[英]Python 3 iterable lazy chunk fetch
是否有標准方法能夠延遲獲取下一個數據塊並通過元素產生它
目前,我正在獲取所有塊並將其與itertools鏈接
def list_blobs(container_name:str, prefix:str):
chunks = []
next_marker=None
while True:
blobs = blob_service.list_blobs(container_name, prefix=prefix, num_results=100, marker=next_marker)
next_marker = blobs.next_marker
chunks.append(blobs)
if not next_marker:
break
return itertools.chain.from_iterable(chunks)
list_blobs提取程序的“懶惰”版本是什么?
您可以只使用yield from
:
def list_blobs(container_name:str, prefix:str):
next_marker = True
while next_marker:
blobs = blob_service.list_blobs(container_name, prefix=prefix, num_results=100, marker=next_marker)
next_marker = blobs.next_marker
yield from blobs
用chunks.append(blobs)
yield from blobs
替換chunks.append(blobs)
,完全擺脫return
和chunks
list
:
def generate_blobs(container_name:str, prefix:str):
next_marker = None
while True:
blobs = blob_service.list_blobs(container_name, prefix=prefix, num_results=100, marker=next_marker)
next_marker = blobs.next_marker
yield from blobs
if not next_marker:
break
它將函數轉換為生成器函數,一次生成一個項。
@ ShadowRanger,@Kasrâmvd非常感謝
@timgeb,可通過Azure BlobStorage進行延遲懶散的完整代碼
from azure.storage.blob import BlockBlobService
from azure.storage.blob import Blob
from typing import Iterable, Tuple
def blob_iterator(account:str, account_key:str, bucket:str, prefix:str)-> Iterable[Tuple[str, str]]:
blob_service = BlockBlobService(account_name=account, account_key=account_key)
def list_blobs(bucket:str, prefix:str)->Blob:
next_marker = None
while True:
blobs = blob_service.list_blobs(bucket, prefix=prefix, num_results=100, marker=next_marker)
yield from blobs
next_marker = blobs.next_marker
if not next_marker:
break
def get_text(bucket:str, name:str)->str:
return blob_service.get_blob_to_text(bucket, name).content
return ( (blob.name, get_text(bucket, blob.name)) for blob in list_blobs(bucket, prefix) )
it = blob_iterator('account', 'account_key', 'container_name', prefix='AA')
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