Loops, loops, loops... clear solution for Pawn Brotherhood by Happiness - python coding challenges

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Loops, loops, loops... solution in Clear category for Pawn Brotherhood by Happiness

def safe_pawns(pawns):
    sPawns = 0
    for p in pawns:
        def1 = chr(ord(p[0]) - 1) + str(int(p[1]) - 1)
        def2 = chr(ord(p[0]) + 1) + str(int(p[1]) - 1)
        if def1 in pawns or def2 in pawns:
            sPawns += 1
    return sPawns

Feb. 14, 2015

Comments:

greenmonke on June 25, 2015, 2:45 p.m.
I tried to do a single if statement, if (def1 or def2) in pawns, but it didn't work. I ended up putting in two if statements, which is ugly.

Happiness on June 26, 2015, 5:29 p.m.
Analysing your statement step by step: <ol> <li>First interpreter executes parenthesis: if (def1 or def2) in pawns</li> </ol> 'Bool1 or Bool2' returns True if one of them True. And False if both False It always returns True in this case because def1 and def2 exists <ol> <li>Second we have: if True in pawns: which doesn't have a sense here</li> </ol>

greenmonke on June 26, 2015, 5:33 p.m.
Yeah. I thought about it for a couple minutes more and it made sense.

Pouf on July 26, 2015, 7:44 p.m.
def1 or def2 would return def1 if bool(def1) == True and def2 otherwise. It would not return True unless def1 is a boolean and is True. if (def1 or def2) in pawns <=> if def1 in pawns

Look up in the Dict()

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