我如何能看到什么是在S3桶与boto3?(例如,写一个“ls”)?
做以下事情:
import boto3
s3 = boto3.resource('s3')
my_bucket = s3.Bucket('some/path/')
返回:
s3.Bucket(name='some/path/')
我如何看到它的内容?
我如何能看到什么是在S3桶与boto3?(例如,写一个“ls”)?
做以下事情:
import boto3
s3 = boto3.resource('s3')
my_bucket = s3.Bucket('some/path/')
返回:
s3.Bucket(name='some/path/')
我如何看到它的内容?
当前回答
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
其他回答
一种更节俭的方法,而不是通过一个for循环来迭代,你也可以只打印原始对象,其中包含S3桶中的所有文件:
session = Session(aws_access_key_id=aws_access_key_id,aws_secret_access_key=aws_secret_access_key)
s3 = session.resource('s3')
bucket = s3.Bucket('bucket_name')
files_in_s3 = bucket.objects.all()
#you can print this iterable with print(list(files_in_s3))
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。
查看内容的一种方法是:
for my_bucket_object in my_bucket.objects.all():
print(my_bucket_object)
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
我以前是这样做的:
import boto3
s3 = boto3.resource('s3')
bucket=s3.Bucket("bucket_name")
contents = [_.key for _ in bucket.objects.all() if "subfolders/ifany/" in _.key]