如何在Python3中从awswrangler中捕获exceptions.NoFilesFound错误

Question

下面是读取存储在S3存储桶路径中的拼图文件的代码。当它在path中找到地块文件时，它可以工作，但是当它找不到任何文件时，会给出exceptions.NoFilesFound。

import boto3
import awswrangler as wr
    
boto3.setup_default_session(profile_name="myAwsProfile", region_name="us-east-1")
    
path_prefix  = 's3://example_bucket/data/parquet_files'
path_suffix = '/y=2021/m=4/d=13/h=17/'
table_path = path_prefix + path_suffix

df = wr.s3.read_parquet(path=table_path)
print(len(df))

输出：

22646

如果S3路径中没有文件，例如，如果我将path_suffix从'/y=2021/m=4/d=13/h=17/'更改为'/y=2021/m=4/d=13/h=170/'，则会得到以下错误：

---------------------------------------------------------------------------
NoFilesFound                              Traceback (most recent call last)
<ipython-input-9-17df460412d8> in <module>
     11 
     12 file_prefix = table_path + date_prefix
---> 13 df = wr.s3.read_parquet(path=file_prefix)

/usr/local/lib/python3.9/site-packages/awswrangler/s3/_read_parquet.py in read_parquet(path, path_suffix, path_ignore_suffix, ignore_empty, ignore_index, partition_filter, columns, validate_schema, chunked, dataset, categories, safe, map_types, use_threads, last_modified_begin, last_modified_end, boto3_session, s3_additional_kwargs)
    602         paths = _apply_partition_filter(path_root=path_root, paths=paths, filter_func=partition_filter)
    603     if len(paths) < 1:
--> 604         raise exceptions.NoFilesFound(f"No files Found on: {path}.")
    605     _logger.debug("paths:\n%s", paths)
    606     args: Dict[str, Any] = {

NoFilesFound: No files Found on: s3://example_bucket/data/parquet_files/y=2021/m=4/d=13/h=170/.

看起来它来自awswrangler Python库，所以botocore.exceptions无法捕获它。我可以简单地使用python的try:和except:来绕过它，但是我需要捕获它才能正确地处理它。我该怎么做呢？

Answer 1

如果您只想捕获异常，

from awswrangler import exceptions
try:
  ...
except exceptions.NoFilesFound:
  ...