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Challenge

Write a program which reads a file containing Morse Code and outputs the conversion.

Specifications

  1. The first argument is a path to a file.
  2. The file contains multiple lines.
  3. Each line is a string of Morse code.
    • Each letter is separated by a space.
    • Each word is separated by two spaces.

Constraints

  1. The input file is correctly formatted.
  2. The Morse Code strings are alphanumeric.

Sample Input

.- ...- ..--- .-- .... .. . -.-. -..-  ....- .....  
-... .... ...--

Sample Output

AV2WHIECX 45
BH3

Source

My Solution:

import sys
import re

morse_alphabet = {
    ''      : ' ',
    '.-'    : 'A',
    '-...'  : 'B',
    '-.-.'  : 'C',
    '-..'   : 'D',
    '.'     : 'E',
    '..-.'  : 'F',
    '--.'   : 'G',
    '....'  : 'H',
    '..'    : 'I',
    '.---'  : 'J',
    '-.-'   : 'K',
    '.-..'  : 'L',
    '--'    : 'M',
    '-.'    : 'N',
    '---'   : 'O',
    '.--.'  : 'P',
    '--.-'  : 'Q',
    '.-.'   : 'R',
    '...'   : 'S',
    '-'     : 'T',
    '..-'   : 'U',
    '...-'  : 'V',
    '.--'   : 'W',
    '-..-'  : 'X',
    '-.--'  : 'Y',
    '--..'  : 'Z',
    '-----' : '0',
    '.----' : '1',
    '..---' : '2',
    '...--' : '3',
    '....-' : '4',
    '.....' : '5',
    '-....' : '6',
    '--...' : '7',
    '---..' : '8',
    '----.' : '9'
}

def decode(morse):
    decoded = ''
    line = re.split('\s', morse)
    for letter in line:
        decoded += morse_alphabet.get(letter)
    return decoded

def main(filename):
    with open(filename) as input_file:
        for line in input_file:
            print(decode(line))

if __name__ == "__main__":
    try:
        main(sys.argv[1])
    except:
        sys.exit("No argument provided / file not found.")

Applying the insight gained from my previous question. It seems fairly succinct but I wonder is it pythonic?

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10
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My remarks are similar to last time. morse_alphabet should be MORSE_ALPHABET. Repeated string concatenation (decoded += …) is bad for performance. The main task could be accomplished using a one-liner substitution.

MORSE_ALPHABET = {
    ' '     : ' ',        # Note: Changed the key from '' to ' '
    '.-'    : 'A',
    '-...'  : 'B',
    '-.-.'  : 'C',
    …
}

def decode(morse):
    return re.sub(' ?( |[.-]*)', lambda m: MORSE_ALPHABET.get(m.group(1)), morse)

The main code would be simpler if you used fileinput. Despite the current Pokémon craze, you should not Catch 'Em All, but only catch the exceptions that you intend to handle.

def main():
    try:
        for line in fileinput.input():
            print(decode(line), end='')
    except IOError as e:
        sys.exit(e)

if __name__ == "__main__":
    main()
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  • \$\begingroup\$ If there are no arguments argv[1] will raise an IndexError \$\endgroup\$ – Fabián Heredia Montiel Jul 25 '16 at 1:37
  • \$\begingroup\$ @FabiánH.jr. I never use argv. If there are not arguments, it will read from sys.stdin. \$\endgroup\$ – 200_success Jul 25 '16 at 1:38
  • \$\begingroup\$ Oh nice, didn't know about that; will def read more about fileinput. \$\endgroup\$ – Fabián Heredia Montiel Jul 25 '16 at 1:40
  • \$\begingroup\$ @Legato Any '\n' that may be in the input is simply passed through unchanged by decode(). Since the newline was never stripped in the first place, it doesn't need to be added, either. \$\endgroup\$ – 200_success Jul 25 '16 at 1:59
  • \$\begingroup\$ Realized the error, thought about it in terms of my original at first. Thank you. \$\endgroup\$ – Legato Jul 25 '16 at 2:03
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No need to do a regex; Python already has str.split() and you can condense it to a ''.join(generator)

def decode(morse):
    return ''.join(morse_alphabet.get(letter)
                    for letter in morse.split(' '))
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  • 3
    \$\begingroup\$ Instead of a list comprehension, omit the [] and make it a generator expression. \$\endgroup\$ – 200_success Jul 25 '16 at 1:09
  • \$\begingroup\$ Done, thanks for the suggestion @200_success \$\endgroup\$ – Fabián Heredia Montiel Jul 25 '16 at 1:24
  • \$\begingroup\$ Is this really valid...? I get an error trying to run it. \$\endgroup\$ – Legato Jul 25 '16 at 1:54
  • \$\begingroup\$ Yes, it is valid python. Could you do me a favour and tell me the output of python --version? \$\endgroup\$ – Fabián Heredia Montiel Jul 25 '16 at 1:59
  • \$\begingroup\$ I'm running version 3.4.3 \$\endgroup\$ – Legato Jul 25 '16 at 2:09

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