如何将嵌套列表转换为字典列表？（支持任意数量子列表）

To convert your nested list into the desired list of dictionaries (and handle any number of sublists), you can use Python's built-in zip() function along with list comprehensions. Here's a concise, flexible approach:

Code Example #

temp = [
    ['header1', '4', '8', '16', '32', '64', '128', '256', '512', '243,6'],
    ['media_range', '1,200', '2,400', '4,800', '4,800', '6,200', '38,400', '76,800', '153,600', '160,000'],
    ['speed', '300', '600', '1,200', '2,000', '2,000', '2,000', '2,000', '2,000', '2,000']
]

# Extract the header keys from each sublist
headers = [sublist[0] for sublist in temp]

# Zip together all the data values (skipping the header in each sublist)
data_rows = zip(*[sublist[1:] for sublist in temp])

# Generate the list of dictionaries
output = [dict(zip(headers, row)) for row in data_rows]

# Print the result to verify
for item in output:
    print(item)

How It Works #

1. Extract Headers: We first grab the first element of each sublist to use as the keys for our dictionaries. This works no matter how many sublists you have.
2. Zip Data Rows: Using zip(*[sublist[1:]...]), we transpose the data—turning columns into rows. Each "row" here is a tuple of corresponding values from each sublist (e.g., ('4', '1,200', '300') for the first entry).
3. Build Dictionaries: For each row of values, we pair them with the headers using dict(zip(headers, row)), creating a dictionary for each set of corresponding entries.

Example Output #

The first few entries of output will look exactly like you expected:

{'header1': '4', 'media_range': '1,200', 'speed': '300'}
{'header1': '8', 'media_range': '2,400', 'speed': '600'}
{'header1': '16', 'media_range': '4,800', 'speed': '1,200'}
...

Notes #

• This approach handles any number of sublists automatically—just add more sublists to temp and the code will adapt (as long as all sublists have the same length after the header element).
• If sublists have mismatched lengths, zip() will stop at the shortest one. For strict handling, you could use itertools.zip_longest, but that's usually unnecessary for structured data like this.

内容的提问来源于stack exchange，提问作者goku