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acc7820574
... since the whole response is huge. We even need to break up the assertions, since kibana otherwise truncates them.
49 lines
1.6 KiB
Python
49 lines
1.6 KiB
Python
# -*- coding: utf-8 -*-
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# Copyright 2014-2016 OpenMarket Ltd
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# Copyright 2020 The Matrix.org Foundation C.I.C.
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#
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# Licensed under the Apache License, Version 2.0 (the "License");
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# you may not use this file except in compliance with the License.
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# You may obtain a copy of the License at
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#
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# http://www.apache.org/licenses/LICENSE-2.0
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#
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# Unless required by applicable law or agreed to in writing, software
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# distributed under the License is distributed on an "AS IS" BASIS,
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# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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# See the License for the specific language governing permissions and
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# limitations under the License.
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from itertools import islice
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from typing import Iterable, Iterator, Sequence, Tuple, TypeVar
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T = TypeVar("T")
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def batch_iter(iterable: Iterable[T], size: int) -> Iterator[Tuple[T]]:
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"""batch an iterable up into tuples with a maximum size
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Args:
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iterable (iterable): the iterable to slice
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size (int): the maximum batch size
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Returns:
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an iterator over the chunks
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"""
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# make sure we can deal with iterables like lists too
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sourceiter = iter(iterable)
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# call islice until it returns an empty tuple
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return iter(lambda: tuple(islice(sourceiter, size)), ())
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ISeq = TypeVar("ISeq", bound=Sequence, covariant=True)
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def chunk_seq(iseq: ISeq, maxlen: int) -> Iterable[ISeq]:
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"""Split the given sequence into chunks of the given size
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The last chunk may be shorter than the given size.
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If the input is empty, no chunks are returned.
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"""
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return (iseq[i : i + maxlen] for i in range(0, len(iseq), maxlen))
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