如何取消格式化文件名？（解析或格式相反）

Question

我想返回已经被这个函数转换的文件的文件名。

from more_itertools import chunked

def file_conversion(input_file, output_file_pattern, chunksize):
    with open(input_file) as fin:
        reader = csv.DictReader(fin, delimiter=',')
        for i, chunk in enumerate(chunked(reader, chunksize)):
            with open(output_file_pattern.format(i), 'w', newline='') as fout:
                writer = csv.DictWriter(fout, reader.fieldnames, delimiter='|')
                writer.writeheader()
                writer.writerows(chunk)
                print("Successfully converted into", output_file)
script_dir = os.path.dirname(os.path.abspath(__file__))
dest_dir = os.path.join(script_dir, 'temp')
try:
     os.makedirs(dest_dir)
except OSError:
     pass
path = os.path.join(dest_dir,'out{:03}.csv' )

#this creates file as out000.csv and more out001.csv or out002.csv depending upon the number of lines

file_conversion('in.csv', path, 7000) 
base_name = os.path.basename(path)
print(base_name)  #returns out{03}.csv

这会打印出{03}.csv。但我想要像out000.csv这样转换的文件名的结果。有什么办法可以退货吗？如果文件较多，如何返回文件名？

Answer 1

好的，你需要在你的函数中创建一个集合，并用所有使用的文件名填充它：

def file_conversion(input_file, output_file_pattern, chunksize):
    output_filenames = []
    with open(input_file) as fin:
        reader = csv.DictReader(fin, delimiter=',')
        for i, chunk in enumerate(chunked(reader, chunksize)):
            output_filename = output_file_pattern.format(i)
            with open(output_filename, 'w', newline='') as fout:
                output_filenames.append(output_filename)
                writer = csv.DictWriter(fout, reader.fieldnames, delimiter='|')
                writer.writeheader()
                writer.writerows(chunk)
                print("Successfully converted into", output_file)
    return output_filenames

你可以这样称呼它：

paths = file_conversion('in.csv', path, 7000) 
base_names = [os.path.basename(path) for path in paths]
print(base_names)