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CodingBat Python string index out of range error

I am currently doing a problem on coding bat called string_bits and have been debugging it using Thonny and putting the code into coding bat to see if it is correct. Right now I am getting an error with my code in codingbat that says string index out of range. The weird thing is when I run it in Thonny I don't get the error. What is happening here?
def string_bits(str):
new_str = [str[0]]
count = 0
for letter in str[1:]:
count += 1
if count % 2 == 0:
new_str.append(letter)
return "".join(new_str)

Maybe the test trying some different kinds of input, and not only the obvious.
for example if your input is an empty string: it will cause such an "out of range" error.
Try to add input check before any operation on the string (which is actually an array)
like so:
def string_bits(str):
# check input for empty string
if str == "":
# quit function if invalid input
return ""
new_str = [str[0]]
count = 0
for letter in str[1:]:
count += 1
if count % 2 == 0:
new_str.append(letter)
return "".join(new_str)

While loop using strings and if statements

Prompt for python program:
5.16 LAB: Password modifier
Many user-created passwords are simple and easy to guess. Write a program that takes a simple password and makes it stronger by replacing characters using the key below, and by appending "!" to the end of the input string.
i becomes 1
a becomes #
m becomes M
B becomes 8
s becomes $
Ex: If the input is:
mypassword
the output is:
Myp#$$word!
Hint: Python strings are immutable, but support string concatenation. Store and build the stronger password in the given password variable.
My code so far:
word = input()
password = ''
i = 0
while i < len(word):
I'm really just confused about changing certain characters in a string using a while loop. I was also trying to use if statements. While loops are hard for me understand. I don't know where to start.

You're doing fine. While loops have this format:
while (condition) == True:
# do this
In your case, you will want to put i += 1 at then end of your loop so it doesn't go on forever, like this:
while i < len(word):
# your code here
i += 1 # this adds 1 to i
# so eventually it will be equal to the length of word, ending the loop
As for if statements, you could try this:
while i < len(word):
if word[i] == 'i':
password += '1'
i += 1
The if statements will only work on the values you want to change, so add an else statement to act as a default.
while i < len(word):
if word[i] == 'i':
password += '1'
else:
password += word[i]
i += 1
Now, if you get a normal letter, like h, it will add it to the password.
Hope this helps you get started!

comparison within loop while maintaining value

I am trying to make a python program that
1. compares the first and last letter of a word
2. tell whether the words inputted to the program are in alphabetical order
e.g.) alaska baobab cadillac => would pass
e.g.) alaska baobab greg => would not pass
my code is shown below
num_words = int(input())
while num_words > 0:
my_word = str(input())
num_words -= 1
alphabet_order = ord(my_word[0].lower())
if my_word[0].lower() != my_word[-1].lower() or alphabet_order != ord(my_word[0].lower()):
print(my_word)
break
alphabet_order += 1
if alphabet_order == ord('z') + 1:
alphabet_order = ord('a')
else:
print('no mistake')
Hi everyone on stack.
I am pretty new to programming and I am starting to find assigning variables within a loop very cumbersome.
It seems like my variable alphabet_order keeps getting renewed every time when the loop takes in a new input.
What would be a great way to steadily increase the alphabet_order variable by 1 while only renewing ord(my_word[0]) every loop?

Few notes, while loops are good for when you don't know how many times you are going to loop. for loops are good for when you know how many items you are looping. In this case we have a known number of words we are to check so using a for loop makes more sense. This way we don't need to take an input on telling us how many words we expect.
Also your code wasn't dealing with the words you were dealing with the sentence, so instead you should split() your input by the space to get a list of the words to operate on. for loops can iterate through a list which is useful for us.
Note that for loops also can take an else, the else section runs after the for loop is finished all the elements it's iterating through and we know that if the for is finish all the words in the list, then the else section will kick in. If we break out, then the else doesn't run.
The rest you more or less had it, just need a starting ord('a') would have made life easier.
my_word = input() #take a sentence such as alaska baobab cadillac
current_ord = ord('a')
for each in my_word.split(' '):
if each[0].lower() != each[-1].lower() or current_ord != ord(each[0].lower()):
print(f"bad word {each}")
break
current_ord += 1
if current_ord == ord('z') + 1:
current_ord = ord('a')
else:
print('no mistake')

Maybe this is helpful to you. Instead of initializing it inside the loop, declare it outside the loop and assign it differently based on the conditions.
Tip: having while num_words > 0 is redundant because it will terminate automatically when it hits 0 as it is treated as False. And there is no need to convert int to a str type as it is str by default.
num_words = int(input("Enter a number"))
alphabet_order = None
while num_words:
my_word = input("Enter a word")
num_words -= 1
if alphabet_order is None: # happens only once
alphabet_order = ord(my_word[0].lower())
if ord(my_word[0].lower()) >= alphabet_order:
print('ok')
alphabet_order = ord(my_word[0].lower()) # update alphabet_order
else:
print('not ok EXITING')
break # closing loop

python help! If/else statement

I have been learning python through code academy. It asked me to create a if else statement that prints the users input, but if there is no input print "empty". I did pass the tutorial but when there is a user input it prints both the users input and "empty".
here is my code:
print "Welcome to the English to Pig Latin translator!"
original = raw_input("what is your name?")
length = len("original")
if length > 0:
print original
else: length < 0
print "empty"
Notice that print under else is not indented. I thought you had to indented it, but when i do it gives me an error.

You seem to have a couple issues with your code. First I believe you should change your assignment of length to:
length = len(original)
That way you get the correct length of the string you binded to the name original. If you use len("original"), you will pass the string "original" to the len() function, and it will just count the characters in that string. Regardless of what the user of your program inputted, your length will always be 8, since that's how many characters are in the string "original"
Also keep in mind that the len() function will never return a negative value. So testing if length < 0 doesn't make sense. You probably wanted to test if it equals 0. It seems as if you were trying to use an elif statement.
Try:
print "Welcome to the English to Pig Latin translator!"
original = raw_input("what is your name?")
length = len(original)
if length > 0:
print original
elif length == 0:
print "empty"
elif statements let you test conditions just like if statements. elif is short for "else if".
Furthermore, as #chepner pointed out in the comments, there is no need to even test the second condition at all. You can just as easily go with:
else:
print "empty"
The else syntax is not followed by any condition. It automatically enters the indented block of code if the other conditions evaluate to False.

Was the length < 0 intended to be a comment? You need the comment character #.
else: # length < 0
print "empty"
It's wrong anyway, it should be length <= 0.
Without a comment character it was being interpreted as the statement to use as the else clause. It didn't create an error since length < 0 just generates a boolean result that you didn't do anything with, but it prevented you from having another statement as part of the else.

else: length < 0
print "empty"
should be
elif length == 0:
print "empty"
Python has significant whitespace, things that are indented the same are in the same scope.

First off, it is not len("original") it is len(original). If you use quotes you are making it a constant value, "original", rather than a variable named original.
Second, instead of checking the length of the string you should use this
if original:
# Type here what happens if the string is not empty
else:
# Here what happens if the string is empty
By Python standard any collection, including strings, equals false if it is empty(aka contains no elements) and true if it contains any elements.

There is a a statement after else.
else: length < 0
print "empty"
Maybe you are looking for an (elif is another way to check another condition after the if if the previous if fails.)
elif length <= 0:
or just a plain
else:
print "empty"
it will never go past zero anyways you could have a == conditional for zero and it would work.
elif length == 0:
this is probably the best way there is no need to check another condition.
if length > 0:
print original
else:
print "empty"
also just a side note length = len("original")
there is not suppose to be quotation marks around original because its a variable :). You will just be getting the string "original" not the actual stuff inside of the variable.
The end result ..
print "Welcome to the English to Pig Latin translator!"
original = raw_input("what is your name?")
length = len(original)
if length > 0:
print original
else:
print "empty"

To check if there is an item in a list, simply do this:
if List_name:
{code here}
...
So, your code should very simply look like this:
print "Welcome to the English to Pig Latin translator!"
original = raw_input("what is your name?")
if original:
print original
else:
print "Empty"
It's that easy :D

Count letters in a word in python debug

I am trying to count the number of times 'e' appears in a word.
def has_no_e(word): #counts 'e's in a word
letters = len(word)
count = 0
while letters >= 0:
if word[letters-1] == 'e':
count = count + 1
letters = letters - 1
print count
It seems to work fine except when the word ends with an 'e'. It will count that 'e' twice. I have no idea why. Any help?
I know my code may be sloppy, I'm a beginner! I'm just trying to figure out the logic behind what's happening.

>>> word = 'eeeooooohoooooeee'
>>> word.count('e')
6
Why not this?

As others mention, you can implement the test with a simple word.count('e'). Unless you're doing this as a simple exercise, this is far better than trying to reinvent the wheel.
The problem with your code is that it counts the last character twice because you are testing index -1 at the end, which in Python returns the last character in the string. Fix it by changing while letters >= 0 to while letters > 0.
There are other ways you can tidy up your code (assuming this is an exercise in learning):
Python provides a nice way of iterating over a string using a for loop. This is far more concise and easier to read than using a while loop and maintaining your own counter variable. As you've already seen here, adding complexity results in bugs. Keep it simple.
Most languages provide a += operator, which for integers adds the amount to a variable. It's more concise than count = count + 1.
Use a parameter to define which character you're counting to make it more flexible. Define a default argument for using char='e' in the parameter list when you have an obvious default.
Choose a more appropriate name for the function. The name has_no_e() makes the reader think the code checks to see if the code has no e, but what it actually does is counts the occurrences of e.
Putting this all together we get:
def count_letter(word, char='e'):
count = 0
for c in word:
if c == char:
count += 1
return count
Some tests:
>>> count_letter('tee')
2
>>> count_letter('tee', 't')
1
>>> count_letter('tee', 'f')
0
>>> count_letter('wh' + 'e'*100)
100

Why not simply
def has_no_e(word):
return sum(1 for letter in word if letter=="e")

The problem is that the last value of 'letters' in your iteration is '0', and when this happens you look at:
word[letters-1]
meaning, you look at word[-1], which in python means "last letter of the word".
so you're actually counting correctly, and adding a "bonus" one if the last letter is 'e'.

It will count it twice when ending with an e because you decrement letters one time too many (because you loop while letters >= 0 and you should be looping while letters > 0). When letters reaches zero you check word[letters-1] == word[-1] which corresponds to the last character in the word.

Many of these suggested solutions will work fine.
Know that, in Python, list[-1] will return the last element of the list.
So, in your original code, when you were referencing word[letters-1] in a while loop constrained by letters >= 0, you would count the 'e' on the end of the word twice (once when letters was the length-1 and a second time when letters was 0).
For example, if my word was "Pete" your code trace would look like this (if you printed out word[letter] each loop.
e (for word[3])
t (for word[2])
e (for word[1])
P (for word[0])
e (for word[-1])
Hope this helps to clear things up and to reveal an interesting little quirk about Python.

#marcog makes some excellent points;
in the meantime, you can do simple debugging by inserting print statements -
def has_no_e(word):
letters = len(word)
count = 0
while letters >= 0:
ch = word[letters-1] # what is it looking at?
if ch == 'e':
count = count + 1
print('{0} <-'.format(ch))
else:
print('{0}'.format(ch))
letters = letters - 1
print count
then
has_no_e('tease')
returns
e <-
s
a
e <-
t
e <-
3
from which you can see that
you are going through the string in reverse order
it is correctly recognizing e's
you are 'wrapping around' to the end of the string - hence the extra e if your string ends in one

If what you really want is 'has_no_e' then the following may be more appropriate than counting 'e's and then later checking for zero,
def has_no_e(word):
return 'e' not in word
>>> has_no_e('Adrian')
True
>>> has_no_e('test')
False
>>> has_no_e('NYSE')
True
If you want to check there are no 'E's either,
def has_no_e(word):
return 'e' not in word.lower()
>>> has_no_e('NYSE')
False

You don't have to use a while-loop. Strings can be used for-loops in Python.
def has_no_e(word):
count = 0
for letter in word:
if letter == "e":
count += 1
print count
or something simpler:
def has_no_e(word):
return sum(1 for letter in word if letter=="e")