First clear solution for Friendly Number by Sim0000 - python coding challenges

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First solution in Clear category for Friendly Number by Sim0000

def friendly_number(number, base=1000, decimals=0, suffix='',
                    powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
    # At first, decompose the number to value and exponent.
    e = 0
    while e + 1 < len(powers) and abs(number) >= base ** (e + 1) : e += 1
    number /= base ** e
    # Then round it.
    number = round(number, decimals) if decimals else int(number)
    # At last, Format it.
    return '{:.0f}'.replace('0', str(decimals)).format(number) + powers[e] + suffix

#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
    assert friendly_number(102) == '102', '102'
    assert friendly_number(10240) == '10k', '10k'
    assert friendly_number(12341234, decimals=1) == '12.3M', '12.3M'
    assert friendly_number(12461, decimals=1) == '12.5k', '12.5k'
    assert friendly_number(1024000000, base=1024, suffix='iB') == '976MiB', '976MiB'

March 27, 2014

Comments:

gyahun_dash on March 28, 2014, 12:33 p.m.
<pre class='brush: python'>return '{:.0f}'.replace('0', str(decimals)).format(number) + powers[e] + suffix </pre> You can use nested brackets for decimals: <pre class='brush: python'>return '{:.{}f}'.replace(number, decimals) + powers[e] + suffix </pre>

Sim0000 on March 28, 2014, 7:54 p.m.
Thank you. I wanted to know it!

veky on March 29, 2014, 3:54 p.m.
Also, while loop can be written as for/else, with else undoing last step.

Sim0000 on March 29, 2014, 4:37 p.m.
I can rewrite line 4-6. It seems good. Thank you. <pre class='brush: python'>for e in range(len(powers)): if abs(number) < base ** (e + 1): break number /= base ** e </pre>

Sim0000 on March 30, 2014, 1:40 p.m.
I can rewrite using enumerate. It seems more pythonic. <pre class='brush: python'> for e, unit in enumerate(powers): if abs(number) < base ** (e + 1): break number /= base ** e </pre>

veky on March 30, 2014, 4:55 p.m.
You can rewrite it in many ways, but I thought about something even simpler. See my solution. ;-]

imtiaz.rahi on Aug. 30, 2019, 12:59 p.m.
This one is much more readable and quick and easy to understand

Negamax on June 10, 2019, 9:37 p.m.
There is an error in your suggestion. Likely, you meant this: <pre class='brush: python'>return '{:.{}f}'.format(number, decimals) + powers[e] + suffix </pre> The `format` instead of `replace`.

apachea on Sept. 22, 2015, 10:05 p.m.
Hello. Thank you for your solution. I have understood everything, except this: <pre class='brush: python'>return '{:.0f}'.replace('0', str(decimals)).format(number) + powers[e] + suffix </pre> Please, can you explain this with comments? Thank you.

Sim0000 on Sept. 23, 2015, 8:16 a.m.
Thank you for your comment. At first, '{:.0f}'.replace('0', str(decimals)) make a format string like '{:.2f}' when decimals is 2. Next format applied. The result of '{:.2f}'.format(number) is '123.40' when number is 123.4. Then power[e] and suffix is merged.

apachea on Sept. 23, 2015, 4:06 p.m.
Thank you!

standrey on June 17, 2016, 3:48 p.m.
My be more pythonic way '{:.{prec}f}{suf}'.format(number, prec = decimals, suff = suffix)

Sim0000 on June 17, 2016, 3:53 p.m.
Thank you!

Nicola_Salari on Nov. 19, 2022, 10:33 a.m.
Heyo only writing cos I need to for quest points and it needs to be at least 30 characters sooooo heyy hope you're doing well and u deserve everything good that's happening in your life :)

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