Re: How can I make this more complex?
On 10/11/2020 10:04, Quentin Bock wrote:
grade = input("Enter your grade: ")
if grade >= 90:
print("You got an A ")
if grade >= 80:
print("You got a B ")
if grade >= 70:
print("You got a C")
if grade >= 60:
print("You got a D ")
if grade >= 50:
print("You failed")
First: open a Python terminal (REPL) and try:
import this
(wrt the post's title, lines three and four apply. This output is known
as "The Zen of Python"!)
Did you ever run the code? Hint: it won't work, and even when 'fixed'
won't work the way you want, either. What if the grade is < 50?
If the grade is 55, which message(s) should be printed?
(compare: syntax errors, semantic errors, and errors in logic)
Others have mentioned elif. Why?
If the code looks like a "ladder" with criteria being applied to the
same variable at every step, only one will be chosen with an elif
structure, but >=1 choice will be made without (as above).
Complexity?
Sure, if that's what you want, we've got complexity, but it'll cost you...
(as I tell my trainees: "just because we can do it, does not make it a
good idea!")
Build a dictionary of "buckets" - with the (lowest point) grade-steps as
keys and the applicable messages as values (this will actually be easier
to maintain than the 'ladder'!), eg
90:"You got an A "
80:"You got a B "
...
0:"You didn't say what should happen"
Take the grade, check, and properly prepare it(!)
(why "check"?)
Loop through the dictionary:
if the grade 'fits into' this bucket:
print the message
break# no point in continuing to loop
# otherwise continue looping
Not to be recommended
- but if you are a 'glutton for punishment', don't bother with the
dictionary's 0/last entry. Instead use a for-else structure...
Such would be an excellent case-study illustration of why 'simple' beats
'complex'!
(sorry for the sardonic humor, disregard the last paragraph - most
professional coders (in fact, all that I know) eschew for-else, or-else!)
--
Regards =dn
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Re: How can I make this more complex?
On 2020-11-09 21:04, Quentin Bock wrote:
grade = input("Enter your grade: ")
if grade >= 90:
print("You got an A ")
if grade >= 80:
print("You got a B ")
if grade >= 70:
print("You got a C")
if grade >= 60:
print("You got a D ")
if grade >= 50:
print("You failed")
How can I make this to say if your grade is 90 or higher BUT less than 100
you got an A,
In addition to other the answers, you should note that the 'input'
function returns a string, so you'll need to convert that string to a
number, most probably by using 'int', and, ideally, print a helpful
message if that raises ValueError because it's not a valid number.
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Re: How can I make this more complex?
On Nov 9, 2020 5:59 PM, "Quentin Bock" wrote:
>
> grade = input("Enter your grade: ")
> if grade >= 90:
> print("You got an A ")
> if grade >= 80:
> print("You got a B ")
> if grade >= 70:
> print("You got a C")
> if grade >= 60:
> print("You got a D ")
> if grade >= 50:
> print("You failed")
>
>
>
>
> How can I make this to say if your grade is 90 or higher BUT less than 100
> you got an A
if 100 > grade <= 90:
Additional suggestions:
change all but the first if to elif
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Re: How can I make this more complex?
if 100 > grade >= 90:
On Mon, Nov 9, 2020 at 6:01 PM Quentin Bock wrote:
> grade = input("Enter your grade: ")
> if grade >= 90:
> print("You got an A ")
> if grade >= 80:
> print("You got a B ")
> if grade >= 70:
> print("You got a C")
> if grade >= 60:
> print("You got a D ")
> if grade >= 50:
> print("You failed")
>
>
>
>
> How can I make this to say if your grade is 90 or higher BUT less than 100
> you got an A,
> Thanks
> --
> https://mail.python.org/mailman/listinfo/python-list
>
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How can I make this more complex?
grade = input("Enter your grade: ")
if grade >= 90:
print("You got an A ")
if grade >= 80:
print("You got a B ")
if grade >= 70:
print("You got a C")
if grade >= 60:
print("You got a D ")
if grade >= 50:
print("You failed")
How can I make this to say if your grade is 90 or higher BUT less than 100
you got an A,
Thanks
--
https://mail.python.org/mailman/listinfo/python-list
