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Counting words containing certain letter

The task was to write a function with name count_letter that takes a list of words and certain letter and returns amount of words where this letter is found at least once. And, we have to use a for loop.
So I did a list of some programming languages and letter "a", and tried to apply everything we learned so far plus some internet tutorials to understand how to translate human logic into lines of code, but obviously I am missing something, because it doesn't work :(
That is how my code looks like at the moment:
mylist = ['fortran', 'basic', 'java', 'python', 'c++']
letter = 'a'
a_list = []
def count_letter(mylist):
count = 0
for i in range(len(mylist)):
if letter in i:
count += 1
a_list.append(i)
return a_list
print(len(a_list))
Result is - no result. Online-python compiler returns response ** process exited - return code: 0 **
My question is - what could I miss or positioned wrongly that loop doesn't work. I want to understand it for myself.
In tutorials I have found one construction which returns correct answer (and looks very elegant and compact), but it has no function, so it is not really what we needed to write:
mylist = ['fortran', 'basic', 'java', 'python', 'c++']
letter = 'a'
res = len ([ele for ele in mylist if letter in ele])
print ('Amount of words containing a: ' +str(res))
Here system response: 3 , as expected.
Please tell me what should I check in code #1.

Few mistakes I found in your code:
When you do for i in range(len(mylist)), you are actually looping through the numbers 1,2,... instead of elements of mylist. So you have to use "for i in mylist" to loop through elements of the array mylist.
When you return from a function, the code that is after the return is not executed. So you have to print it first and then return from the function.
Don't forget to call the function. Otherwise the function won't be executed.
No need count variable as you can access the length using len method.
mylist = ['fortran', 'basic', 'java', 'python', 'c++']
letter = 'a'
a_list = []
def count_letter(mylist):
for i in mylist:
if letter in i:
a_list.append(i)
print(len(a_list))
return a_list
print(count_letter(mylist))
All the best in your journey!

Personally, I find Python source code easier to read when things are indented by four spaces. So, here is your function again with a wider indentation:
def count_letter(mylist):
count = 0
for i in range(len(mylist)):
if letter in i:
count += 1
a_list.append(i)
return a_list
print(len(a_list))
for i in range(...) will iterate over a collection of integers. So, i will take on a new integer value for each iteration of the loop. First i will be 0, then 1 on the next iteration, and so on.
You then ask if letter in i:. This can never be true. letter is a string, and i is an integer. A string can never be "in" an integer - so this if-statement will never execute. Rather, you want to check if letter is in the current word (the ith word in the list). The if-statement should read:
if letter in mylist[i]:
...
Where mylist[i] is the current word.
You then increment count and append i to a_list. You probably meant to append mylist[i] to a_list, but I don't see why you even need a_list. You just need count, since that keeps track of how many words you've encountered so far for which the condition is true. count is also the variable you should be returning in the end, since that is the purpose of the function: to return the number of words (not the words themselves) which contain a certain letter.
Also, the way your final print statement is indented makes it part of the function's body. It's after the return, though, which means it will never actually get a chance to print. When you use return inside a function, it ends the function, and flow of execution returns to the place from which the function was originally invoked.
One final change that needs to be applied, is that your function should accept a letter to look for as a parameter. Right now, your function only takes one parameter - the list of words through which to search.
Here are the changes I would apply to your code:
def count_letter(words, letter): # I find 'words' is a better name than 'mylist'.
count = 0
for i in range(len(words)):
current_word = words[i]
if letter in current_word:
count += 1
return count
Here is how you might use the function:
words = ["hello", "world", "apple", "sauce"]
letter = "e"
count = count_letter(words, letter)
print("The letter '{}' appeared in {} words.".format(letter, count))
Output:
The letter 'e' appeared in 3 words.

I think that 'takes' means that function must be defined with two parameters: words_list and letter:
def count_letter(words_list, letter):
Algorithm in natural language could be: give me sum of words where letter is present for every word in words list.
In Python it can be expressed as:
def count_letter(words_list, letter):
return sum(letter in word for word in words_list)
Some explanation: letter in word returns boolean value (True or False) and in Python bools are subclass of integer (True is 1 and False is 0). If letter is in word the result would be 1 and if not it would be 0. Summing up of results gives number of words where letter is present

I have read all your answers, some important points I wrote down, sat today again with my code, and after some more tries it worked...
So final version looks like:
words = ['fortran', 'basic', 'java', 'python', 'c++']
letter = "a"
def count_letter(words, letter):
count = 0
for word in words:
if letter in word:
count += 1
return count
print(count_letter((words),letter))
System response:
3
What is not yet obvious for me: correct indents (they were also part of a problem), and additional pair of parentheses around words in print line. But it comes with learning.
Thank you once again!

How to check if a word is a palindrome (Python)

Please help...
So the instruction says to program the computer to check whether a word is a palindrome or not. I inputted this code:
def is_palindrome(word):
counter_from_first_letter=0
counter_from_last_letter=-1
from_first_letter = word[counter_from_first_letter]
from_last_letter = word[counter_from_last_letter]
max_index_from_first= len(word)
max_index_from_last= (len(word))*-1
while from_first_letter == from_last_letter:
from_first_letter = word[counter_from_first_letter]
from_last_letter = word[counter_from_last_letter]
counter_from_first_letter += 1
counter_from_last_letter -= 1
return True
The problem is the computer only checks whether the first and last letters are the same, and if they are, it just returns true. How do I make sure the computer checks every single letter? Thanks

Maybe something like this:
def is_palindrome(word):
if word == word[::-1]:
return True
else:
return False

in python-3
name = 'madam'
print(name.find(name[::-1]) == 0)

Maybe you can try this: first convert your string into a list, then reverse the list and convert it back into a string. compare both the strings and if they match? they are palindromes, if not, they aren't.
'''checking whether a word is a palindrome
we first convert the word into a list then join it;
use an if statement to compare the two strings'''
def palindrome(string):#you need the input(string)
palindrome_check=[]#create an empty list
for character in string [::-1]:#create a list from the input
#(use a for loop because you now know the range)
#the [::-1] analyzes the characters in reverse
palindrome_check.append(character)#add each character to the new empty list
#print(palindrome_check)
rev_string= ''.join(palindrome_check)#.join -creates a string from the created list
print(rev_string)
#REMOVE SPECIAL CHARACTERS- IM THINKING OF A LOOPING THROUGH, BUT NOT SURE HOW TO IMPLEMENT IT
string=string.replace(' ', '')
rev_string=rev_string.replace(' ', '')
string=string.replace(',', '')
rev_string=rev_string.replace(',', '')
string=string.replace('.', '')
rev_string=rev_string.replace('.', '')
#THIS IS THE LOGIC: IT CHECKS BOTH STRINGS, if they are equal, it is a palindrome;
if string.lower()==rev_string.lower():
return True, print('It is a Palindrome')
else:
return False, print('It isnt a palindrome')
#call the function; key in the parameters-
palindrome= palindrome("No, Mel Gibson Is A Casinos Big Lemon")
#maybe we can try having a user key in the parameter? lets try
#palindrome=palindrome(input('kindly enter your word/phrase '))-wrong
#print('Kindly enter your word or phrase')
#user_palindrome=input('')
#palindrome=palindrome(user_palindrome)
#it wont work this way either
If you can have the user define the parameter(string), the better, if you know how to do this, kindly share.

To check whether a word or phrase is a palindrome, it be necessary to check if the original sentence is equal to the original sentence reversed.
word = "Eva can I see bees in a cave"
word_lower = word.lower().replace(" ", "")
if word_lower == word_lower[::-1]:
print("It's a palindrome")
else:
print("This is not a palindrome")

Cannot remove two vowels in a row

I have to enter a string, remove all spaces and print the string without vowels. I also have to print a string of all the removed vowels.
I have gotten very close to this goal, but for some reason when I try to remove all the vowels it will not remove two vowels in a row. Why is this? Please give answers for this specific block of code, as solutions have helped me solve the challenge but not my specific problem
# first define our function
def disemvowel(words):
# separate the sentence into separate letters in a list
no_v = list(words.lower().replace(" ", ""))
print no_v
# create an empty list for all vowels
v = []
# assign the number 0 to a
a = 0
for l in no_v:
# if a letter in the list is a vowel:
if l == "a" or l == "e" or l == "i" or l == "o" or l == "u":
# add it to the vowel list
v.append(l)
#print v
# delete it from the original list with a
del no_v[a]
print no_v
# increment a by 1, in order to keep a's position in the list moving
else:
a += 1
# print both lists with all spaces removed, joined together
print "".join(no_v)
print "".join(v)
disemvowel(raw_input(""))

Mistakes
So there are a lot of other, and perhaps better approaches to solve this problem. But as you mentioned I just discuss your failures or what you can do better.
1. Make a list of input word
There are a lot of thins you could do better
no_v = list(words.lower().replace(" ", ""))
You don't replaces all spaces cause of " " -> " " so just use this instead
no_v = list(words.lower().translate( None, string.whitespace))
2. Replace for loop with while loop
Because if you delete an element of the list the for l in no_v: will go to the next position. But because of the deletion you need the same position, to remove all the vowels in no_v and put them in v.
while a < len(no_v):
l = no_v[a]
3. Return the values
Cause it's a function don't print the values just return them. In this case replace the print no_v print v and just return and print them.
return (no_v,v) # returning both lists as tuple
4. Not a mistake but be prepared for python 3.x
Just try to use always print("Have a nice day") instead of print "Have a nice day"
Your Algorithm without the mistakes
Your algorithm now looks like this
import string
def disemvowel(words):
no_v = list(words.lower().translate( None, string.whitespace))
v = []
a = 0
while a < len(no_v):
l = no_v[a]
if l == "a" or l == "e" or l == "i" or l == "o" or l == "u":
v.append(l)
del no_v[a]
else:
a += 1
return ("".join(no_v),"".join(v))
print(disemvowel("Stackoverflow is cool !"))
Output
For the sentence Stackoverflow is cool !\n it outputs
('stckvrflwscl!', 'aoeoioo')
How I would do this in python
Not asked but I give you a solution I would probably use. Cause it has something to do with string replacement, or matching I would just use regex.
def myDisemvowel(words):
words = words.lower().translate( None, string.whitespace)
nv = re.sub("[aeiou]*","", words)
v = re.sub("[^a^e^i^o^u]*","", words)
return (nv, v)
print(myDisemvowel("Stackoverflow is cool !\n"))
I use just a regular expression and for the nv string I just replace all voewls with and empty string. For the vowel string I just replace the group of all non vowels with an empty string. If you write this compact, you could solve this with 2 lines of code (Just returning the replacement)
Output
For the sentence Stackoverflow is cool !\n it outputs
('stckvrflwscl!', 'aoeoioo')

You are modifying no_v while iterating through it. It'd be a lot simpler just to make two new lists, one with vowels and one without.
Another option is to convert it to a while loop:
while a < len(no_v):
l = no_v[a]
This way you have just a single variable tracking your place in no_v instead of the two you currently have.

For educational purposes, this all can be made significantly less cumbersome.
def devowel(input_str, vowels="aeiou"):
filtered_chars = [char for char in input_str
if char.lower() not in vowels and not char.isspace()]
return ''.join(filtered_chars)
assert devowel('big BOOM') == 'bgBM'
To help you learn, do the following:
Define a function that returns True if a particular character has to be removed.
Using that function, loop through the characters of the input string and only leave eligible characters.
In the above, avoid using indexes and len(), instead iterate over characters, as in for char in input_str:.
Learn about list comprehensions.
(Bonus points:) Read about the filter function.

exercise 12.4 from "think python", the script doesnt get to the end

What is the longest English word, that remains a valid English word, as you remove its
letters one at a time?
Now, letters can be removed from either end, or the middle, but you can’t rearrange any
of the letters. Every time you drop a letter, you wind up with another English word. If
you do that, you’re eventually going to wind up with one letter and that too is going
to be an English word—one that’s found in the dictionary. I want to know what’s the
longest word and how many letters does it have?
I’m going to give you a little modest example: Sprite. Ok? You start off with sprite,
you take a letter off, one from the interior of the word, take the r away, and we’re left
with the word spite, then we take the e off the end, we’re left with spit, we take the s off,
we’re left with pit, it, and I.
I wrote it by defining two functions:
reduced - for a given word returning all the reduced words that are
real words. if the word is "a" or "i" that means the word is the
minimum length word so it returns True. if the word does not have any
real reducible words it returns False.
create - for a given word (it actually gets a one word string), returns True if the word is reducible till it gets to "a" or "i"
def reduced(words):
''' creating a list of reduced words True : 'a' or 'i' False : no reduced words'''
words = list()
if word == 'a' or word == 'i':
return True
for letter in range(len(word)):
reduced_word = word[:letter] + word[letter+1:]
if reduced_word in world_list:
words.append(reduced_word)
if len(words) == 0:
return False
return words
def create(root):
'''
getting a list type!
return True : root reducable till 0
return False: else'''
if root == True:
return True
elif root == False:
return False
else:
for word in root:
word = reduced(word)
return create(word)
fin = open("words.txt")
world_list = list() #world list
reducable_words = list() #list of reducable words
longest = "" # the longest reducable word
# Creating world_list
for line in fin:
word = line.strip()
world_list.append(word)
# Creating tuples list
for word in world_list:
if (create([word])) == True:
reducable_words.append(word)
print(reducable_words)
The problem is, the script never gets to the last line, there is a problem with the second for loop. The world_list is correctly appended, so I can't see why this isn't working.

Your create() function seems of no value, I suggest tossing it altogether and focusing on your reduced() function which is very close. I've made some small changes to it: world_list -> word_list; instead of True or False, it returns an empty or non-empty list; it only returns non-empty if a complete reduction can be found (but ignores multiple possible reductions.)
def reduced(word):
''' returns a list of reduced words or an empty list if no reduced words '''
if word == 'a' or word == 'i':
return list(word)
words = list()
for letter in range(len(word)):
reduced_word = word[:letter] + word[letter + 1:]
if reduced_word in word_list:
words = reduced(reduced_word)
if words:
return [word] + words
return words
Using this slight rework of reduce(), you should be able to finish off the remainder of your program.
('daunt', '->', ['daunt', 'aunt', 'ant', 'at', 'a'])

Recursive Function palindrome in Python [closed]

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Closed 10 years ago.
I need help writing a recursive function which detects whether a string is a palindrome. But i can't use any loops it must be recursive. Can anyone help show me how this is done . Im using Python.

def ispalindrome(word):
if len(word) < 2: return True
if word[0] != word[-1]: return False
return ispalindrome(word[1:-1])
And here is the best one liner
def ispalindrome(word):
return word == word[::-1]

From a general algorithm perspective, the recursive function has 3 cases:
1) 0 items left. Item is a palindrome, by identity.
2) 1 item left. Item is a palindrome, by identity.
3) 2 or more items. Remove first and last item. Compare. If they are the same, call function on what's left of string. If first and last are not the same, item is not a palindrome.
The implementation of the function itself is left as an exercise to the reader :)

If a string is zero or one letters long, it's a palindrome.
If a string has the first and last letters the same, and the remaining letters (I think it's a [1: -1] slice in Python, but my Python is a bit rusty) are a palindrome, it's a palindrome.
Now, write that as a palindrome function that takes a string. It will call itself.

Since we're posting code anyway, and no one-liner has been posted yet, here goes:
def palindrome(s):
return len(s) < 2 or s[0] == s[-1] and palindrome(s[1:-1])

Here's another viewpoint
A palindromic string is
Some letter, x.
Some palindromic substrinng.
The same letter, x, repeated.
Also, note that you may be given a proper English sentence "Able was I ere I saw Elba." with punctuation. Your palindrome checker may have to quietly skip punctuation. Also, you may have to quietly match without considering case. This is slightly more complex.
Some leading punctuation. Some letter, x.
Some palindromic substring.
Some letter, x, repeated without regard to case. Some trailing punctuation.
And, by definition, a zero-length string is a palindrome. Also a single-letter string (after removing punctuation) is a palindrome.

Here's a way you can think of simple recursive functions... flip around the problem and think about it that way. How do you make a palindrome recursively? Here's how I would do it...
def make_palindrome():
maybe:
return ""
elsemaybe:
return some_char()
else:
c = some_char()
return c + make_palindrome() + c
Then you can flip it around to build the test.

The function should expect a string. If there is more then one letter in the string compare the first and the last letter. If 1 or 0 letters, return true. If the two letters are equal call the function then again with the string, without the first and the last letter. If they are not equal return false.
palindrom( word):
IF length of word 1 or 0 THEN
return 0;
IF last and first letter equal THEN
word := remove first and last letter of word;
palindrom( word);
ELSE
return false;

My solution
#To solve this I'm using the stride notation within a slice [::]
def amazonPalindrome(input):
inputB = input
input = input[::-1]
#print input
noPalindrome = inputB + " is not a palindrome"
isPalindrome = inputB + " is a palindrome"
#compare the value of the reversed string to input string
if input[0]!= input[-1]:
print noPalindrome
else:
print isPalindrome
#invoking the def requires at least 1 value or else it fails
#tests include splitting the string,mixing integers, odd amount palindromes.
#call the def
amazonPalindrome('yayay')

a=raw_input("enter the string:")
b=len(a)
c=0
for i in range(b):
if a[i]==a[-(i+1)]:
c=c+1
if c==b:
print a,"is polindrome"
else:
print a,"is not polindrome"

n=raw_input("Enter a number===>")
n=str(n)
l=len(n)
s=""
for i in range(1,l+1):
s=s+n[l-i]
if s==n:
print "Given number is polindrom"
else:
print "Given number is not polindrom"