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Loop through a string with variable and replace an uppercase it with # in Python

I want to loop through a string and when it finds an uppercase letter, I want to replace it with #. Like this:
string = "hOw Are yOu?"
for x in string:
if x.isupper():
string.replace(x, "#")
print(string)
else:
print(string)
However, its not working as intended and is instead outputting the same string. Do tell me if there is a way to fix this or if you'd suggest another way.

Use list comprehension with join:
In [4]: ''.join([i if not i.isupper() else '#' for i in string])
Out[4]: 'h#w #re y#u?'

You just want to put the result again in string see below
string = "hOw Are yOu?"
for x in string:
if x.isupper():
string = string.replace(x, "#")
print(string)
else:
print(string)

Strings are immutable in Python. string.replace(x, "#") must thus be string = string.replace(x, "#") to have an effect on the string.
Note that currently your code has quadratic complexity as each replace operation has to loop over the entire string in linear time. A more efficient approach would be to perform the replacements yourself, as you're already looping over every character:
string = "".join(["#" if c.isupper() else c for c in "hOw Are yOu?"])
it would be even more concise (and possibly faster) to use a very simple RegEx for this:
import re
string = re.sub("[A-Z]", "#", "hOw Are yOu?")
this will fail for non-ASCII alphabets however; you'd have to use unicode properties & regex there.

This should do the trick!
string = "hOw Are yOu?"
for x in string:
if x.isupper():
string = string.replace(x, "#")
else:
pass
print(string)
I'm a novice as well! But from what I learned: when doing loops or if statements, you want to specify the value you are changing as I did in line 4 with: string = string.replace(x,'#') If not the change will not take effect!
Example:
my_list = [1,2,3,4]
for x in my_list:
my_list[x-1] = x + 1
print(my_list)
This is a poor example coding wise but it exemplifies the concept. If you don't address the variable it wont have any effect on it!
Hope this helps!

Python 3, receive a string as an argument without any return value

I'm learning Python and have been taking an online class. This class was very basic and I am know trying to continue my studies elsewhere. Stackoverflow.com has helped me a great deal. In the online course we didn't cover a lot about return statements, which I am now trying to learn. I would like to do something very basic, so I was thinking of creating a program that would receive a string as an argument without having any return value. I want the user to type a word that will be shown with characters or symbols between every letter.
Example
User types in the word Python.
The word will be shown as =P=y=t=h=o=n= or -P-y-t-h-o-n- or maybe with * between every letter.
Is this an easy task? Can someone help me how to go about doing this?
Thank you.
Joel

If you want to do it yourself, you can go through your string like this:
my_string = "Python"
for letter in my_string:
# do something with the letter
print(letter)
This will print each letter in your word. What you want to do is having a new string with your desired character. You probably know you can concatenate (append) two strings in this way :
str1 = "hello"
str2 = "world"
str3 = str1 + str2
print(str3) #helloworld
So to do what you'd like to do, you can see each letter as a substring of your main string, and your desired character (for example *) as another string, and build a result string in that way.
inputString = "Python"
result = ""
myChar = "*"
for letter in inputString:
# build your result
build = build + letter
print(build)
This will just copy inputString into result, though I think you'll have understood how to use it in order to add your custom chars between the letters.

Yes python makes this sort of string manipulation very easy (some other languages... not so much). Look up the standard join function in the python docs.

def fancy_print(s, join_char='-'):
# split string into a list of characters
letters = list(s)
# create joined string
output = join_char + join_char.join(letters) + join_char
# show it
print(output)
then
>>> fancy_print("PYTHON")
-P-Y-T-H-O-N-
>>> fancy_print("PYTHON", "*")
*P*Y*T*H*O*N*

Find multiple list items within a string

Working on a problem in which I am trying to get a count of the number of vowels in a string. I wrote the following code:
def vowel_count(s):
count = 0
for i in s:
if i == 'a' or i == 'e' or i == 'i' or i == 'o' or i == 'u':
count += 1
print count
vowel_count(s)
While the above works, I would like to know how to do this more simply by creating a list of all vowels, then looping my If statement through that, instead of multiple boolean checks. I'm sure there's an even more elegant way to do this with import modules, but interested in this type of solution.
Relative noob...appreciate the help.

No need to create a list, you can use a string like 'aeiou' to do this:
>>> vowels = 'aeiou'
>>> s = 'fooBArSpaM'
>>> sum(c.lower() in vowels for c in s)
4

You can actually treat a string similarly to how you would a list in python (as they are both iterables), for example
vowels = 'aeiou'
sum(1 for i in s if i.lower() in vowels)
For completeness sake, others suggest vowels = set('aeiou') to allow not matching checks such as 'eio' in vowels. However note if you are iterating over your string in a for loop one character at a time, you won't run into this problem.

A weird way around this is the following:
vowels = len(s) - len(s.translate(None, 'aeiou'))
What you are doing with s.translate(None, 'aeiou') is creating a copy of the string removing all vowels. And then checking how the length differed.
Special note: the way I'm using it is even part of the official documentation
What is a vowel?
Note, though, that method presented here only replaces exactly the characters present in the second parameter of the translate string method. In particular, this means that it will not replace uppercase versions characters, let alone accented ones (like áèïôǔ).
Uppercase vowels
Solving the uppercase ones is kind of easy, just do the replacemente on a copy of the string that has been converted to lowercase:
vowels = len(s) - len(s.lower().translate(None, 'aeiou'))
Accented vowels
This one is a little bit more convoluted, but thanks to this other SO question we know the best way to do it. The resulting code would be:
from unicodedate import normalize
# translate special characters to unaccented versions
normalized_str = normalize('NFD', s).encode('ascii', 'ignore')
vowels = len(s) - len(normalized_str.lower().translate(None, 'aeiou'))

You can filter using a list comprehension, like so:
len([letter for letter in s if letter in 'aeiou'])

How do I calculate the number of times a word occurs in a sentence?

So I've been learning Python for some months now and was wondering how I would go about writing a function that will count the number of times a word occurs in a sentence. I would appreciate if someone could please give me a step-by-step method for doing this.

Quick answer:
def count_occurrences(word, sentence):
return sentence.lower().split().count(word)
'some string.split() will split the string on whitespace (spaces, tabs and linefeeds) into a list of word-ish things. Then ['some', 'string'].count(item) returns the number of times item occurs in the list.
That doesn't handle removing punctuation. You could do that using string.maketrans and str.translate.
# Make collection of chars to keep (don't translate them)
import string
keep = string.lowercase + string.digits + string.whitespace
table = string.maketrans(keep, keep)
delete = ''.join(set(string.printable) - set(keep))
def count_occurrences(word, sentence):
return sentence.lower().translate(table, delete).split().count(word)
The key here is that we've constructed the string delete so that it contains all the ascii characters except letters, numbers and spaces. Then str.translate in this case takes a translation table that doesn't change the string, but also a string of chars to strip out.

wilberforce has the quick, correct answer, and I'll give the long winded 'how to get to that conclusion' answer.
First, here are some tools to get you started, and some questions you need to ask yourself.
You need to read the section on Sequence Types, in the python docs, because it is your best friend for solving this problem. Seriously, read it. Once you have read that, you should have some ideas. For example you can take a long string and break it up using the split() function. To be explicit:
mystring = "This sentence is a simple sentence."
result = mystring.split()
print result
print "The total number of words is: " + str(len(result))
print "The word 'sentence' occurs: " + str(result.count("sentence"))
Takes the input string and splits it on any whitespace, and will give you:
["This", "sentence", "is", "a", "simple", "sentence."]
The total number of words is 6
The word 'sentence' occurs: 1
Now note here that you do have the period still at the end of the second 'sentence'. This is a problem because 'sentence' is not the same as 'sentence.'. If you are going to go over your list and count words, you need to make sure that the strings are identical. You may need to find and remove some punctuation.
A naieve approach to this might be:
no_period_string = mystring.replace(".", " ")
print no_period_string
To get me a period-less sentence:
"This sentence is a simple sentence"
You also need to decide if your input going to be just a single sentence, or maybe a paragraph of text. If you have many sentences in your input, you might want to find a way to break them up into individual sentences, and find the periods (or question marks, or exclamation marks, or other punctuation that ends a sentence). Once you find out where in the string the 'sentence terminator' is you could maybe split up the string at that point, or something like that.
You should give this a try yourself - hopefully I've peppered in enough hints to get you to look at some specific functions in the documentation.

Simplest way:
def count_occurrences(word, sentence):
return sentence.count(word)

text=input("Enter your sentence:")
print("'the' appears", text.count("the"),"times")
simplest way to do it

Problem with using count() method is that it not always gives the correct number of occurrence when there is overlapping, for example
print('banana'.count('ana'))
output
1
but 'ana' occurs twice in 'banana'
To solve this issue, i used
def total_occurrence(string,word):
count = 0
tempsting = string
while(word in tempsting):
count +=1
tempsting = tempsting[tempsting.index(word)+1:]
return count

You can do it like this:
def countWord(word):
numWord = 0
for i in range(1, len(word)-1):
if word[i-1:i+3] == 'word':
numWord += 1
print 'Number of times "word" occurs is:', numWord
then calling the string:
countWord('wordetcetcetcetcetcetcetcword')
will return: Number of times "word" occurs is: 2

def check_Search_WordCount(mySearchStr, mySentence):
len_mySentence = len(mySentence)
len_Sentence_without_Find_Word = len(mySentence.replace(mySearchStr,""))
len_Remaining_Sentence = len_mySentence - len_Sentence_without_Find_Word
count = len_Remaining_Sentence/len(mySearchStr)
return (int(count))

I assume that you just know about python string and for loop.
def count_occurences(s,word):
count = 0
for i in range(len(s)):
if s[i:i+len(word)] == word:
count += 1
return count
mystring = "This sentence is a simple sentence."
myword = "sentence"
print(count_occurences(mystring,myword))
explanation:
s[i:i+len(word)]: slicing the string s to extract a word having the same length with the word (argument)
count += 1 : increase the counter whenever matched.

How to delete a character from a string using Python

There is a string, for example. EXAMPLE.
How can I remove the middle character, i.e., M from it? I don't need the code. I want to know:
Do strings in Python end in any special character?
Which is a better way - shifting everything right to left starting from the middle character OR creation of a new string and not copying the middle character?

In Python, strings are immutable, so you have to create a new string. You have a few options of how to create the new string. If you want to remove the 'M' wherever it appears:
newstr = oldstr.replace("M", "")
If you want to remove the central character:
midlen = len(oldstr) // 2
newstr = oldstr[:midlen] + oldstr[midlen+1:]
You asked if strings end with a special character. No, you are thinking like a C programmer. In Python, strings are stored with their length, so any byte value, including \0, can appear in a string.

To replace a specific position:
s = s[:pos] + s[(pos+1):]
To replace a specific character:
s = s.replace('M','')

This is probably the best way:
original = "EXAMPLE"
removed = original.replace("M", "")
Don't worry about shifting characters and such. Most Python code takes place on a much higher level of abstraction.

Strings are immutable. But you can convert them to a list, which is mutable, and then convert the list back to a string after you've changed it.
s = "this is a string"
l = list(s) # convert to list
l[1] = "" # "delete" letter h (the item actually still exists but is empty)
l[1:2] = [] # really delete letter h (the item is actually removed from the list)
del(l[1]) # another way to delete it
p = l.index("a") # find position of the letter "a"
del(l[p]) # delete it
s = "".join(l) # convert back to string
You can also create a new string, as others have shown, by taking everything except the character you want from the existing string.

How can I remove the middle character, i.e., M from it?
You can't, because strings in Python are immutable.
Do strings in Python end in any special character?
No. They are similar to lists of characters; the length of the list defines the length of the string, and no character acts as a terminator.
Which is a better way - shifting everything right to left starting from the middle character OR creation of a new string and not copying the middle character?
You cannot modify the existing string, so you must create a new one containing everything except the middle character.

Use the translate() method:
>>> s = 'EXAMPLE'
>>> s.translate(None, 'M')
'EXAPLE'

def kill_char(string, n): # n = position of which character you want to remove
begin = string[:n] # from beginning to n (n not included)
end = string[n+1:] # n+1 through end of string
return begin + end
print kill_char("EXAMPLE", 3) # "M" removed
I have seen this somewhere here.

card = random.choice(cards)
cardsLeft = cards.replace(card, '', 1)
How to remove one character from a string:
Here is an example where there is a stack of cards represented as characters in a string.
One of them is drawn (import random module for the random.choice() function, that picks a random character in the string).
A new string, cardsLeft, is created to hold the remaining cards given by the string function replace() where the last parameter indicates that only one "card" is to be replaced by the empty string...

On Python 2, you can use UserString.MutableString to do it in a mutable way:
>>> import UserString
>>> s = UserString.MutableString("EXAMPLE")
>>> type(s)
<class 'UserString.MutableString'>
>>> del s[3] # Delete 'M'
>>> s = str(s) # Turn it into an immutable value
>>> s
'EXAPLE'
MutableString was removed in Python 3.

Another way is with a function,
Below is a way to remove all vowels from a string, just by calling the function
def disemvowel(s):
return s.translate(None, "aeiouAEIOU")

Here's what I did to slice out the "M":
s = 'EXAMPLE'
s1 = s[:s.index('M')] + s[s.index('M')+1:]

To delete a char or a sub-string once (only the first occurrence):
main_string = main_string.replace(sub_str, replace_with, 1)
NOTE: Here 1 can be replaced with any int for the number of occurrence you want to replace.

You can simply use list comprehension.
Assume that you have the string: my name is and you want to remove character m. use the following code:
"".join([x for x in "my name is" if x is not 'm'])

If you want to delete/ignore characters in a string, and, for instance, you have this string,
"[11:L:0]"
from a web API response or something like that, like a CSV file, let's say you are using requests
import requests
udid = 123456
url = 'http://webservices.yourserver.com/action/id-' + udid
s = requests.Session()
s.verify = False
resp = s.get(url, stream=True)
content = resp.content
loop and get rid of unwanted chars:
for line in resp.iter_lines():
line = line.replace("[", "")
line = line.replace("]", "")
line = line.replace('"', "")
Optional split, and you will be able to read values individually:
listofvalues = line.split(':')
Now accessing each value is easier:
print listofvalues[0]
print listofvalues[1]
print listofvalues[2]
This will print
11
L
0

Two new string removal methods are introduced in Python 3.9+
#str.removeprefix("prefix_to_be_removed")
#str.removesuffix("suffix_to_be_removed")
s='EXAMPLE'
In this case position of 'M' is 3
s = s[:3] + s[3:].removeprefix('M')
OR
s = s[:4].removesuffix('M') + s[4:]
#output'EXAPLE'

from random import randint
def shuffle_word(word):
newWord=""
for i in range(0,len(word)):
pos=randint(0,len(word)-1)
newWord += word[pos]
word = word[:pos]+word[pos+1:]
return newWord
word = "Sarajevo"
print(shuffle_word(word))

Strings are immutable in Python so both your options mean the same thing basically.