本文整理汇总了Python中ufora.native.FORA.createTestingVDMM方法的典型用法代码示例。如果您正苦于以下问题：Python FORA.createTestingVDMM方法的具体用法？Python FORA.createTestingVDMM怎么用？Python FORA.createTestingVDMM使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类ufora.native.FORA的用法示例。
在下文中一共展示了FORA.createTestingVDMM方法的3个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: measureMmapPerformance
# 需要导入模块: from ufora.native import FORA [as 别名]
# 或者: from ufora.native.FORA import createTestingVDMM [as 别名]
def measureMmapPerformance(self, measurementTime, blockSize, threadCount, initializeData):
t0 = time.time()
results = []
vdmm = ForaNative.createTestingVDMM()
def runTest():
results.append(
ForaNative.executeMMapAndFreeForTimePeriod(
vdmm,
measurementTime,
blockSize,
initializeData
)
)
threads = [threading.Thread(target=runTest) for ix in range(threadCount)]
for t in threads:
t.start()
for t in threads:
t.join()
return (time.time() - t0) / (sum(results) / 1024.0 ** 3)示例2: test_verifySmallAllocNotRemapping
# 需要导入模块: from ufora.native import FORA [as 别名]
# 或者: from ufora.native.FORA import createTestingVDMM [as 别名]
def test_verifySmallAllocNotRemapping(self):
vdmm = ForaNative.createTestingVDMM()
ForaNative.executeBigVectorHandleAllocationsForTimePeriod(
vdmm,
.25,
1024 * 1024
)
bytes1 = vdmm.totalBytesMmappedCumulatively()
ForaNative.executeBigVectorHandleAllocationsForTimePeriod(
vdmm,
1.0,
1024 * 1024
)
bytes2 = vdmm.totalBytesMmappedCumulatively()
self.assertTrue(bytes2 <= 1024 * 1024 * 4)示例3: measureMemoryAllocationPerformance
# 需要导入模块: from ufora.native import FORA [as 别名]
# 或者: from ufora.native.FORA import createTestingVDMM [as 别名]
def measureMemoryAllocationPerformance(self, measurementTime, blockSize, threadCount):
t0 = time.time()
results = []
vdmm = ForaNative.createTestingVDMM()
def runTest():
results.append(
ForaNative.executeBigVectorHandleAllocationsForTimePeriod(
vdmm,
measurementTime,
blockSize
)
)
threads = [threading.Thread(target=runTest) for ix in range(threadCount)]
for t in threads:
t.start()
for t in threads:
t.join()
return (time.time() - t0) / (sum(results) / 1024.0 ** 3)