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how to add a dot before each letter in a string in python

we get a string from user and want to lowercase it and remove vowels and add a '.' before each letter of it. for example we get 'aBAcAba' and change it to '.b.c.b' . two early things are done but i want some help with third one.
str = input()
str=str.lower()
for i in range(0,len(str)):
str=str.replace('a','')
str=str.replace('e','')
str=str.replace('o','')
str=str.replace('i','')
str=str.replace('u','')
print(str)
for j in range(0,len(str)):
str=str.replace(str[j],('.'+str[j]))
print(str)

A few things:
You should avoid the variable name str because this is used by a builtin, so I've changed it to st
In the first part, no loop is necessary; replace will replace all occurrences of a substring
For the last part, it is probably easiest to loop through the string and build up a new string. Limiting this answer to basic syntax, a simple for loop will work.
st = input()
st=st.lower()
st=st.replace('a','')
st=st.replace('e','')
st=st.replace('o','')
st=st.replace('i','')
st=st.replace('u','')
print(st)
st_new = ''
for c in st:
st_new += '.' + c
print(st_new)
Another potential improvement: for the second part, you can also write a loop (instead of your five separate replace lines):
for c in 'aeiou':
st = st.replace(c, '')
Other possibilities using more advanced techniques:
For the second part, a regular expression could be used:
st = re.sub('[aeiou]', '', st)
For the third part, a generator expression could be used:
st_new = ''.join(f'.{c}' for c in st)

You can use str.join() to place some character in between all the existing characters, and then you can use string concatenation to place it again at the end:
# st = 'bcb'
st = '.' + '.'.join(st)
# '.b.c.b'
As a sidenote, please don't use str as a variable name. It's the name of the "string" datatype, and if you make a variable named it then you can't properly work with other strings any more. string, st, s, etc. are fine, as they're not the reserved keyword str.

z = "aBAcAba"
z = z.lower()
newstring = ''
for i in z:
if not i in 'aeiou':
newstring+='.'
newstring+=i
print(newstring)
Here I have gone step by step, first converting the string to lowercase, then checking if the word is not vowel, then add a dot to our final string then add the word to our final string.

You could try splitting the string into an array and then build a new string with the indexes of the array appending an "."
not too efficient but will work.

thanks to all of you especially allani. the bellow code worked.
st = input()
st=st.lower()
st=st.replace('a','')
st=st.replace('e','')
st=st.replace('o','')
st=st.replace('i','')
st=st.replace('u','')
print(st)
st_new = ''
for c in st:
st_new += '.' + c
print(st_new)

This does everything.
import re
data = 'KujhKyjiubBMNBHJGJhbvgqsauijuetystareFGcvb'
matches = re.compile('[^aeiou]', re.I).finditer(data)
final = f".{'.'.join([m.group().lower() for m in matches])}"
print(final)
#.k.j.h.k.y.j.b.b.m.n.b.h.j.g.j.h.b.v.g.q.s.j.t.y.s.t.r.f.g.c.v.b

s = input()
s = s.lower()
for i in s:
for x in ['a','e','i','o','u']:
if i == x:
s = s.replace(i,'')
new_s = ''
for i in s:
new_s += '.'+ i
print(new_s)

def add_dots(n):
return ".".join(n)
print(add_dots("test"))
def remove_dots(a):
return a.replace(".", "")
print(remove_dots("t.e.s.t"))

To print the count of occurrences of a word ending with "on" in python

my_string = """Strings are gameon amongst gameon the most popular data types in Python. We can create the strings by enclosing characters briton in quotes. Python treats briton single quotes the same as double quotes."""
def count_words(string):
for word in string.split():
if word.endswith("on") == True:
print(word,":",string.count(word))
string = string.replace(word,'')
count_words(my_string)
I want to print all the words and their occurences in a word if they end with "on". I am getting something like
gameon : 2
gameon : 0
briton : 2
Python : 2
briton : 0
this even after removing the word.
Why it is repeating?
Edit: I can't use any module. Only logic.

You do not need to modify your string while you are counting.
Instead, you can use collections.Counter with a generator expression. It's also worth, as below, converting to lowercase and removing punctuation.
from collections import Counter
from string import punctuation
table = str.maketrans(punctuation, ' ' * len(punctuation))
x = my_string.translate(table).lower()
c = Counter(i for i in x.split() if i.endswith('on'))
print(c)
Counter({'gameon': 2, 'python': 2, 'briton': 2})

my_string = """Strings are gameon amongst gameon the most popular data types in Python. We can create the strings by enclosing characters briton in quotes. Python treats briton single quotes the same as double quotes."""
di={}
def count_words(string):
for word in string.split():
if word.endswith("on") == True:
if word in di:
di[word]+=1
else:
di[word]=1
string = string.replace(word,'')
#print(string)
count_words(my_string)
for i in di:
print(i,di[i])
You can use a dictionary to achieve the same.

Using collections.Counter
Ex:
import collections
my_string = """Strings are gameon amongst gameon the most popular data types in Python. We can create the strings by enclosing characters briton in quotes. Python treats briton single quotes the same as double quotes."""
def count_words(string):
for word, v in collections.Counter(string.split()).items():
if word.endswith("on"):
print(word,":",v)
count_words(my_string)
Output:
('Python', ':', 1)
('briton', ':', 2)
('gameon', ':', 2)

You can use pandas.Series to value_counts() these words
from string import punctuation
my_string = ''.join(w for w in my_string if w not in set(punctuation))
pd.Series([i for i in my_string.split(" ") if i.endswith("on")]).value_counts()
>> (gameon, 2), (briton, 2), (Python, 2)

Python string replace

I have this code:
ALPHABET1 = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
key = "TES"
ALPHABET2 = key + ALPHABET1
count_result = ALPHABET2.count("T")
if (count_result > 1):
ALPHABET3 = ALPHABET1.replace("T","")
ALPHABET2 = key + ALPHABET3
print(ALPHABET2)
I want to be able to put the keyword at the start of the alphabet string to create a new string without repeating the letters in the keyword. I'm having some problems doing this though. I need the keyword to work for all letters as it will be user input in my program. Any suggestions?

Two things:
You don't need to make the alphabet yourself, import string and use string.ascii_uppercase; and
You can use a for loop to work through the characters in your key.
To illustrate the latter:
for c in key:
alphabet = alphabet.replace(c, "")
Better yet, a list is mutable, so you can do:
alpha = [c for c in string.ascii_uppercase if c not in key]
alpha.extend(set(key))

its easy and clean to do this with a regex
import re
ALPHABET1 = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
key = "TES"
newalphabet = key.upper() + re.sub(r'%s'%'|'.join(key.upper()), '', ALPHABET1)
or with a list comprehension like #jonrsharpe suggested

Persistent index in python string

I'm trying to get string.index() to ignore instances of a character that it has already located within a string. Here is my best attempt:
alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
def save_alphabet(phrase):
saved_alphabet = ""
for item in phrase:
if item in alphabet:
saved_alphabet = saved_alphabet + str(phrase.index(item))
return saved_alphabet
print save_alphabet("aAaEaaUA")
The output I'd like is "1367" but, as it only finds the first instance of item it is outputting "1361".
What's the best way to do this? The returned value should be in string format.

>>> from string import ascii_uppercase as alphabet
>>> "".join([str(i) for i, c in enumerate("aAaEaaUA") if c in alphabet])
'1367'
regex solution (do not prefer regex in this case)
>>> import re
>>> "".join([str(m.start()) for m in re.finditer(r'[A-Z]', "aAaEaaUA")])
'1367'

`' '.join(e)` isn't changing value of e

I'm creating a function to create all 26 combinations of words with a fixed suffix. The script works except for the JOIN in the second-to-last line.
def create_word(suffix):
e=[]
letters="abcefghijklmnopqrstuvwxyz"
t=list(letters)
for i in t:
e.append(i)
e.append(suffix)
' '.join(e)
print e
Currently, it is printing ['a', 'suffix', 'b', 'suffix, ...etc]. And I want it to print out as one long string: 'aSuffixbSuffixcSuffix...etc.' Why isn't the join working in this? How can I fix this?
In addition, how would I separate the characters once I have the string? For example to translate "take the last character of the suffix and add a space to it every time ('aSuffixbSuffixcSuffix' --> 'aSuffix bSuffix cSuffix')". Or, more generally, to replace the x-nth character, where x is any integer (e.g., to replace the 3rd, 6th, 9th, etc. character some something I choose).

str.join returns the new value, not transform the existing one. Here's one way to accomplish it.
result = ' '.join(e)
print result
But if you're feeling clever, you can streamline a lot of the setup.
import string
def create_word(suffix):
return ' '.join(i + suffix for i in string.ascii_lowercase)

join doesn't change its arguments - it just returns a new string:
result = ' '.join(e)
return result

If you really want the output you specified (all of the results concatenated together):
>>> import string
>>> string.ascii_lowercase
'abcdefghijklmnopqrstuvwxyz'
>>> letters = string.ascii_lowercase
>>> suffix = 'Suffix'
>>> ''.join('%s%s' % (l, suffix) for l in letters)
'aSuffixbSuffixcSuffixdSuffixeSuffixfSuffixgSuffixhSuffixiSuffixjSuffixkSuffixlSuffixmSuffixnSuffixoSuffixpSuffixqSuffixrSuffixsSuffixtSuffixuSuffixvSuffixwSuffixxSuffixySuffixzSuffix'

Beside the problem already mentioned by rekursive, you should have a look at list comprehension:
def create_word(suffix):
return ''.join(
[i+suffix for i in "abcefghijklmnopqrstuvwxyz"]
)
print create_word('suffix')