Python自定义排序列表：能否无需转字典实现？附转字典方法

Great question! Let's break this down step by step to address both of your questions.

1. Can we sort without converting to a dictionary? #

Absolutely! You can work directly with the original string in your Data list to build the ordered result. Here's a straightforward approach:

• First, split the single string into individual key-value pairs.
• Iterate through your keyorder list, and for each key, find the matching pair from the split string.
• Build your final ordered list by appending these matched pairs in the desired sequence.

Here's the code implementation:

Data = ['ID:123 GE:m FN:Amir LN:Maleki EM:a@b.c MN:0400101010 ST:VIC']
keyorder = ['ID', 'GE','FN','LN','MN','EM', 'ST']

# Split the string into separate key-value pairs
kv_pairs = Data[0].split()
final_ordered = []

for key in keyorder:
    # Find the pair that matches the current key
    for pair in kv_pairs:
        if pair.startswith(f"{key}:"):
            k, v = pair.split(':', 1)  # Split only once to handle values with colons (if any)
            # Convert ID to integer like your original dictionary example
            if key == 'ID':
                v = int(v)
            final_ordered.append((k, v))
            break

print(final_ordered)

This will output exactly the same result as your dictionary-based approach:

[('ID', 123), ('GE', 'm'), ('FN', 'Amir'), ('LN', 'Maleki'), ('MN', '0400101010'), ('EM', 'a@b.c'), ('ST', 'VIC')]

2. How to convert the original list to your target dictionary? #

If you do want to convert the Data list to the dictionary you showed, it's simple with a loop or dictionary comprehension. Here are two ways to do it:

Option 1: Using a basic loop #

Data = ['ID:123 GE:m FN:Amir LN:Maleki EM:a@b.c MN:0400101010 ST:VIC']
d = {}

# Split the string into key-value pairs, then process each pair
for pair in Data[0].split():
    key, value = pair.split(':', 1)
    # Convert ID to integer as in your example
    if key == 'ID':
        value = int(value)
    d[key] = value

print(d)

Option 2: Using a concise dictionary comprehension #

Data = ['ID:123 GE:m FN:Amir LN:Maleki EM:a@b.c MN:0400101010 ST:VIC']

d = {
    key: int(value) if key == 'ID' else value
    for pair in Data[0].split()
    for key, value in [pair.split(':', 1)]
}

print(d)

Both methods will output your target dictionary:

{'ID': 123, 'GE': 'm', 'FN': 'Amir', 'LN': 'Maleki', 'EM': 'a@b.c', 'MN': '0400101010', 'ST': 'VIC'}

内容的提问来源于stack exchange，提问作者KenjiChan1212