我的s3键实用函数本质上是@Hephaestus的答案的优化版本:

import boto3


s3_paginator = boto3.client('s3').get_paginator('list_objects_v2')


def keys(bucket_name, prefix='/', delimiter='/', start_after=''):
    prefix = prefix.lstrip(delimiter)
    start_after = (start_after or prefix) if prefix.endswith(delimiter) else start_after
    for page in s3_paginator.paginate(Bucket=bucket_name, Prefix=prefix, StartAfter=start_after):
        for content in page.get('Contents', ()):
            yield content['Key']

在我的测试(boto3 1.9.84)中，它比等效的(但更简单)代码要快得多:

import boto3


def keys(bucket_name, prefix='/', delimiter='/'):
    prefix = prefix.lstrip(delimiter)
    bucket = boto3.resource('s3').Bucket(bucket_name)
    return (_.key for _ in bucket.objects.filter(Prefix=prefix))

由于S3保证UTF-8二进制排序结果，因此在第一个函数中添加了start_after优化。

我假设您已经单独配置了身份验证。

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

my_bucket = s3.Bucket('bucket_name')

for file in my_bucket.objects.all():
    print(file.key)

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

#To print all filenames in a bucket
import boto3

s3 = boto3.client('s3')

def get_s3_keys(bucket):

    """Get a list of keys in an S3 bucket."""
    resp = s3.list_objects_v2(Bucket=bucket)
    for obj in resp['Contents']:
      files = obj['Key']
    return files

  
filename = get_s3_keys('your_bucket_name')

print(filename)

#To print all filenames in a certain directory in a bucket
import boto3

s3 = boto3.client('s3')

def get_s3_keys(bucket, prefix):

    """Get a list of keys in an S3 bucket."""
    resp = s3.list_objects_v2(Bucket=bucket, Prefix=prefix)
    for obj in resp['Contents']:
      files = obj['Key']
      print(files)
    return files

  
filename = get_s3_keys('your_bucket_name', 'folder_name/sub_folder_name/')

print(filename)

更新: 最简单的方法是使用awswrangler

import awswrangler as wr
wr.s3.list_objects('s3://bucket_name')

ObjectSummary:

有两个标识符附加到ObjectSummary:

bucket_name 关键

boto3 S3: ObjectSummary

有关AWS S3文档中的对象键的更多信息:

Object Keys: When you create an object, you specify the key name, which uniquely identifies the object in the bucket. For example, in the Amazon S3 console (see AWS Management Console), when you highlight a bucket, a list of objects in your bucket appears. These names are the object keys. The name for a key is a sequence of Unicode characters whose UTF-8 encoding is at most 1024 bytes long. The Amazon S3 data model is a flat structure: you create a bucket, and the bucket stores objects. There is no hierarchy of subbuckets or subfolders; however, you can infer logical hierarchy using key name prefixes and delimiters as the Amazon S3 console does. The Amazon S3 console supports a concept of folders. Suppose that your bucket (admin-created) has four objects with the following object keys: Development/Projects1.xls Finance/statement1.pdf Private/taxdocument.pdf s3-dg.pdf Reference: AWS S3: Object Keys

下面是一些示例代码，演示如何获取桶名和对象键。

例子:

import boto3
from pprint import pprint

def main():

    def enumerate_s3():
        s3 = boto3.resource('s3')
        for bucket in s3.buckets.all():
             print("Name: {}".format(bucket.name))
             print("Creation Date: {}".format(bucket.creation_date))
             for object in bucket.objects.all():
                 print("Object: {}".format(object))
                 print("Object bucket_name: {}".format(object.bucket_name))
                 print("Object key: {}".format(object.key))

    enumerate_s3()


if __name__ == '__main__':
    main()

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