return False
else: return True
Using string subscripting
def is_palindrome(string):
return all(char == string[-i - 1] for i, char in enumerate(string))
Using list reversing
def is_palindrome(string):
to_list = list(string)
# reverse to_list now
to_list.reverse()
# our reversed list should be equal string as a list
# if string is a palindrome
return to_list == list(string)
This is mine:
def palindrome_checker(word):
return True if len(word) < 2 else (word[0] == word[-1]) and palindrome_checker(word[1:-1])
This should work.
return (s[:1+len(s)//2]==s[len(s)//2:][::-1])
For the people doing the palindrome test from testdome.com that takes into account casing, here is my solution:
def is_palindrome(word):
wordoriginal = []
wordreversed = []
for i in reversed(word):
i = i.lower()
wordreversed.append(i)
for i in word:
i = i.lower()
wordoriginal.append(i)
return wordoriginal == wordreversed
This is what I came up with, hope it works for you:
def palindrome_(word):
word = input("enter your word Here ")
return word == word[::-1]
print palindrome_("word")