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[英]What is the most efficient way to convert numpy arrays to Shapely Points?
[英]What is the most efficient way to deal with a loop on NumPy arrays?
問題很簡單:這是我目前的算法。 由於陣列上的循環,這非常慢。 有沒有辦法改變它以避免循環並利用 NumPy 數組類型?
import numpy as np
def loopingFunction(listOfVector1, listOfVector2):
resultArray = []
for vector1 in listOfVector1:
result = 0
for vector2 in listOfVector2:
result += np.dot(vector1, vector2) * vector2[2]
resultArray.append(result)
return np.array(resultArray)
listOfVector1x = np.linspace(0,0.33,1000)
listOfVector1y = np.linspace(0.33,0.66,1000)
listOfVector1z = np.linspace(0.66,1,1000)
listOfVector1 = np.column_stack((listOfVector1x, listOfVector1y, listOfVector1z))
listOfVector2x = np.linspace(0.33,0.66,1000)
listOfVector2y = np.linspace(0.66,1,1000)
listOfVector2z = np.linspace(0, 0.33, 1000)
listOfVector2 = np.column_stack((listOfVector2x, listOfVector2y, listOfVector2z))
result = loopingFunction(listOfVector1, listOfVector2)
我應該處理非常大的數組,每個數組都有超過 1000 個向量。 所以如果你有什么建議,我會采納的。
強制性np.einsum
基准
r2 = np.einsum('ij, kj, k->i', listOfVector1, listOfVector2, listOfVector2[:,2], optimize=['einsum_path', (1, 2), (0, 1)])
#%timeit result: 10000 loops, best of 5: 116 µs per loop
np.testing.assert_allclose(result, r2)
只是為了好玩,我編寫了一個優化的 Numba 實現,它優於所有其他實現。 它基於@MichaelSzczesny 答案的einsum
優化。
import numpy as np
import numba as nb
# This decorator ask Numba to eagerly compile the code using
# the provided signature string (containing the parameter types).
@nb.njit('(float64[:,::1], float64[:,::1])')
def loopingFunction_numba(listOfVector1, listOfVector2):
n, m = listOfVector1.shape
assert m == 3
result = np.empty(n)
s1 = s2 = s3 = 0.0
for i in range(n):
factor = listOfVector2[i, 2]
s1 += listOfVector2[i, 0] * factor
s2 += listOfVector2[i, 1] * factor
s3 += listOfVector2[i, 2] * factor
for i in range(n):
result[i] = listOfVector1[i, 0] * s1 + listOfVector1[i, 1] * s2 + listOfVector1[i, 2] * s3
return result
result = loopingFunction_numba(listOfVector1, listOfVector2)
以下是我的 i5-9600KF 處理器的計時:
Initial: 1052.0 ms
ymmx: 5.121 ms
MichaelSzczesny: 75.40 us
MechanicPig: 3.36 us
Numba: 2.74 us
Optimal lower bound: 0.66 us
此解決方案比原始解決方案快約 384_000 倍。 請注意,它甚至不使用處理器的 SIMD 指令,這會導致我的機器加速約 4 倍。 這只有通過轉置輸入比當前輸入更易於 SIMD 來實現。 轉置還可以加快其他答案,例如 MechanicPig 的答案,因為 BLAS 通常可以從中受益。 生成的代碼將達到符號 1_000_000 加速因子!
你至少可以去掉兩個forloop來節省很多時間,直接使用矩陣計算
import time
import numpy as np
def loopingFunction(listOfVector1, listOfVector2):
resultArray = []
for vector1 in listOfVector1:
result = 0
for vector2 in listOfVector2:
result += np.dot(vector1, vector2) * vector2[2]
resultArray.append(result)
return np.array(resultArray)
def loopingFunction2(listOfVector1, listOfVector2):
resultArray = np.sum(np.dot(listOfVector1, listOfVector2.T) * listOfVector2[:,2], axis=1)
return resultArray
listOfVector1x = np.linspace(0,0.33,1000)
listOfVector1y = np.linspace(0.33,0.66,1000)
listOfVector1z = np.linspace(0.66,1,1000)
listOfVector1 = np.column_stack((listOfVector1x, listOfVector1y, listOfVector1z))
listOfVector2x = np.linspace(0.33,0.66,1000)
listOfVector2y = np.linspace(0.66,1,1000)
listOfVector2z = np.linspace(0, 0.33, 1000)
listOfVector2 = np.column_stack((listOfVector2x, listOfVector2y, listOfVector2z))
import time
t0 = time.time()
result = loopingFunction(listOfVector1, listOfVector2)
print('time old version',time.time() - t0)
t0 = time.time()
result2 = loopingFunction2(listOfVector1, listOfVector2)
print('time matrix computation version',time.time() - t0)
print('Are results are the same',np.allclose(result,result2))
這使
time old version 1.174513578414917
time matrix computation version 0.011968612670898438
Are results are the same True
基本上,循環越少越好。
避免嵌套循環,調整計算順序,比優化后的np.einsum
快 20 倍,比原程序快近 400_000 倍:
>>> out = listOfVector1.dot(listOfVector2[:, 2].dot(listOfVector2))
>>> np.allclose(out, loopingFunction(listOfVector1, listOfVector2))
True
測試:
>>> timeit(lambda: loopingFunction(listOfVector1, listOfVector2), number=1)
1.4389081999834161
>>> timeit(lambda: listOfVector1.dot(listOfVector2[:, 2].dot(listOfVector2)), number=400_000)
1.3162514999858104
>>> timeit(lambda: np.einsum('ij, kj, k->i', listOfVector1, listOfVector2, listOfVector2[:, 2], optimize=['einsum_path', (1, 2), (0, 1)]), number=18_000)
1.3501517999975476
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