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How to count words that end with a letter? (python)

I am a beginner and this is what I came up with so far. However, it does not output the correct number of words that end with "a" or "b." Any tips on how to correct this code?
names = input("Enter list of names: ")
name = names.split(" ")
num = len(name)
ab = ""
print("Number of words:", num)
for i in range(num):
if name[i] == ' ':
if name[i-1] == a:
ab.append() + " "
elif name[i-1] == b:
ab.append() + " "
a_b = ab.split(' ')
print("Number of words that end with a or b: ",len(a_b))

In Python boolean (True and False) are the same as the integers 1 and 0. This means you can use sum() to count True booleans. str.endswith() returns a boolean. This means you can just do this:
words = ["stab", "drama", "hello", "magma", "test"]
a_b = sum(word.endswith(('a', 'b')) for word in words)
# 3
z_c = sum(word.endswith(('z', 'c')) for word in words)
# 0

Any tips on how to correct this code?
Others have answered alternative, working solutions, but I want to try to point out the specific things wrong in your code snippet and tips to correct.
First here's a copy of your snippet with some simple debugging prints along with their output (running in python 3.10).
names = "Barry Lima Bab"
name = names.split(" ")
print(f"{name=}") # name=['Barry', 'Lima', 'Bab']
num = len(name)
print(f"{num=}") # num=3
ab = ""
print("Number of words:", num)
for i in range(num):
print(name[i]) # Barry -> Lima -> Bab
if name[i] == ' ':
if name[i-1] == a:
ab.append() + " "
elif name[i-1] == b:
ab.append() + " "
print(f"{ab=}") # ab=''
a_b = ab.split(' ')
print(f"{a_b=}") # a_b=['']
Breaking things down step by step like this is a great starting point for better understanding what's going on.
if name[i] == ' ': Notice the value of name[i] so this check never resolves to True meaning the code inside never runs
if the inner code did run, you'd hit a NameError: name 'a' is not defined. Did you mean: 'ab'? because a is undefined. Probably meant 'a' here. same for b
In my example, name[i-1] would be 'Bab' -> 'Barry' -> 'Lima' which is probably not what you're expecting. This is because you're getting the -1th, 0th, 1st items in name. to get the last letter from the name, you want name[i][-1] here
if you then get into either of the furthest inside conditions, you'd encounter AttributeError: 'str' object has no attribute 'append' which happens because append is for list not str. You couldchange ab to be a list to use this in which case you'd want ab.append(name[i]) or you can use string concatenation like ab += " " + name[i] or using str.concat
1 last note of interest, you may have noticed your code as-is really likes to return and say that there's 1 item. This is because the above issues always (if the program doesn't break) leaves ab == '' and ''.split(' ') => [''] and thus len(ab.split(" ")) => 1
1 tip that I think would help in code comprehension is that the name variable here is not a single name string like it implies. It's actually a list[str]. I'd probably denote the variables something more like names_str: str vs names: list[str] or raw_names vs names. Then just use something like for name in names: and not worry about indexes. You can also use name.endswith('a') instead of name[-1] == 'a' for better readability.
Eventually you can combine these into a list comprehension for maximal succinctness -> len([name for name in names if name.endswith('a') or name.endswith('b')]).

words = ["ab", "bab", "pa", "pap"]
result = 0
for word in words:
if word[-1] in "ab":
result += 1
print(result)

As a list comprehension:
words = ["word1a", "word2b", "word3", "word4", "word5a"] # ['word1a', 'word2b', 'word5a']
filtered_words = [word for word in words if word[-1] in "ab"]
filtered_words = [word for word in words if word.endswith(("a", "b"))] # better, if you have multiple endings you want to search for with different lengths
len(filtered_words) # 3

name[i-1] is the previous name in the list, not the last character of the current name in the loop.
There's no need to append the matching names to a string. If you just need the count of matches, increment a counter variable.
You need to put a and b in quotes to make them strings, not variables.
names = input("Enter list of names: ")
name = names.split(" ")
matches = 0
print("Number of words:", len(name))
for cur_name in name:
if cur_name.endswith('a') or cur_name.endswith('b'):
matches += 1
print("Number of words that end with a or b: ", matches)

Use the endswith string method to detect if the last letter is "a" or "b"
names = ["bob","boba","larry","curly","moe"]
count = 0
for name in names:
if name.endswith("a") or name.endswith("b"):
count += 1
print(f"{count=}") # count=2

How do I separate words in python?

I have tried to make this work with a phrase, but I am having issues with getting it to the ie to go at the end of the words. For example HankTIE ouYIE would be the output of the input Thank You.
Here is what I have:
string=input("Please input a word: ")
def silly_encrypter(string):
strr = string.split()
for strr in string:
first_letter_at_the_end = strr[1:] + strr[0]
ie_at_the_end = first_letter_at_the_end + "IE"
print (ie_at_the_end)
silly_encrypter(string)

You can do this:
string=input("Please input a word: ")
def silly_encrypter(string):
splitspace = string.split() # first split the string into spaces.
for s in splitspace: # then looping over each element,
strlist = list(s) # turn the element into a list
strlist.append(strlist[0]) # first number to the last
del strlist[0] # delete the first number
strlist[0] = strlist[0].capitalize() # Capitalize the first letter
strlist.append('IE') # add IE
print(''.join(strlist), end=" ") # join the list
silly_encrypter(string)

Upon reading the accepted answer, I had to provide a cleaner solution:
def silly_encryptor(phrase, suffix="IE"):
new_phrase = []
for word in phrase.split():
new_phrase.append(word[1:]+word[:1]+suffix)
return " ".join(new_phrase)
phrase = input("Please enter your phrase: ")
print (silly_encryptor(phrase))

Alternatives to index()

So for my project I have to allow the user to input a sentence and then input a word and find all the occourunces of the word and print the numbers. Here's what I have
found = 0
sen = input("Enter the sentence you would like to break down!")
sen1 = sen.upper()
list = sen1.split()
search=input("Enter the word you want to search")
search1 = search.upper()
for search1 in list:
found = found + 1
position=list.index(search1)
if position == 0:
print("First word in the sentence")
if position == 1:
print("Second word in the sentence")
if position == 2:
print("Third word in the sentence")
if position == 3:
print("Fourth word in the sentence")
if position == 4:
print("Fifth word in the sentence")
if position == 5:
print("6th word in the sentence")
else:
position1 = position + 1
print(position1, "th word in the sentence")
but it only prints the first occurunce of the word and rarely works. Any solutions?

Replace list with a_list.
List of positions of a search1 occurances:
positions = [idx for idx, el in enumerate(a_list) if el == search1]

You have a great alternative which is re.finditer:
import re
sen = input("Enter the sentence you would like to break down!")
search = input("Enter the word you want to search")
for match in re.finditer(search, sen):
print (match.start())

Several comments have mentioned the danger of using list as a variable name. It's not actually a reserved word, but it is the name of a built-in type, and shadowing it by using it as a variable name can lead to mysterious bugs if you later wish to use this type to construct a list or test the type of an object.
A major problem with the code you posted is here:
search1 = search.upper()
for search1 in list:
The first line saves the upper-case version of the string search to the name search1. But the next line simply clobbers that with the words in list; it does not perform any searching operation. At the end of the for loop, search1 will be equal to the last item in list, and that's why your code isn't doing what you expect it to when it executes position=list.index(search1): you're telling it to find the position of the last word in list.
You could use .index to do what you want. To find multiple occurences you need to use a loop and pass .index a starting position. Eg,
def find_all(wordlist, word):
result = []
i = 0
while True:
try:
i = wordlist.index(word, i) + 1
result.append(i)
except ValueError:
return result
However, there's really not much benefit in using .index here..index performs its scan at C speed, so it's faster than scanning in a Python loop but you probably won't notice much of a speed difference unless the list you're scanning is large.
The simpler approach is as given in Tomasz's answer. Here's a variation I wrote while Tomasz was writing his answer.
def ordinal(n):
k = n % 10
return "%d%s" % (n, "tsnrhtdd"[(n // 10 % 10 != 1) * (k < 4) * k::4])
def find_all(wordlist, word):
return [i for i, s in enumerate(wordlist, 1) if s == word]
sen = 'this has this like this'
wordlist = sen.upper().split()
words = 'this has that like'
for word in words.split():
pos = find_all(wordlist, word.upper())
if pos:
pos = ', '.join([ordinal(u) for u in pos])
else:
pos = 'Not found'
print('{0}: {1}'.format(word, pos))
output
this: 1st, 3rd, 5th
has: 2nd
that: Not found
like: 4th
The code for ordinal was "borrowed" from this answer.

Storing multiple inputs in one variable

I have a while loop, which will keep asking a user to input words until they type stop. The input is stored in a variable called sentence.
My question is how do I store multiple inputs into one variable.
My current code is
stop = "stop"
sentence = []
while sentence != stop:
sentence = input("Enter a word: ")
sentence = sentence
print(sentence)
I don't understand how I would keep storing variables from one input and print out all the variable stored separated by commas/spaces etc

All you need to do is append() your new variables to the array:
>>> a = []
>>> for x in range(5):
... a.append("Hello!")
...
>>> a
['Hello!', 'Hello!', 'Hello!', 'Hello!', 'Hello!']
At the end, if you need everything in a single variable you can use join():
>>> ",".join(a)
'Hello!,Hello!,Hello!,Hello!,Hello!'

stop = "stop"
# okay --- 'sentence' is a list. Good start.
sentence = []
while sentence != stop:
# ...but now you've replaced the list 'sentence' with the word that was just input
# NOTE that in Python versions < 3, you should use raw_input below.
sentence = input("Enter a word: ")
# ...and this does nothing.
sentence = sentence
print(sentence)
Works better if you do something like this:
stop = "stop"
sentence = []
# create a new variable that just contains the most recent word.
word = ''
while word != stop:
word = input("Enter a word: ")
# stick the new word onto the end of the list
sentence.append(word)
print(sentence)
# ...and convert the list of words into a single string, each word
# separated by a space.
print " ".join(sentence)
...or to re-design a bit to omit the 'stop', something like:
stop = "stop"
sentence = []
while True:
word = input("Enter a word: ")
if word == stop:
# exit the loop
break
sentence.append(word)
# ...and convert the list of words into a single string, each word
# separated by a space.
print " ".join(sentence)

Its pretty simple
stop = "stop"
sentence = []
all = ""
while sentence != stop:
sentence = input("Enter a word: ")
all += sentence + ","
print(all)

One of your problems is that you are constantly writing over your sentence variable.
What you want to do is make use of the list append method. Documentation on lists:
https://docs.python.org/3/tutorial/datastructures.html
Example:
a = []
a.append(1)
a.append(2)
a.append(3)
print(a)
[1, 2, 3]
Next, you are looking to end your code if the user enters "stop". So what you should do is check in your loop if "stop" was written, and make use of Python's break, to break out of the loop.
What this means is that you should change your loop to loop indefinitely until you get that stop, using while True.
Your code can now simply look like this:
sentence = []
while True:
entry = input("Enter a word: ")
if entry == "stop":
break
sentence.append(entry)
print(sentence)

You probably want something like this:
sentence = []
while True:
word = input("Enter a word: ")
if word == "stop":
break
sentence.append(word)
print " ".join(sentence) + "."

Creating a word scrambler but it won't work, need help as a beginner

Beginner python coder here, keep things simple, please.
So, I need this code below to scramble two letters without scrambling the first or last letters. Everything seems to work right up until the scrambler() function.
from random import randint
def wordScramble(string):
stringArray = string.split()
for word in stringArray:
if len(word) >= 4:
letter = randint(1,len(word)-2)
point = letter
while point == letter:
point = randint(1, len(word)-2)
word = switcher(word,letter,point)
' '.join(stringArray)
return stringArray
def switcher(word,letter,point):
word = list(word)
word[letter],word[point]=word[point],word[letter]
return word
print(wordScramble("I can't wait to see how this turns itself out"))
The outcome is always:
I can't wait to see how this turns itself out

Since you are a beginner, I tried to change your code as little as possible. Mostly you are expecting changes to word to change the contents or your list stringArray. The comments mark the changes and reasons.
from random import randint
def wordScramble(myString): # avoid name clashes with python modules
stringArray = myString.split()
for i, word in enumerate(stringArray): # keep the index so we can update the list
if len(word) >= 4:
letter = randint(1,len(word)-2)
point = letter
while point == letter:
point = randint(1, len(word)-2)
stringArray[i] = switcher(word,letter,point) # update the array
return ' '.join(stringArray) # return the result of the join
def switcher(word,letter,point):
word = list(word)
word[letter],word[point]=word[point],word[letter]
return ''.join(word) # return word back as a string
print(wordScramble("I can't wait to see how this turns itself out"))

Because there had to be a cleaner (and better documented) way to do this:
from random import sample
def wordScramble(sentence):
# Split sentence into words; apply switcher to each; rejoin into a sentence
return ' '.join([switcher(x) for x in sentence.split()])
def switcher(word):
if len(word) <= 3: # Don't bother if not enough letters to scramble
return word
# Pick 2 positions from interior of word
a,b = sorted(sample( xrange(1,len(word)-1), 2 ))
# Re-assemble word with out 2 positions swapped using bits before, between & after them
return word[:a] + word[b] + word[a+1:b] + word[a] + word[b+1:]
print wordScramble("I can't wait to see how this turns itself out")