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這篇文章主要為大家展示了“Python3如何實(shí)現(xiàn)打格點(diǎn)算法的GPU加速實(shí)例”，內(nèi)容簡(jiǎn)而易懂，條理清晰，希望能夠幫助大家解決疑惑，下面讓小編帶領(lǐng)大家一起研究并學(xué)習(xí)一下“Python3如何實(shí)現(xiàn)打格點(diǎn)算法的GPU加速實(shí)例”這篇文章吧。
在數(shù)學(xué)和物理學(xué)領(lǐng)域，總是充滿了各種連續(xù)的函數(shù)模型。而當(dāng)我們用現(xiàn)代計(jì)算機(jī)的技術(shù)去處理這些問題的時(shí)候，事實(shí)上是無法直接處理連續(xù)模型的，絕大多數(shù)的情況下都要轉(zhuǎn)化成一個(gè)離散的模型再進(jìn)行數(shù)值的計(jì)算。比如計(jì)算數(shù)值的積分，計(jì)算數(shù)值的二階導(dǎo)數(shù)（海森矩陣）等等。這里我們所介紹的打格點(diǎn)的算法，正是一種典型的離散化方法。這個(gè)對(duì)空間做離散化的方法，可以在很大程度上簡(jiǎn)化運(yùn)算量。比如在分子動(dòng)力學(xué)模擬中，計(jì)算近鄰表的時(shí)候，如果不采用打格點(diǎn)的方法，那么就要針對(duì)整個(gè)空間所有的原子進(jìn)行搜索，計(jì)算出來距離再判斷是否近鄰。而如果采用打格點(diǎn)的方法，我們只需要先遍歷一遍原子對(duì)齊進(jìn)行打格點(diǎn)的離散化，之后再計(jì)算近鄰表的時(shí)候，只需要計(jì)算三維空間下鄰近的27個(gè)格子中的原子是否滿足近鄰條件即可。在這篇文章中，我們主要探討如何用GPU來實(shí)現(xiàn)打格點(diǎn)的算法。
我們先來用一個(gè)例子說明一下什么叫打格點(diǎn)。對(duì)于一個(gè)給定所有原子坐標(biāo)的系統(tǒng)，也就是已知了[x,y,z]，我們需要得到的是這些原子所在的對(duì)應(yīng)的格子位置[nx,ny,nz]。我們先看一下在CPU上的實(shí)現(xiàn)方案，是一個(gè)遍歷一次的算法：
# cuda_grid.py
from numba import jit
from numba import cuda
import numpy as np
def grid_by_cpu(crd, rxyz, atoms, grids):
"""Transform coordinates [x,y,z] into grids [nx,ny,nz].
Args:
crd(list): The 3-D coordinates of atoms.
rxyz(list): The list includes xmin,ymin,zmin,grid_num.
atoms(int): The total number of atoms.
grids(list): The transformed grids matrix.
"""
for i in range(atoms):
grids[i][0] = int((crd[i][0]-rxyz[0])/rxyz[3])
grids[i][1] = int((crd[i][1]-rxyz[1])/rxyz[3])
grids[i][2] = int((crd[i][2]-rxyz[2])/rxyz[3])
return grids
if __name__=='__main__':
np.random.seed(1)
atoms = 4
grid_size = 0.1
crd = np.random.random((atoms,3)).astype(np.float32)
xmin = min(crd[:,0])
ymin = min(crd[:,1])
zmin = min(crd[:,2])
xmax = max(crd[:,0])
ymax = max(crd[:,1])
zmax = max(crd[:,2])
xgrids = int((xmax-xmin)/grid_size)+1
ygrids = int((ymax-ymin)/grid_size)+1
zgrids = int((zmax-zmin)/grid_size)+1
rxyz = np.array([xmin,ymin,zmin,grid_size], dtype=np.float32)
grids = np.ones_like(crd)*(-1)
grids = grids.astype(np.float32)
grids_cpu = grid_by_cpu(crd, rxyz, atoms, grids)
print (crd)
print (grids_cpu)
import matplotlib.pyplot as plt
plt.figure()
plt.plot(crd[:,0], crd[:,1], 'o', color='red')
for grid in range(ygrids+1):
plt.plot([xmin,xmin+grid_size*xgrids], [ymin+grid_size*grid,ymin+grid_size*grid], color='black')
for grid in range(xgrids+1):
plt.plot([xmin+grid_size*grid,xmin+grid_size*grid], [ymin,ymin+grid_size*ygrids], color='black')
plt.savefig('Atom_Grids.png')輸出結(jié)果如下，
$ python3 cuda_grid.py
[[4.17021990e-01 7.20324516e-01 1.14374816e-04]
[3.02332580e-01 1.46755889e-01 9.23385918e-02]
[1.86260208e-01 3.45560730e-01 3.96767467e-01]
[5.38816750e-01 4.19194520e-01 6.85219526e-01]]
[[2. 5. 0.]
[1. 0. 0.]
[0. 1. 3.]
[3. 2. 6.]]
上面兩個(gè)打印輸出就分別對(duì)應(yīng)于[x,y,z]和[nx,ny,nz]，比如第一個(gè)原子被放到了編號(hào)為[2,5,0]的格點(diǎn)。那么為了方便理解打格點(diǎn)的方法，我們把這個(gè)三維空間的原子系統(tǒng)和打格點(diǎn)以后的標(biāo)號(hào)取前兩個(gè)維度來可視化一下結(jié)果，作圖以后效果如下：
我們可以看到，這些紅色的點(diǎn)就是原子所處的位置，而黑色的網(wǎng)格線就是我們所標(biāo)記的格點(diǎn)。在原子數(shù)量比較多的時(shí)候，有可能出現(xiàn)在一個(gè)網(wǎng)格中存在很多個(gè)原子的情況，所以如何打格點(diǎn)，格點(diǎn)大小如何去定義，這都是不同場(chǎng)景下的經(jīng)驗(yàn)參數(shù)，需要大家一起去摸索。
在上面這個(gè)算法實(shí)現(xiàn)中，我們主要是用到了一個(gè)for循環(huán)，這時(shí)候我們可以想到numba所支持的向量化運(yùn)算，還有GPU硬件加速，這里我們先對(duì)比一下三種實(shí)現(xiàn)方案的計(jì)算結(jié)果：
# cuda_grid.py from numba import jit from numba import cuda import numpy as np def grid_by_cpu(crd, rxyz, atoms, grids): """Transform coordinates [x,y,z] into grids [nx,ny,nz]. Args: crd(list): The 3-D coordinates of atoms. rxyz(list): The list includes xmin,ymin,zmin,grid_num. atoms(int): The total number of atoms. grids(list): The transformed grids matrix. """ for i in range(atoms): grids[i][0] = int((crd[i][0]-rxyz[0])/rxyz[3]) grids[i][1] = int((crd[i][1]-rxyz[1])/rxyz[3]) grids[i][2] = int((crd[i][2]-rxyz[2])/rxyz[3]) return grids @jit def grid_by_jit(crd, rxyz, atoms, grids): """Transform coordinates [x,y,z] into grids [nx,ny,nz]. Args: crd(list): The 3-D coordinates of atoms. rxyz(list): The list includes xmin,ymin,zmin,grid_num. atoms(int): The total number of atoms. grids(list): The transformed grids matrix. """ for i in range(atoms): grids[i][0] = int((crd[i][0]-rxyz[0])/rxyz[3]) grids[i][1] = int((crd[i][1]-rxyz[1])/rxyz[3]) grids[i][2] = int((crd[i][2]-rxyz[2])/rxyz[3]) return grids @cuda.jit def grid_by_gpu(crd, rxyz, grids): """Transform coordinates [x,y,z] into grids [nx,ny,nz]. Args: crd(list): The 3-D coordinates of atoms. rxyz(list): The list includes xmin,ymin,zmin,grid_num. atoms(int): The total number of atoms. grids(list): The transformed grids matrix. """ i,j = cuda.grid(2) grids[i][j] = int((crd[i][j]-rxyz[j])/rxyz[3]) if __name__=='__main__': np.random.seed(1) atoms = 4 grid_size = 0.1 crd = np.random.random((atoms,3)).astype(np.float32) xmin = min(crd[:,0]) ymin = min(crd[:,1]) zmin = min(crd[:,2]) xmax = max(crd[:,0]) ymax = max(crd[:,1]) zmax = max(crd[:,2]) xgrids = int((xmax-xmin)/grid_size)+1 ygrids = int((ymax-ymin)/grid_size)+1 zgrids = int((zmax-zmin)/grid_size)+1 rxyz = np.array([xmin,ymin,zmin,grid_size], dtype=np.float32) crd_cuda = cuda.to_device(crd) rxyz_cuda = cuda.to_device(rxyz) grids = np.ones_like(crd)*(-1) grids = grids.astype(np.float32) grids_cpu = grid_by_cpu(crd, rxyz, atoms, grids) grids = np.ones_like(crd)*(-1) grids_jit = grid_by_jit(crd, rxyz, atoms, grids) grids = np.ones_like(crd)*(-1) grids_cuda = cuda.to_device(grids) grid_by_gpu[(atoms,3),(1,1)](crd_cuda, rxyz_cuda, grids_cuda) print (crd) print (grids_cpu) print (grids_jit) print (grids_cuda.copy_to_host())
輸出結(jié)果如下：
$ python3 cuda_grid.py
/home/dechin/anaconda3/lib/python3.8/site-packages/numba/cuda/compiler.py:865: NumbaPerformanceWarning: Grid size (12) < 2 * SM count (72) will likely result in GPU under utilization due to low occupancy.
warn(NumbaPerformanceWarning(msg))
[[4.17021990e-01 7.20324516e-01 1.14374816e-04]
[3.02332580e-01 1.46755889e-01 9.23385918e-02]
[1.86260208e-01 3.45560730e-01 3.96767467e-01]
[5.38816750e-01 4.19194520e-01 6.85219526e-01]]
[[2. 5. 0.]
[1. 0. 0.]
[0. 1. 3.]
[3. 2. 6.]]
[[2. 5. 0.]
[1. 0. 0.]
[0. 1. 3.]
[3. 2. 6.]]
[[2. 5. 0.]
[1. 0. 0.]
[0. 1. 3.]
[3. 2. 6.]]
我們先看到這里面的告警信息，因?yàn)镚PU硬件加速要在一定密度的運(yùn)算量之上才能夠有比較明顯的加速效果。比如說我們只是計(jì)算兩個(gè)數(shù)字的加和，那么是完全沒有必要使用到GPU的。但是如果我們要計(jì)算兩個(gè)非常大的數(shù)組的加和，那么這個(gè)時(shí)候GPU就能夠發(fā)揮出非常大的價(jià)值。因?yàn)檫@里我們的案例中只有4個(gè)原子，因此提示我們這時(shí)候是體現(xiàn)不出來GPU的加速效果的。我們僅僅關(guān)注下這里的運(yùn)算結(jié)果，在不同體系下得到的格點(diǎn)結(jié)果是一致的，那么接下來就可以對(duì)比一下幾種不同實(shí)現(xiàn)方式的速度差異。
# cuda_grid.py
from numba import jit
from numba import cuda
import numpy as np
def grid_by_cpu(crd, rxyz, atoms, grids):
"""Transform coordinates [x,y,z] into grids [nx,ny,nz].
Args:
crd(list): The 3-D coordinates of atoms.
rxyz(list): The list includes xmin,ymin,zmin,grid_num.
atoms(int): The total number of atoms.
grids(list): The transformed grids matrix.
"""
for i in range(atoms):
grids[i][0] = int((crd[i][0]-rxyz[0])/rxyz[3])
grids[i][1] = int((crd[i][1]-rxyz[1])/rxyz[3])
grids[i][2] = int((crd[i][2]-rxyz[2])/rxyz[3])
return grids
@jit
def grid_by_jit(crd, rxyz, atoms, grids):
"""Transform coordinates [x,y,z] into grids [nx,ny,nz].
Args:
crd(list): The 3-D coordinates of atoms.
rxyz(list): The list includes xmin,ymin,zmin,grid_num.
atoms(int): The total number of atoms.
grids(list): The transformed grids matrix.
"""
for i in range(atoms):
grids[i][0] = int((crd[i][0]-rxyz[0])/rxyz[3])
grids[i][1] = int((crd[i][1]-rxyz[1])/rxyz[3])
grids[i][2] = int((crd[i][2]-rxyz[2])/rxyz[3])
return grids
@cuda.jit
def grid_by_gpu(crd, rxyz, grids):
"""Transform coordinates [x,y,z] into grids [nx,ny,nz].
Args:
crd(list): The 3-D coordinates of atoms.
rxyz(list): The list includes xmin,ymin,zmin,grid_num.
atoms(int): The total number of atoms.
grids(list): The transformed grids matrix.
"""
i,j = cuda.grid(2)
grids[i][j] = int((crd[i][j]-rxyz[j])/rxyz[3])
if __name__=='__main__':
import time
from tqdm import trange
np.random.seed(1)
atoms = 100000
grid_size = 0.1
crd = np.random.random((atoms,3)).astype(np.float32)
xmin = min(crd[:,0])
ymin = min(crd[:,1])
zmin = min(crd[:,2])
xmax = max(crd[:,0])
ymax = max(crd[:,1])
zmax = max(crd[:,2])
xgrids = int((xmax-xmin)/grid_size)+1
ygrids = int((ymax-ymin)/grid_size)+1
zgrids = int((zmax-zmin)/grid_size)+1
rxyz = np.array([xmin,ymin,zmin,grid_size], dtype=np.float32)
crd_cuda = cuda.to_device(crd)
rxyz_cuda = cuda.to_device(rxyz)
cpu_time = 0
jit_time = 0
gpu_time = 0
for i in trange(100):
grids = np.ones_like(crd)*(-1)
grids = grids.astype(np.float32)
time0 = time.time()
grids_cpu = grid_by_cpu(crd, rxyz, atoms, grids)
time1 = time.time()
grids = np.ones_like(crd)*(-1)
time2 = time.time()
grids_jit = grid_by_jit(crd, rxyz, atoms, grids)
time3 = time.time()
grids = np.ones_like(crd)*(-1)
grids_cuda = cuda.to_device(grids)
time4 = time.time()
grid_by_gpu[(atoms,3),(1,1)](crd_cuda,
rxyz_cuda,
grids_cuda)
time5 = time.time()
if i != 0:
cpu_time += time1 - time0
jit_time += time3 - time2
gpu_time += time5 - time4
print ('The time cost of CPU calculation is: {}s'.format(cpu_time))
print ('The time cost of JIT calculation is: {}s'.format(jit_time))
print ('The time cost of GPU calculation is: {}s'.format(gpu_time))輸出結(jié)果如下：
$ python3 cuda_grid.py
100%|███████████████████████████| 100/100 [00:23<00:00, 4.18it/s]
The time cost of CPU calculation is: 23.01943016052246s
The time cost of JIT calculation is: 0.04810166358947754s
The time cost of GPU calculation is: 0.01806473731994629s
在100000個(gè)原子的體系規(guī)模下，普通的for循環(huán)實(shí)現(xiàn)效率就非常的低下，需要23s，而經(jīng)過向量化運(yùn)算的加速之后，直接飛升到了0.048s，而GPU上的加速更是達(dá)到了0.018s，相比于沒有GPU硬件加速的場(chǎng)景，實(shí)現(xiàn)了將近2倍的加速。但是這還遠(yuǎn)遠(yuǎn)不是GPU加速的上限，讓我們?cè)贉y(cè)試一個(gè)更大的案例：
# cuda_grid.py
from numba import jit
from numba import cuda
import numpy as np
def grid_by_cpu(crd, rxyz, atoms, grids):
"""Transform coordinates [x,y,z] into grids [nx,ny,nz].
Args:
crd(list): The 3-D coordinates of atoms.
rxyz(list): The list includes xmin,ymin,zmin,grid_num.
atoms(int): The total number of atoms.
grids(list): The transformed grids matrix.
"""
for i in range(atoms):
grids[i][0] = int((crd[i][0]-rxyz[0])/rxyz[3])
grids[i][1] = int((crd[i][1]-rxyz[1])/rxyz[3])
grids[i][2] = int((crd[i][2]-rxyz[2])/rxyz[3])
return grids
@jit
def grid_by_jit(crd, rxyz, atoms, grids):
"""Transform coordinates [x,y,z] into grids [nx,ny,nz].
Args:
crd(list): The 3-D coordinates of atoms.
rxyz(list): The list includes xmin,ymin,zmin,grid_num.
atoms(int): The total number of atoms.
grids(list): The transformed grids matrix.
"""
for i in range(atoms):
grids[i][0] = int((crd[i][0]-rxyz[0])/rxyz[3])
grids[i][1] = int((crd[i][1]-rxyz[1])/rxyz[3])
grids[i][2] = int((crd[i][2]-rxyz[2])/rxyz[3])
return grids
@cuda.jit
def grid_by_gpu(crd, rxyz, grids):
"""Transform coordinates [x,y,z] into grids [nx,ny,nz].
Args:
crd(list): The 3-D coordinates of atoms.
rxyz(list): The list includes xmin,ymin,zmin,grid_num.
atoms(int): The total number of atoms.
grids(list): The transformed grids matrix.
"""
i,j = cuda.grid(2)
grids[i][j] = int((crd[i][j]-rxyz[j])/rxyz[3])
if __name__=='__main__':
import time
from tqdm import trange
np.random.seed(1)
atoms = 5000000
grid_size = 0.1
crd = np.random.random((atoms,3)).astype(np.float32)
xmin = min(crd[:,0])
ymin = min(crd[:,1])
zmin = min(crd[:,2])
xmax = max(crd[:,0])
ymax = max(crd[:,1])
zmax = max(crd[:,2])
xgrids = int((xmax-xmin)/grid_size)+1
ygrids = int((ymax-ymin)/grid_size)+1
zgrids = int((zmax-zmin)/grid_size)+1
rxyz = np.array([xmin,ymin,zmin,grid_size], dtype=np.float32)
crd_cuda = cuda.to_device(crd)
rxyz_cuda = cuda.to_device(rxyz)
jit_time = 0
gpu_time = 0
for i in trange(100):
grids = np.ones_like(crd)*(-1)
time2 = time.time()
grids_jit = grid_by_jit(crd, rxyz, atoms, grids)
time3 = time.time()
grids = np.ones_like(crd)*(-1)
grids_cuda = cuda.to_device(grids)
time4 = time.time()
grid_by_gpu[(atoms,3),(1,1)](crd_cuda,
rxyz_cuda,
grids_cuda)
time5 = time.time()
if i != 0:
jit_time += time3 - time2
gpu_time += time5 - time4
print ('The time cost of JIT calculation is: {}s'.format(jit_time))
print ('The time cost of GPU calculation is: {}s'.format(gpu_time))在這個(gè)5000000個(gè)原子的案例中，因?yàn)槠胀ǖ膄or循環(huán)已經(jīng)實(shí)在是跑不動(dòng)了，因此我們就干脆不統(tǒng)計(jì)這一部分的時(shí)間，最后輸出結(jié)果如下：
$ python3 cuda_grid.py
100%|███████████████████████████| 100/100 [00:09<00:00, 10.15it/s]
The time cost of JIT calculation is: 2.3743042945861816s
The time cost of GPU calculation is: 0.022843599319458008s
在如此大規(guī)模的運(yùn)算下，GPU實(shí)現(xiàn)100倍的加速，而此時(shí)作為對(duì)比的CPU上的實(shí)現(xiàn)方法是已經(jīng)用上了向量化運(yùn)算的操作，也已經(jīng)可以認(rèn)為是一個(gè)極致的加速了。
以上是“Python3如何實(shí)現(xiàn)打格點(diǎn)算法的GPU加速實(shí)例”這篇文章的所有內(nèi)容，感謝各位的閱讀！相信大家都有了一定的了解，希望分享的內(nèi)容對(duì)大家有所幫助，如果還想學(xué)習(xí)更多知識(shí)，歡迎關(guān)注億速云行業(yè)資訊頻道！
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