119 creative solution for Digit Stack by DiZ - python coding challenges

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119 solution in Creative category for Digit Stack by DiZ

digit_stack=d=lambda c,a='':len(c)and(d(c[1:],c[0][5]+a)if' 'in c[0]else(len(a)and int(a[0]))+d(c[1:],a['O'in c[0]:]))

July 9, 2014

Comments:

Mycroft0121 on July 18, 2014, 4:12 p.m.
good job

xtofl on Dec. 30, 2014, 12:31 p.m.
`len(c)` can be replaced by just `c`. You must have lost your enter key somewhere :)

DiZ on Jan. 3, 2015, 5:05 p.m.
Wrong. Have you tried your suggestion? It raises a TypeError exception (unsupported operand type(s) for +: 'int' and 'list' for the first len and Can't convert 'int' object to str implicitly for the second one). As an hint, look at digitstack([])_ ;-)

xtofl on Jan. 4, 2015, 3:16 p.m.
You're right of course. I thought the shorthand `if len(c) == 0` --> `if c` could also be used in any boolean expression. I reconstructed your enter-strokes, too! <pre class='brush: python'>digit_stack=d=lambda c,a='':\ len(c)\ and \ ( d(c[1:],c[0][5]+a) if ' ' in c[0] else (len(a) and int(a[0])) + d(c[1:],a['O'in c[0]:]) ) </pre>

StefanPochmann on April 26, 2015, 5:35 p.m.
You can shorten it a bit by using `sum(a[:1])` and `c[0]>'PF'`: <pre class='brush: python'>digit_stack=d=lambda c,a=[]:len(c)and(d(c[1:],[int(c[0][5])]+a)if' 'in c[0]else sum(a[:1])+d(c[1:],a[c[0]>'PF':])) </pre>

StefanPochmann on April 26, 2015, 7:54 p.m.
And further with `if c[0][4:]` instead of `if' 'in c[0]`.

V.Shkaberda on March 6, 2020, 10:04 p.m.
`' 'in c[0]` well noticed :)

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