i'm trying to allow the user to change the value for a dictionary, my dictionary is d as follows:
d = {'gary:' : 'B','john:': 'C',}
Whenever I type in 'gary:'(gary with a semicolon) for the nameedit input, the why variable will always result in true but regardless, it will read the value as none and never reach the second input for the letter grade(inpu)
nameedit = str(input("Which student do you wish to edit the grade of?"))
why = print(nameedit in d)
if why == None:
print("That student doesn't exist")
else:
inpu = str(input("Enter new letter grade: A,B,C,D,F,"))
d[nameedit] = inpu
print(d)
I also tried some variations on this such as if nameedit == True: and an else with the same problem, the print statement will yield True but it will just continue to the else statement. I also tried a elif nameedit in d:
why = print(nameedit in d)
if nameedit in d == True:
inpu = str(input("Enter new letter grade: A,B,C,D,F,"))
d[nameedit] = inpu
print(d)
else:
print("That student doesn't exist")
, but with no avail. Is it impossible to pick up the value the print statement is reading? My ultimate goal is to simply check if the name is in the dictionary, if it is, continue, if it isn't, stop
python v 3.5
Use:
why = nameedit in d
if why:
inpu = str(input("Enter new letter grade: A,B,C,D,F,"))
d[nameedit] = inpu
print(d)
else:
print('Does not exists')
print() just puts output to the screen; it does not return what it is printing. If you want to assign why to the result of nameedit in d and print it; do it on separate lines:
why = nameedit in d
print(why)
Also, if ... is seeing if ... is True. Using if ... == True is seeing if ... == True == True. That is redundant. Just do this:
why = nameedit in d
print(why)
if why:
inpu = input("Enter new letter grade: A,B,C,D,F,")
d[nameedit] = inpu
print(d)
else:
print("That student doesn't exist")
I also removed your conversion of input(...) to a string. It already returns a string, so the conversion is redundant.
nameedit in d returns True or False, never None. And wrapping in print (which always returns None) is just nonsensical.
Remove the print wrapping, and change the test to if not why: or just move the test into the condition itself:
nameedit = input("Which student do you wish to edit the grade of?")
if nameedit not in d:
print("That student doesn't exist")
else:
d[nameedit] = input("Enter new letter grade: A,B,C,D,F,")
print(d)
I removed unnecessary str wrapping (input already returns str in Py3), and unnecessary intermediate locals.
Use just like:
if why in d:
or
if why not in d:
You don't need use two strings for this. Also, you can get value with get:
d = {'gary:' : 'B','john:': 'C',}
name = str(input("Which student do you wish to edit the grade of?\n"))
value = d.get(name, None)
if not value:
print("That student doesn't exist: {0}: [1}".format(name, value))
else:
value = str(input("Enter new letter grade: A,B,C,D,F \n"))
d[name] = value.upper()
print(d)
Related
For the get_letter_from_user function, while using the while loop for validation, it keeps repeating the invalid input; I want to make sure that it is a single letter and lower case, and I want to make sure that it doesn't equal the second parameter of the function. I'm not sure what I'm doing wrong, though. (and how to get gud at coding if u have tips)
def get_text_from_user(prompt):
return input(prompt).lower()
def get_letter_from_user(prompt, not_allowed):
not_allowed = ''
allowed = input(prompt).lower()
while not allowed == not_allowed or allowed.isalpha() or len(allowed) > 1:
allowed = str(input('Invalid letter, try again:'))
return allowed
def main():
text = get_text_from_user("Enter some text: ")
ltr1 = get_letter_from_user("Enter a letter: ", '')
ltr2 = get_letter_from_user("Enter another letter: ", ltr1)
new_text = text.replace(ltr1,ltr2)
print("The new text is", new_text)
if __name__ == "__main__":
main()
Suggestion for the function:
def get_letter_from_user(prompt, not_allowed):
allowed = input(prompt).lower()
while allowed == not_allowed or len(allowed) > 1:
print('not_allowed:',not_allowed)
allowed = str(input('Invalid letter, try again:'))
return allowed
ltr1 = get_letter_from_user("Enter a letter: ", '')
ltr2 = get_letter_from_user("Enter another letter: ", ltr1)
Sample output:
Enter a letter: d
Enter another letter: d
not_allowed: d
Invalid letter, try again:d
not_allowed: d
Invalid letter, try again:a
To replace a letter or sequence of letters in a string, you might want to take a look at the string.replace() function:
text = input('Enter some text: ')
find = input('Enter a letter to replace: ')
replace_with = input(f'Enter a letter to replace \'{find}\' with: ')
replaced = text.replace(find, reolace_with)
print('The new text is:', replaced)
To add another little detail because you asked how to get better at coding:
I would never make a function with a parameter that is immediately changed to an empty string. Like:
def get_letter_from_user(prompt, not_allowed):
not_allowed = ''
Rather use a default value like this:
def get_letter_from_user(prompt, not_allowed=''):
...
Simply question making phone dictionary
What I want to do is putting person's name and number and finding them in dictionary!
Examples what I want to do
Enter command (a, f, d, or q).: a
Enter new name................: Perry
Enter new phone number........: 229-449-9683
Enter command (a, f, d, or q).: f
Enter name to look up...:
I would like to find full name and number when I type
Phone dictionary code what I wrote so far:
phone_dict = {}
command = input('Enter command (a, f, d, or q).: ')
newname = input('Enter new name................: ')
newphone = input('Enter new phone number........: ')
while True:
if command == 'a':
newname
newphone
phone_dict[newname] = newphone
print(phone_dict)
# In here, 'while-loop' does not work.
In there, if I enter 'a' command, and type the name
The dictionary is supposed to be { Perry: 229-449-9683}
Thanks, The question might be little confused, but if you can help this out, I am very happy!
To find the result from the dictionary, you can loop through the items and check if the key contains the string you want to find. If you want to get all values which satisfy your query, you can create another list or dictionary and store the items you find:
phone_dict = {
"Han Perry": "1234",
"Harry Gildong": "2345",
"Hanny Test": "123",
}
find_str = "Han"
result = {}
for key, value in phone_dict.items():
# Converting it to lower makes it case insensitive
if find_str.lower().strip() in key.lower():
result[key] = value
print(result)
# {'Han Perry': '1234', 'Hanny Test': '123'}
Take note that this will run through all of the values of the dictionary: O(n)
To find the number using the first o last name of the person you could do:
a = 'Add a new phone number'
d = 'Delete a phone number'
f = 'Find a phone number'
q = 'Quit'
phone_dict = {}
while True:
# Gets the user command every loop
command = input('Enter command (a, f, d, or q).: ')
# Add a new registry to the directory
if command == 'a':
newname = input('Enter new name................: ')
newphone = input('Enter new phone number........: ')
phone_dict[newname] = newphone
print(phone_dict)
# Find a registry on the directory
elif command == "f"
query = input("Enter name to look up...: ")
match = None
for key in phone_dict.keys():
if query.strip() in key:
match = phone_dict[key]
break
if match is None:
print(f"The name {query} could not be found on the directory")
else:
print(f"The phone number of {query} is {match}")
elif command == "d":
# Delete registry
elif command == "q":
# Quits program
else:
print(f"The command {command} was not found, please try again!")
In this case, I am using query.strip() to remove any extra start/end spaces that could cause to not find the person.
I'm trying to make a program where I input a name and a surname and the code checks if the name is invalid (list of invalidiations below). If it has any invalidations, it asks me to say the name again and presents me a list of all the invalidations.
Invalidations list (I'll show the code version too):
- The name has digits
- The name has symbols
- The name has no spaces
- It has more then one space
- One of the names is either too short or too long
- The first letter of the name is a space
- The last letter of the name is a space
I can't use exceptions here, because these are not code erros. I've made it with Ifs, but it got to a point where there a simply lots of Ifs for it to be viable.
def has_digits(name):
digits = any(c.isdigit() for c in name)
if digits == True:
return True
print("Your name has digits.")
else:
return False
def has_symbols(name):
symbols = any(not c.isalnum() and not c.isspace() for c in name)
if symbols == True:
return True
print("Your name has symbols.")
else:
return False
def has_no_spaces(name):
spaces = any(c.isspace() for c in name)
if not spaces == True:
return True
print("You only gave me a name.")
else:
return False
def many_spaces(name):
m_s = name.count(' ') > 1
if m_s == True:
return True
print("Your name has more than one space.")
else:
return False
def unrealistic_length(name, surname):
length= (float(len(name)) < 3 or float(len(name)) > 12) or float(len(surname)) < 5 or float(len(surname) > 15)
if length == True:
return True
print("Your name has an unrealistic size.")
else:
return False
def first_space(name):
f_s = name[0] == " "
if f_s == True:
return True
print("The first letter of your name is a space.")
else:
return False
def last_space(name):
l_s = name[-1] == " "
if l_s == True:
return True
print("The last letter of your name is a space.")
else:
return False
name = "bruh browski"
namesplit = name.split(" ")
name1 = namesplit[0]
name2 = namesplit[1]
print(has_digits(name))
print(has_symbols(name))
print(has_no_spaces(name))
print(many_spaces(name))
print(unrealistic_length(name1, name2))
print(first_space(name))
print(last_space(name))
Maybe the prints shouldn't be in the defs themselves. I don't know. I'm almost sure doing a for loop is the way to go, but I just can't imagine how to do it.
Result:
False
False
False
False
False
False
False
The methods you've used to define exactly what counts as each "invalidation" will have to stay, unless you can replace them with something else that does the same thing. But you can check all of those conditions at once using a generator expression:
if any(is_invalid(name) for is_invalid in [
has_digits, has_symbols, has_no_spaces, many_spaces, unrealistic_length, first_name, last_name
]):
# then this string is invalid
# otherwise, all of those returned false, meaning the string is valid.
You can then use that condition to determine when to stop asking the user, or however else you need to.
If you wanted to not individually define all those functions, you could also maybe use lambdas to do the same thing.
As a sidenote, before actually using this in production for checking the validity of names, I advise having a look at the list of Falsehoods Programmers Believe about Names. It's a fun read even if it's not relevant to your use case, though.
You could have a single function which calls all of your other functions and handles it appropriately.
def master_verify(name):
# Put all your verify functions in the list below.
verify_funcs = [has_digits, has_symbols, has_no_spaces, many_spaces,
unrealistic_length, first_space, last_space]
# It will return True if any your functions return True. In this case,
# returning True means the name is invalid (matching your other
# function design). Returning False means the name is valid.
return any(is_invalid(name) for is_invalid in verify_funcs)
Since you mentioned you want the program to find any name errors and ask the user to try again, we can write a loop to handle this.
def get_name():
while True:
# Loop until you get a good name
name = input("Enter your name: ").strip()
if master_verify(name):
# Remember, if True this means invalid
print("Invalid name. Try again.")
continue # continue jumps to the top of a loop, skipping everything else.
return name # Will only get here if the name is valid.
I also suggest you should do the name and surname split inside your unrealistic_length function.
Then, all you need to do is
name = get_name()
# All of the validation has already happened.
print(f"The correct and validated name is: {name}")
Last but not least, anything in a function after a return is unreachable. So a lot of your prints will never happen. Put the print statements before your return.
Alright. I've managed to do it by myself. I still fill there's a better way to do it, but this is the way I found.
errors_list = []
print("Hi. Tell me your first and last name.")
def choose_name(name):
global fname
global sname
fname = ""
sname = ""
global errors_list
try:
no_letters = any(c.isalpha() for c in name)
no_spaces = name.count(" ") == 0
digits = any(c.isdigit() for c in name)
symbols = any(not c.isalnum() and not c.isspace() for c in name)
many_spaces = name.count(" ") > 1
first_space = name[0] == " "
last_space = name[-1] == " "
if no_letters == False:
errors_list.append("It has no letters")
if no_spaces == True:
errors_list.append("It has no spaces")
else:
namesplit = name.split(" ")
fname = namesplit[0]
sname = namesplit[1]
pass
if fname and sname is not "":
bad_length = (float(len(fname)) < 3 or float(len(fname)) > 12) or float(len(sname)) < 4 or float(len(sname) > 15)
if bad_length == True:
errors_list.append("One of your names has an unrealistic size")
pass
else:
bad_length = (float(len(name)) < 3 or float(len(name)) > 12)
if bad_length == True:
errors_list.append("It has an unrealistic size")
pass
if digits == True:
errors_list.append("It has digits")
pass
if symbols == True:
errors_list.append("It has symbols")
pass
if many_spaces == True:
errors_list.append("It has more than one space")
pass
if first_space == True:
errors_list.append("The first letter is a space")
pass
if last_space == True:
errors_list.append("The last letter is a space")
pass
except IndexError:
print("You must write something. Try again.")
name = input("My name is ").title()
choose_name(name)
name = input("My name is ").title()
choose_name(name)
while True:
if len(errors_list) != 0:
print("Your name has these errors:")
for i in errors_list:
print(" " + str(errors_list.index(i) + 1) + "- " + i + ".")
print("Try again.")
errors_list.clear()
name = input("My name is ").title()
choose_name(name)
else:
print("Nice to meet you, " + fname + " " + sname + ".")
break
Result when I type the name '----... '
Hi. Tell me your first and last name.
My name is ----...
Your name has these errors:
1- It has no letters.
2- It has symbols.
3- It has more than one space.
4- The last letter is a space.
Try again.
My name is
I have a list of dictionaries l1:
l1 = [
{'name':'Omer','fname':'ghaffar','sid':1},
{'name':'zoya','fname':'sattar','sid':2}
]
I'm trying to verify whether some content exists in l1. For example, I want to check if the name and sid is the same as the input values (a1 and a2), then it should display "valid user" along with the values.
My code:
a1 = input("name ")
a2 = input("sid ")
for e in l1:
if e['name'] == a1 and e['sid'] == a2:
print("valid user")
break
else:
print("invalid user")
The result I'm getting with the above code is:
name Omer
sid 1
invalid user
invalid user
I expect the result to be:
name Omer
sid 1
valid user
Change your condition to:
if e['name'] == a1 and e['sid'] == int(a2):
This is because by default input reads as string, so when you compare string with an integer, it's always a False.
Alternatively, you could also cast at the time you read from user also:
a2 = int(input("sid "))
Also note that your code keeps on printing 'invalid user' for an invalid entry or a valid entry that comes later in the list. To avoid this, you can add else to for instead of if:
a1 = input("name ")
a2 = input("sid ")
for e in l1:
if e['name'] == a1 and e['sid'] == int(a2):
print("valid user")
break
else:
print("invalid user")
While working on my program I have run into a problem where the information stored in Menu option 1 is not being transferred to Menu option 2. As you can see it is correctly stored when in menu one. When it returns to go to menu option 2 its like it never went to option 1.
update #1:
some suggestions I've had is to understand scope? from what I can tell the program is not passing the data along to its parent program even though I've typed out return in each of the definitions.
#Must be able to store at least 4 grades
#Each class can have up to 6 tests and 8 hw's
#Weighted 40%*testavg 40% hw average attendance is 20%
#User must be able to input a minimum grade warning
#after each test the your program must calculate the students average and issue warning if necessary
##Define the Modules##
import math
def menu (a): #2nd thing to happen
menuend = 'a'
while menuend not in 'e':
menuend = raw_input("Type anything other then 'e' to continue:\n")
print "What would you like to do ?"
menudo = 0
print "1 - Enter Courses\n2 - Select Course to Edit\n3 - Save File\n4 - Load File\n5 - Exit\n"
menudo = input("Enter Selection:")
if (menudo == 1):
menuchck = 0
menuchck = raw_input("\nYou have entered #1 (y/n)?:\n")
if menuchck in ["Yes","yes","y","Y"]:
x = m1()
else:
print "I'm sorry,",nam,",for the confusion, lets try again\n"
menu()
elif (menudo == 2):
menuchck1 = 0
menuchck1 = raw_input("\nYou have entered #2 (y/n)?:\n")
if menuchck1 in ["Yes","yes","y","Y"]:
x = m2()
else:
print "I'm sorry,",nam,",for the confusion, lets try again\n"
menu()
elif (menudo == 3):
print "Entered 3"
elif (menudo == 4):
print "Entered 4"
else:
print "Anything Else Entered"
def course(): #3rd thing to happen
b = {}
while True:
while True:
print "\n",name,", please enter your courses below ('e' to end):"
coursename = raw_input("Course Name:")
if (coursename == 'e'):
break
will = None
while will not in ('y','n'):
will = raw_input('Ok for this name : %s ? (y/n)' % coursename)
if will=='y':
b[coursename] = {}
print "\n",name,", current course load:\n",b
coursechck = None
while coursechck not in ('y','n'):
coursechck = raw_input("Are your courses correct (y/n)")
if coursechck =='y':
return b
else:
b = {}
print
##Menu Options##
def m1():
a = course()
return a
def m2():
print "Excellent",name,"lets see what courses your enrolled in\n"
print x
return x
###User Input Section###
name = raw_input("Enter Students Name:\n")
a = {}
menu(a)
raw_input("This is the end, my only friend the end")
In your if-elif blocks in the do==1 case, you write m1(), but for the last case, you write x=m1(). You should have the latter everywhere (by typing m1() you only run the function, but do not store the returned x anywhere).
By the way, you can avoid this if-elif confusion using if chck in ["Yes","yes","Y","y"]: