Django中UnboundLocalError求助：摩尔斯电码解密函数报错

Hey there! Let's tackle that UnboundLocalError you're hitting in your Morse code decryption function. The error happens because the space variable is only defined inside the if letter != ' ' block—so if the first character in your input is a space, the code tries to run space += 1 before space has ever been assigned a value. Using global didn't work here because this is a local variable initialization issue, not a scope problem with global variables.

The Solution: Initialize the Local Variable Upfront #

The fix is straightforward: initialize space as a local variable before your for loop runs, so it exists regardless of which branch of the conditional executes first. Here's how to update your decrypt function:

def decrypt(message):
    # extra space added at the end to access the
    # last morse code
    message += ' '
    decipherMsg = ''
    citext = ''
    space = 0  # Initialize space here, outside the loop
    for letter in message:
        # checks for space
        if letter != ' ':
            # counter to keep track of space: reset to 0 when we hit a non-space
            space = 0
            # storing morse code of a single character
            citext += letter
        # in case of space
        else:
            # increment space counter
            space += 1
            # if space == 2 that indicates a new word
            if space == 2:
                # adding space to separate words
                decipherMsg += ' '
            else:
                # accessing the keys using their values (reverse of encryption)
                decipherMsg += list(MORSE_CODE_DICT.keys())[list(MORSE_CODE_DICT.values()).index(citext)]
                citext = ''
    return decipherMsg

What Changed? #

• We moved space = 0 to the top of the decrypt function, right after setting up other local variables. This ensures the variable exists in the function's local scope from the start.
• When we hit a non-space character, we still reset space to 0 (preserving your original logic for tracking consecutive spaces).
• Now, even if the first character is a space, space has already been assigned a value, so space += 1 won't throw an error.

Optional Optimization #

As a side note, looking up keys by values with list(MORSE_CODE_DICT.keys())[list(MORSE_CODE_DICT.values()).index(citext)] is inefficient for longer messages. You can pre-create a reverse dictionary once (outside the decrypt function) to speed this up:

# Inside the decipher function, before defining decrypt
REVERSE_MORSE_DICT = {v: k for k, v in MORSE_CODE_DICT.items()}

# Then replace the lookup line in decrypt with:
decipherMsg += REVERSE_MORSE_DICT[citext]

Dictionary lookups are O(1), so this will make your decryption much faster, especially with large inputs.

内容的提问来源于stack exchange，提问作者Din