FREE Programming/Coding Help Desk

When I first started programming I preferred to use a simple editor, like Atom or Sublime Text, since I was forced to learn APIs and didn't have to rely on IDE autocorrections/suggestions. Although, it's not necessarily a bad idea to start with an IDE and I agree with Async that Jetbrains' products are usually very good.

For the first example, its
if text in word:
not the other way around.

For the second one, do
if food.lower() == "y":

.lower() basically makes it all lowercase, so it doesnt matter if the person inputs Y or y, it will work.

Hello!

I need help with validation for my python code.
text=input("What is you favourite animal?" )
word=['wolf' , 'cat' , 'lion' , 'zebra']
if word in text:
print("Your favourite animal is a mammal." )

Here is another example

print(" y is yes and n is no" )
food= input("Do you like chocolate cake?" )
if food== 'y' : ## what if the user inputs Y or yes instead of y, it won't work that's why i need help.
print(" Me too" )

I basically need help with validation so if the user input is in capital letters or has punctuation or i guess even typos it should still work. I need help doing this.



Thanks

hey i need help on task 2 and 3 its getting the speeding reg plates into a text file that i am having trouble with i will pm the code to anyone who is willing to help

m = 1
my_list_1 = [2 , 4, 1]
for x in my_list_1:
for y in range(1,3):
if (x + y) % 3:
m = m * x
print (m)the part i highlighted is my problem
this is the first time i have seen sometime different to the normal whereby the conditon is to see if the remainder is 0 or not.

but for the highlighted if condition there is no such comparison
i wanted to ask what is the if statement doing exactly

thank you

Basically, specifying that the modulus have to be equal to zero is redundant because python knows it has to be anyway.

The if statement is seeing if x+y is divisible by 3 - seeing if the remainder is 0 or not. You don't need to specify "X MOD = 0" in the if statement because % essentially means divisible which makes you deduce the fact that remainder will indeed be 0. Basically, specifying that the modulus have to be equal to zero is redundant because python knows it has to be anyway.

Hi! I need help with python, i am a beginner so please help! I need some help in recognizing multiple words.

w1=['wolf' , 'lion' , 'elephant']
w2=['chicken' , 'cuckoo']

question=str(input("What is you favourite animal?")

if question.lower()in w1:
print("your fav animal is a mammal"
elif question.lower() in w2:
print("your fav animal is a bird"

If i type in single words as the input like wolf or chicken , the code works but my problem is if i enter 'my favourite animal is a lion', it won't work. So python isn't recognizing the words if its with other words.why? and how do i fix this? if you can think of any other working methods that recognize multiple words in python, please tell me!
Please help , thanks!

The easiest way to do this would be splitting the "question" variable to make a FOR loop for each word, and then check if that word is in the "w1" or "w2" list, like so:
w1=['wolf' , 'lion' , 'elephant']
w2=['chicken' , 'cuckoo']

question=str(input("What is you favourite animal?")

for word in question.split():
if word in w1:
print("your fav animal is a mammal"
elif word in w2:
print("your fav animal is a bird"

Hi again!

y=['yes' , 'yeah' , 'yep' , 'y' , 'yup']
n=['no' , 'nah' , 'nope']

def work():
a=input("is this correct?"
if a==y:
print("Have a great day,Good bye!"
elif a==n:
print("We are sorry."
else:
work()

so my problem here is with my data validation
if the user doesn't enter y or n then the work will be called again

bit its stuck in an infinite loop!
Even if i input y it still calls the function again.

and after that if the user keeps entering the incorrect answer for 10 times i wanna automatically exit the programme?
I can use a for loop? but how?
Please help!

You're checking if the string the user has input into a is equal to your list of strings. You need to check if their input is in your list instead.

Easiest way to exit would be to have a variable to count the number of times they have been asked, and just before calling your work() function increment it and check if it is greater than 10.