(Python) For loop syntax - execute for only one item? - python

Pretty new to python/programming in general, this is my biggest project yet.
I am writing a program that will do SUVAT equations for you. (SUVAT equations are used to find the displacement, start/end velocity, acceleration and time travelled by an object with constant velocity, you may call them something different.)
I made this list:
variables = ["Displacement", "Start Velocity", "End Velocity", "Acceleration", "Time"]
which is used in the following while/for loop:
a = 0
while a==0:
for variable in variables:
# choice1 is what the user is looking to calculate
choice1 = raw_input("Welcome to Mattin's SVUVAT Simulator! Choose the value you are trying to find. You can pick from " + str(variables))
# will execute the following code when the for loop reaches an item that matches the raw_input
if choice1 == variable:
print "You chave chosen", choice1
variables.remove(variable) #Removes the chosen variable from the list, so the new list can be used later on
a = 1 # Ends the for loop by making the while loop false
# This part is so that the error message will not show when the raw_input does not match with the 4 items in the list the user has not chosen
else:
if choice1 == "Displacement":
pass
elif choice1 == "Start Velocity":
pass
elif choice1 == "End Velocity":
pass
elif choice1 == "Acceleration":
pass
# This error message will show if the input did not match any item in the list
else:
print "Sorry, I didn't understand that, try again. Make sure your spelling is correct (Case Sensitive), and that you did not inlcude the quotation marks."
Hopefully the comments I have written in the code should explain my intentions, if not, feel free to ask anything.
The problem is that when I run the code, and input choice1, the for loop activates the last line of code:
else:
print "Sorry, I didn't understand that, try again. Make sure your spelling is correct (Case Sensitive), and that you did not inlcude the quotation marks."
and then prompts me to enter the input again, and will do this as many times as it needs to get to the item on the list that I am typing.
However, I specifically coded that if what I input does not match the item on the list the for loop is currently checking, but does match one of the other items on the list, then it should pass and loop round to checking the next item.
I am probably doing something stupid, but I don't see it, so please help me figure out what I have to do to get my desired result? I assumed it was the syntax I had wrong so that is why that is the title.
Thanks for any help, I appreciate it.

Besides the problem with the indentation in your pasted code, I would rewrite it as such:
while True:
choice = raw_input('...')
if choice in variables:
print "You chave chosen", choice
# Remove the chosen member from the list
variables = [v for v in variables if v != choice]
# Break out of loop
break
# Print error messages etc.
Also remember that string comparisons are case sensitive. I.e 'Displacement' != 'displacement'.

Related

Comparing raw input to an specific element in an array using Python

I have a little piece of code in Python where I'm trying to compare a user input to a specific element in an array. Here is the code:
movies = ["movie 1", "movie2", "movie3"];
answer = raw_input("What is your guess: ")
if answer == movies[1]
then print ("yes that is correct")
else:
print ("no that is incorrect")
I know the indentation above looks wrong becasue I typed it out in the text box and I'm new to this site as well as python.
I also know that I probably need to use some sort of conditional loop, maybe a while loop, but I'm having trouble finding where I can compare user input string value to a string value in my array. Any ideas how I might accomplish this?
Have fun with Python! I guess you are trying to make a loop which keeps receiving inputs from user to compare with the desired input until user types the correct input. If so, one way, it can be implemented as following (but think of adding a break condition, like input == "Bored" , to avoid infinite loop and hard stopping your code):
movies = ["movie 1", "movie2", "movie3"]
correctAnswer = movies[1]
is_notCorrect = True
while(is_notCorrect):
answer = raw_input("What is your guess: ")
if answer == correctAnswer:
print("Yes, that is correct")
is_notCorrect = False
else:
print("No, that is incorrect")
In the code above, when is_notCorrect turns into False. At next condition checking, it will break condition, and done with the loop.
Your code has some issues
movies = ["movie 1", "movie2", "movie3"]; # No need the semi-colon in Python
answer = raw_input("What is your guess: ")
# Need a colon here after if condition, new line, and indent.
#If you don't like the colon, you need to write a different way with one line of code Eg: <Do A> if <Condition happens> else <Do B>
if answer == movies[1]
then print ("yes that is correct") # No then in if-else statement in Python
else:
print ("no that is incorrect")

Pulling from a list

My question is about getting a user to pull and item from a list. If the item from the list isn't pulled from the list I want to tell the user that he is incorrect. So my code looks like this:
Body_Type = ['Large', 'Medium', 'Small']
print('Create a Character-')
print('Body Type Choices: ' + str(Body_Type))
bt = input('Enter your Body Type: ')
while bt != Body_Type:
if bt == Body_Type[0:]:
print('Your Body Type is: ' + bt)
else:
print('Invalid Body Type')
What I'm trying to do is make my user create a character. This is just the first part of my first simple project. I want to have him pull from one of the items on the list, being "Large, Medium, Small" respectively. I want it to repeat until the user chooses one of the three. I tried to use or but it seems to feel unorganized and I'd have to break up the list and assign each individual variable.
Thanks in advance!
Several errors here like comparing a string to a list, or random slicing hoping that it would work. And the fact that your input statement is before the loop creates an infinite loop because you're comparing 2 variables of a different type again and again (bt != Body_Type is always True regardless of the content of bt since left hand is a string, right hand is a list).
But it shouldn't be so complex to write some working code.
I would create an infinite loop and break only if choice is in the list:
while True:
bt = input('Enter your Body Type: ')
if bt in Body_Type:
print('Your Body Type is: ' + bt)
break
else:
print('Invalid Body Type')
simpler and clearer (and repeats input if fails). The infinite loop (with an always true condition) allows to avoid double input call & test. Just loop, input the string, and break from the loop if matches.
The key statement you were looking for was bt in Body_Type which tests if the string is within the list.

Python user defined list not being recognised within a while true loop

Thanks firstly for bearing with me as a relative newcomer to the world of Python. I'm working on a simple set of code and have been racking my brain to understand where I am going wrong. I suspect it is a relatively simple thing to correct but all searches so far have been fruitless. If this has been covered before then please be gentle, I have looked for a couple of days!
I'm working on the following and after catching and correcting a number of issues I suspect that I'm on the last hurdle:-
def main():
our_list = []
ne = int(input('How many numbers do you wish to enter? '))
for i in range(0, (ne)): # set up loop to run user specified number of time
number=int(input('Choose a number:- '))
our_list.append(number) # append to our_list
print ('The list of numbers you have entered is ')
print (our_list)
main()
while True:
op = input ('For the mean type <1>, for the median type <2>, for the mode type <3>, to enter a new set of numbers type <4> or 5 to exit')
import statistics
if op == "1":
mn = statistics.mean(our_list)
print ("The mean of the values you have entered is:- ",mn)
if op == "2":
me = statistics.median(our_list)
print ("The median of the values you have entered is:- ",me)
if op == "3":
mo = statistics.mode(our_list)
print ("The mode of the values you have entered is:- ",mo)
if op == "5":
main()
else:
print("Goodbye")
break`
For some reason the appended (our_list) is not being recognised within the while true loop rendering the statistics calculation void. Any steer would be really appreciated as to where I am missing the obvious, thanks in advance.
Cheers
Bryan
I'm not sure exactly what you mean by "not being recognized", but our_list is a local variable inside main, so it can't be used anywhere but inside main.
So, if you try to use it elsewhere, you should get a NameError.
If your code actually has a global variable with the same name as the local variable that we aren't seeing here, things can be more confusing—you won't get a NameError, you'll get the value of the global variable, which isn't what you want.
The best solution here is to return the value from the function, and then have the caller use the returned value. For example:
def main():
our_list = []
ne = int(input('How many numbers do you wish to enter? '))
for i in range(0, (ne)): # set up loop to run user specified number of time
number=int(input('Choose a number:- '))
our_list.append(number) # append to our_list
print ('The list of numbers you have entered is ')
print (our_list)
return our_list
the_list = main()
while True:
op = input ('For the mean type <1>, for the median type <2>, for the mode type <3>, to enter a new set of numbers type <4> or 5 to exit')
import statistics
if op == "1":
mn = statistics.mean(the_list)
print ("The mean of the values you have entered is:- ",mn)
if op == "2":
me = statistics.median(the_list)
print ("The median of the values you have entered is:- ",me)
if op == "3":
mo = statistics.mode(the_list)
print ("The mode of the values you have entered is:- ",mo)
if op == "5":
the_list = main()
else:
print("Goodbye")
break
There are other options—you could pass in an empty list for main to fill, or use a global variable (or, better, a more restricted equivalent like an attribute on a class instance or a closure variable), or refactor your code so everyone who needs to access our_list is inside the same function… but I think this is the cleanest way to do what you're trying to do here.
By the way, this isn't quite the last hurdle—but you're very close:
After any mean, median, or mode, it's going to hit the "Goodbye" and exit instead of going back through the loop. Do you know about elif?
You mixed up '5' and '4' in the menu.
If the user enters 2 and 3 and asks for the mode, your code will dump a ValueError traceback to the screen; probably not what you want. Do you know try/except?
That's all I noticed, and they're all pretty simple things to add, so congrats in advance.
The issue is that our_list was defined in the main() function, and is not visible outside of the main() function scope.
Since you're doing everything in one chunk, you could remove line 1 and 6, taking the code from your main() function and putting it on the same indentation level as the code which follows.
This seems to be because you defined our_list within the main() function. You should probably define it as a global variable by creating it outside the main() function.
You could also put the while loop inside a function and pass in our_list as a parameter to the list.

Error Tapping a section of code

I have a while statement which works well and I have a whole section of code that asks the user to input how many names they have which will then ask them for a name that amount of times and then each time a name will be entered.
I need the section of the names entered to be error tapped but I don't know how to do it, as I have a while statement and I may need to put another while statement in, although I have error tapped the section for amount of names in numbers.
Also there is code further on with a dictionary and sorts but I need help with the one section of error tapping started at while currentnum part
print("Please enter each name when asked without any spaces.") #The program will post this
print("Please enter each of your names individually also.") #Program will again post this
names = [] #This is the value of names which will be changed depending on the input
currentnum = 0 #Currentnum value is 0
while True: #While loop as it will revert to the start if question answered incorrectly
try:
numofnames = int(input("How many names do you have? "))
except ValueError: #if the input is not an integer or a whole number it will
print("Sorry that was not a valid input please retry")
continue #it will loop back and ask the question again as it says that the unput was not valid
else:
break #If the input is correct then the loop will break and continue to the next section of the program
while currentnum < numofnames: #This means that while currentnum is smaller than input for numofnames it will continue to ask question. This is another loop
currentnum = currentnum + 1 # every time the question is asked it means that currentnum gets 1 added to it and will continue to ask untill it is the same as the input for numofnames
name = str(input("Enter your name: ")) #Name asked to be entered in string
name = name.upper() #This changes all letters to upper case no matter what so there is no error for upper and lower case or a bigger dictionary showing lower and upper case values.
names.append(name)
Yep. The easiest way to describe what you're doing is to use the .isalpha attribute in an if statement. First you will have to def your while loop
Like the following:
def Loop():
name_input_complete = False
while name_input_complete != True:
string = input("Please input :")
string = str(string)
if string.isalpha():
name_input_complete = True
else:
print("Please do not use numbers and spaces")
Loop()
Loop()
Baisically you have to define the loop and then run it. The if statement(which is the part you should add to your loop) then checks if there is nothing other than letters. If true, your done with error trapping and the loop is exited. The program continues outside the loop. If not then the while is repeated because the Loop() function is called again.
Your code looks very good to me. I dont see what input can the user put that could cause an error, since most data in python can be stringed AND names can be pretty much anything!. If you could comment exactly what error could be caused i might be able to help you

In this very basic code i can't figure out what's the sytax error here in line 6 is (python)

myName = input("Hey there, what's your name?")
print("Hello",myName,"!")
print("Here's a game called ''Guess my number'', in this game you will have to guess my number in 5 tips, I will think of a number between 1 and 20.")
ready = input("Are you readyyyy!?")
if ready = "yes" or "yeah" or "totally" or "hell yeah" or "yupp" or "yepp" or "uhumm" or "sure": <-- here's the problem it says, at "sure"'s 1st "-sign
print("Let's go!")
loop = "y"
else:
print("I'm sorry to hear that.")
loop "n"
Could please anyone help, beginner here. I tried to delete and add new word, I restared the program and the computer because there's something clearly wrong. If I delete a word like "sure" the pointer will still point to the same exact place but there's nothing there...
You're using a single = sign in your if statement. That's not allowed. If you want to check for equality, you'll need to use ==. The = operator is only for assignment statements.
While changing = to == will fix the syntax error, your code still won't work exactly right. That's because == will not be distributed over all the or options you show. The expression a == b or c gets interpreted as (a == b) or c, and if c is "truthy" (as any non-empty string will be), the expression will be considered true.
Instead, you probably want to use something like if ready in {"yes", "yeah", "totally"}. This creates a constant set object and tests if the value of the ready variable is in the set (which is a fast check).
You are using a = instead of a == in your if statement. However, I would recommend doing if ready.lower() in {"yes", "yeah", "totally", "hell yeah", "yupp", "yepp"} to account for them using all uppercase.
Also, you seem to be missing your actual loop statements. I noticed you had variables named loop that are 'y' and 'n' but don't actually use them. You should also do something like this:
myName = input("Hey there, what's your name?")
print("Hello",myName,"!")
print("Here's a game called ''Guess my number'', in this game you will have to guess my number in 5 tips, I will think of a number between 1 and 20.")
loop = True
while loop:
ready = input("Are you readyyyy!?")
if ready.lower() in {"yes", "yeah", "totally", "hell yeah", "yupp", "yepp", "uhumm", "sure"}:
print("Let's go!")
loop = False
#To break out of the while loop that will keep asking them when they are ready
else:
print("I'm sorry to hear that.")

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