我只是这样做的，包括身份验证方法:

s3_client = boto3.client(
                's3',
                aws_access_key_id='access_key',
                aws_secret_access_key='access_key_secret',
                config=boto3.session.Config(signature_version='s3v4'),
                region_name='region'
            )

response = s3_client.list_objects(Bucket='bucket_name', Prefix=key)
if ('Contents' in response):
    # Object / key exists!
    return True
else:
    # Object / key DOES NOT exist!
    return False

import boto3
s3 = boto3.resource('s3')

## Bucket to use
my_bucket = s3.Bucket('city-bucket')

## List objects within a given prefix
for obj in my_bucket.objects.filter(Delimiter='/', Prefix='city/'):
  print obj.key

输出:

city/pune.csv
city/goa.csv

这类似于'ls'，但它没有考虑到前缀文件夹约定，并将列出bucket中的对象。由读取器来过滤掉作为Key名称一部分的前缀。

在Python 2中:

from boto.s3.connection import S3Connection

conn = S3Connection() # assumes boto.cfg setup
bucket = conn.get_bucket('bucket_name')
for obj in bucket.get_all_keys():
    print(obj.key)

在Python 3中:

from boto3 import client

conn = client('s3')  # again assumes boto.cfg setup, assume AWS S3
for key in conn.list_objects(Bucket='bucket_name')['Contents']:
    print(key['Key'])

所以你在boto3中要求等同于aws s3 ls。这将列出所有顶级文件夹和文件。这是我能得到的最接近的结果;它只列出所有顶级文件夹。这么简单的操作居然这么难。

import boto3

def s3_ls():
  s3 = boto3.resource('s3')
  bucket = s3.Bucket('example-bucket')
  result = bucket.meta.client.list_objects(Bucket=bucket.name,
                                           Delimiter='/')
  for o in result.get('CommonPrefixes'):
    print(o.get('Prefix'))

下面是一个简单的函数，它返回所有文件的文件名或具有特定类型的文件，如'json'， 'jpg'。

def get_file_list_s3(bucket, prefix="", file_extension=None):
            """Return the list of all file paths (prefix + file name) with certain type or all
            Parameters
            ----------
            bucket: str
                The name of the bucket. For example, if your bucket is "s3://my_bucket" then it should be "my_bucket"
            prefix: str
                The full path to the the 'folder' of the files (objects). For example, if your files are in 
                s3://my_bucket/recipes/deserts then it should be "recipes/deserts". Default : ""
            file_extension: str
                The type of the files. If you want all, just leave it None. If you only want "json" files then it
                should be "json". Default: None       
            Return
            ------
            file_names: list
                The list of file names including the prefix
            """
            import boto3
            s3 = boto3.resource('s3')
            my_bucket = s3.Bucket(bucket)
            file_objs =  my_bucket.objects.filter(Prefix=prefix).all()
            file_names = [file_obj.key for file_obj in file_objs if file_extension is not None and file_obj.key.split(".")[-1] == file_extension]
            return file_names

使用cloudpathlib

cloudpathlib提供了一个方便的包装器，这样您就可以使用简单的pathlib API与AWS S3(以及Azure blob存储、GCS等)进行交互。你可以用pip install "cloudpathlib[s3]"来安装。

像pathlib一样，你可以使用glob或iterdir来列出目录的内容。

下面是一个带有公共AWS S3桶的示例，您可以复制并过去运行该桶。

from cloudpathlib import CloudPath

s3_path = CloudPath("s3://ladi/Images/FEMA_CAP/2020/70349")

# list items with glob
list(
    s3_path.glob("*")
)[:3]
#> [ S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0001_5a63d42e-27c6-448a-84f1-bfc632125b8e.jpg'),
#>   S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0002_a89f1b79-786f-4dac-9dcc-609fb1a977b1.jpg'),
#>   S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0003_02c30af6-911e-4e01-8c24-7644da2b8672.jpg')]

# list items with iterdir
list(
    s3_path.iterdir()
)[:3]
#> [ S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0001_5a63d42e-27c6-448a-84f1-bfc632125b8e.jpg'),
#>   S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0002_a89f1b79-786f-4dac-9dcc-609fb1a977b1.jpg'),
#>   S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0003_02c30af6-911e-4e01-8c24-7644da2b8672.jpg')]

创建于2021-05-21 20:38:47 PDT由reprexlite v0.4.2创建