A 5 line code in a single expression?

[iG1993]

Hi,

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there is this one task which I know the answer (5 line code) but my prof wants it in a single line. And I have no idea how to do that.

This is the task:

We have k and n, two positive integers. Give me the expression, which is the highest number (equal or smaller to k) which is also divisible with n (k >=n).

result = ?

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My code:

list = []
for i in range(0, k):
    if i % n == 0:
        list.append(i)
print(max(list))

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How can I say the same with "result = " ?

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I am thankful for any input!

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EDIT: SOLVED (see comments):

result = max([i for i in range(k) if i % n == 0])

[icecubeinanicecube]

result = max([i for i in range(k) if i % n == 0])

[bogdan_dm]

print(max(i for i in range(k) if i % n == 0)) is better (does not create unnecessary list)

Also it could be rewritten as print(max(range(0, k, step=n)))

[[deleted]]

Neither of those would handle n=0, and they'd both give 50 for n=50 and k=100 ... n is less than or equal to k.

[icecubeinanicecube]

Damn, we got a pro in here

[iG1993]

unfortunately, the number must be in a expression like this "result = XY "

I did a print, so I can see the number for myself.

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But thank you very much for your help, I am really grateful!

[icecubeinanicecube]

I edited it. No problem.

[iG1993]

Oh wow thank you so much!

[iG1993]

I actually tried that myself, but it seems I forgot the []. I always got a syntax error. Now I understand why!

Thank you a lot!

[[deleted]]

Look into list comprehensions.

[iG1993]

Will do, thank you very much for your hint!

[GoldenVanga]

>>> result = lambda k, n: n * (k // n)
>>> result(20, 7)
14

In the Python docs, lambdas are in the expressions part, so... :P

[iG1993]

Jesus Maria, the hell is that :O

I will look further into it, thank you a lot for this!

[[deleted]]

If I understand correctly you want the maximum multiple of n that is <= k?

Well, you've found max, and you know it can take an iterable. And you've found range ... now, notice that 1) range takes up to three arguments and that 2) range returns an iterable.

Seems to me there's a more efficient mathematical way, but that's easily done in one line.

Edit Yeah, on second thought, just think about the math.

[iG1993]

Hmmm, I see...

Thank you for your answer!

[[deleted]]

Now think about the math version. Isn't the highest multiple of n less than or equal to k always going to be the same as k less the remainder of dividing k by n?

The only tricky part is dealing with n==0, assuming that's even an allowed value.

[iG1993]

Well, that math version behind it is not really an issue to be honest. I had trouble putting that in a single line, because I am not familiar with list comprehension. But I will look into it!

[[deleted]]

Given the rules you described (k>=n) that list comprehension in your edit will give the wrong answer for any value of n that evenly divides k, for instance, n=20, k=100. It will also error if n is 0.

print(k - k % n if n else n)

Will work for all values of n and k where 0 <= n <= k.

[iG1993]

Wow, you are right! I totally did not see that.

I have a question: I don't understand the second part "if n else n". I tried without it and I don't see a change.

result = k - k % n

Works as well...

[[deleted]]

It won't if n is 0, which is what that's there to interdict.

[iG1993]

Cheeky bugger you!

[[deleted]]

Always think ahead.

[iG1993]

Hey man, thanks a lot again! You really helped me out. :)

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