dict of a function clear solution for Digit Stack by Olpag - python coding challenges

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dict of a function solution in Clear category for Digit Stack by Olpag

def digit_stack(commands):
    func = {'PUSH': lambda stack, arg: stack.append(arg),
            'POP': lambda stack: stack.pop() if stack else 0,
            'PEEK': lambda stack: stack[-1] if stack else 0}
    stack, summ = [],0
    for command in commands:
        com, arg = '',''
        if command.find('PUSH') != -1:
            com, arg = command.split()
            func[com](stack, arg)
        else:
            com = command
            summ += int(func[com](stack))
    return summ

April 3, 2019

Comments:

obone on Aug. 3, 2019, 11:35 a.m.
Looking clear. But it may be done more nicer. I guess you can use only one func() call, i.e.: comm, arg = command.split() if command.find('PUSH') != -1 else command, '' summ += int(func<a href="stack, arg">com</a>)

Olpag on Aug. 5, 2019, 8:12 a.m.
Thanks for the code review. It is possible. But then I need to rewrite all three functions. In the POP and PEEK functions, a second empty argument must be added. And in the PUSH function need to wrap the append() method so that an integer or string value is returned (the append() method returns None). The result code: <pre class='brush: python'>def digit_stack(commands): func = {'PUSH': lambda stack, arg: bool(stack.append(arg)), 'POP': lambda stack, arg: stack.pop() if stack else 0, 'PEEK': lambda stack, arg: stack[-1] if stack else 0} stack, summ = [],0 for command in commands: com, arg = (command.split()) if command.find('PUSH') != -1 else (command,'') summ += int(func[com](stack, arg)) return summ </pre>

obone on Aug. 10, 2019, 5:21 p.m.
Hello! I guess that this looks more elegant. You also can use _ for arg (if it unused in lambda). Good job!

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