First uncategorized solution for Morse Clock by bunnychai - python coding challenges

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First solution in Uncategorized category for Morse Clock by bunnychai

def checkio(data):
    ret = []
    for i, d in enumerate(data.split(':')):
        r, d = '', int(d)
        f, s = d // 10, d % 10
        if i == 0:
            r += '{0:02b} '.format(f)
        else:
            r += '{0:03b} '.format(f)
        r += '{0:04b}'.format(s)
        ret.append(r)
    ret = ' : '.join(ret)
    return ret.replace('0', '.').replace('1', '-')

#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':    
    assert checkio("10:37:49") == ".- .... : .-- .--- : -.. -..-", "First Test"
    assert checkio("21:34:56") == "-. ...- : .-- .-.. : -.- .--.", "Second Test"
    assert checkio("11:10:12") == ".- ...- : ..- .... : ..- ..-.", "Third Test"
    assert checkio("23:59:59") == "-. ..-- : -.- -..- : -.- -..-", "Fourth Test"

July 19, 2013

Comments:

Listjj on Sept. 6, 2013, 7:04 a.m.
nice

alexche_8 on Sept. 6, 2013, 12:55 p.m.
clear

shantanoo on Oct. 4, 2013, 9:26 p.m.
<pre class='brush: python'>if __name__ == '__main__': </pre> Invalid time also works. e.g. assert checkio("30:37:49") == "-- .... : .-- .--- : -.. -..-", "First Test"

PTM on Oct. 15, 2013, 4:55 a.m.
<pre class='brush: python'> assert checkio("11:10:12") == ".- ...- : ..- .... : ..- ..-.", "Third Test" </pre> assert checkio("11:1:12") == ".- ...- : ..- .... : ..- ..-.", works?

bunnychai on Oct. 15, 2013, 8:15 a.m.
refer to example 3, shouldn't it be assert checkio("11:1:12") == ".- ...- : ... ...- : ..- ..-." ?

PTM on Oct. 15, 2013, 11:05 a.m.
yep.

GHfan on Oct. 20, 2013, 2:48 a.m.
format is amazing

BpLife on Oct. 23, 2013, 9:06 a.m.
<pre class='brush: python'> assert checkio("10:37:49") == ".- .... : .-- .--- : -.. -..-", "First Test" </pre> Perfect

junxxx on Nov. 4, 2013, 5:24 a.m.
awesome

rmidyanyy on Nov. 20, 2013, 5:31 p.m.
Honestly? I consider this solution correct, but a bit ugly :) But still algo is probably the same in all solutions.

bylijinnan on Nov. 29, 2013, 9:17 a.m.
I think "Table-Driven-Approach" is better

rupadhya on Dec. 6, 2013, 12:16 a.m.
Like your solution. r, d = '', int(d) is new to me .

Victor_Weng on Dec. 7, 2013, 1:50 p.m.
line 5 -- f, s = d / 10, d % 10 if d==21, f will get 2.1. Obviously, f is float. Python IDE will show " ValueError: Unknown format code 'b' for object of type 'float' "

bunnychai on Dec. 7, 2013, 2:33 p.m.
I think you are using python 3.x, but in 2.7, it would be an integer division.

romalops on Dec. 10, 2013, 10:44 a.m.
My god ! Nice work !

marshall.zheng on Dec. 10, 2013, 1:24 p.m.
Nice work!

forrest1988 on Dec. 11, 2013, 11:42 a.m.
<pre class='brush: python'> r += '{0:03b} '.format(f) </pre> wow, I didn't know that it can be so simple.

datruth29 on Dec. 16, 2013, 4:06 a.m.
Sexy. I was thinking of something along the lines with using format to display the binary digits that are needed, but I couldn't figure out how to format up to a specific number of digits. Little did I know it was so simple, lol.

sapamja on Dec. 17, 2013, 2:35 p.m.
Damn this is awesome, never thought i can solve this way :) but I learnt.

shreekar on Dec. 20, 2013, 10:49 a.m.
Did almost the same way, but using functions.

qbi on Dec. 22, 2013, 11:49 p.m.
<pre class='brush: python'> f, s = d / 10, d % 10 </pre> Seems like a perfect case for divmod().

Strings Theory

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