ProfRose - 27 days ago 5x
Python Question

# Round to whole numbers without using conditional statements in Python - Logic

I'm taking a Python course at Udacity, and I'm trying to work this out for myself without looking at the answer. Perhaps you can give me a hint for my logic?

Below are the instructions and what I have so far. We haven't learned conditional statements yet, so I can't use those. We've only learned how to assign/print a variable, strings, indexing strings, sub-sequences, and .find. They just introduced the str command in this final exercise.

``````# Given a variable, x, that stores the
# value of any decimal number, write Python
# code that prints out the nearest whole
# number to x.
# If x is exactly half way between two
# whole numbers, round up, so
# 3.5 rounds to 4 and 2.5 rounds to 3.
# You may assume x is not negative.

# Hint: The str function can convert any number into a string.
# eg str(89) converts the number 89 to the string '89'

# Along with the str function, this problem can be solved
# using just the information introduced in unit 1.

# x = 3.14159
# >>> 3 (not 3.0)
# x = 27.63
# >>> 28 (not 28.0)
# x = 3.5
# >>> 4 (not 4.0)

x = 3.54159

#ENTER CODE BELOW HERE

x = str(x)
dec = x.find('.')
tenth = dec + 1

print x[0:dec]
``````

////

So this gets me to print the characters up to the decimal point, but I can't figure out how to have the computer check whether "tenth" is > 4 or < 5 and print out something according to the answer.

I figured I could probably get far enough for it to return a -1 if "tenth" wasn't > 4, but I don't know how I can get it to print x[0:dec] if it's < 5 and x[0:dec]+1 if it's > 4.

:/

Could someone please give me a nudge in the right direction?

This is a weird restriction, but you could do this:

``````x = str(x)
dec_index = x.find('.')
tenth_index = dec_index + 1
tenth_place = x[tenth_index] # will be a string of length 1
should_round_up = 5 + tenth_place.find('5') + tenth_place.find('6') + tenth_place.find('7') + tenth_place.find('8') + tenth_place.find('9')

print int(x[0:dec_index]) + should_round_up
``````

What we do is look at the tenths place. Since `.find()` returns -1 if the argument isn't found, the sum of the `.find()` calls will be -4 if if the tenths place is 5, 6, 7, 8, or 9 (since one of the `.find()` calls will succeed and return `0`), but will be -5 if the tenths place is 0, 1, 2, 3, or 4. We add 5 to that, so that `should_round_up` equals 1 if we should round up, and 0 otherwise. Add that to the whole number part, and we're done.

That said, if you weren't subject to this artificial restriction, you would do:

``````print round(x)
``````

And move on with your life.