[英]Converting values to np.nan in a numpy array if a condition for value is met
[英]Sum values from numpy array if condition on value in another array is met
我在對 function 進行矢量化以使其有效地應用於 numpy 陣列時遇到問題。
我的節目條目:
然后是需要時間的計算,因為我無法找到一種方法來正確處理 arrays 的 numpy 操作。 我想做的是:
到目前為止我的想法:
所以,我在這里。 可能與排序和累積總和有關,但我不知道該怎么做,所以任何幫助將不勝感激。 請在下面找到代碼以使事情更清楚
提前致謝。
import numpy as np
import time
import matplotlib.pyplot as plt
# Creation of particles' positions
Nb_part = 150_000
pos_part = 10*np.random.rand(Nb_part,3)
pos_part[:,0] = pos_part[:,1] = 0
#usefull property creation
beta = 1/1.5
prop_part = (1/beta)*np.exp(-pos_part[:,2]/beta)
z_distances = np.arange(0,10,0.1)
#my version 0
t0=time.time()
result = np.empty(len(z_distances))
for index_dist, val_dist in enumerate(z_distances):
positions = np.where(pos_part[:,2]<val_dist)[0]
result[index_dist] = sum(prop_part[i] for i in positions)
print("v0 :",time.time()-t0)
#A graph to help understand
plt.figure()
plt.plot(z_distances,result, c="red")
plt.ylabel("Sum of particles' usefull property for particles with z-pos<d")
plt.xlabel("d")
#version 1 ??
t1=time.time()
combi = np.column_stack((pos_part[:,2],prop_part))
result2 = np.empty(len(z_distances))
for index_dist, val_dist in enumerate(z_distances):
mask = (combi[:,0]<val_dist)
result2[index_dist]=sum(combi[:,1][mask])
print("v1 :",time.time()-t1)
plt.plot(z_distances,result2, c="blue")
#version 2
t2=time.time()
def themask(a):
mask = (combi[:,0]<a)
return sum(combi[:,1][mask])
thefunc = np.vectorize(themask)
result3 = thefunc(z_distances)
print("v2 :",time.time()-t2)
plt.plot(z_distances,result3, c="green")
### This does not work so far
# version 3
# =============================
# t3=time.time()
# def thesum(a):
# mask = combi[combi[:,0]<a]
# return sum(mask[:,1])
# result4 = thesum(z_distances)
# print("v3 :",time.time()-t3)
# =============================
通過完全在 numpy 中編寫您的第一個版本,您可以獲得更多的性能。 將 python 的sum
替換為np.sum
。 而不是for i in positions
列表理解中,只需傳遞您正在創建的positions
掩碼。 事實上, np.where
不是必需的,我最好的版本如下所示:
#my version 0
t0=time.time()
result = np.empty(len(z_distances))
for index_dist, val_dist in enumerate(z_distances):
positions = pos_part[:, 2] < val_dist
result[index_dist] = np.sum(prop_part[positions])
print("v0 :",time.time()-t0)
# out: v0 : 0.06322097778320312
如果 z_distances 很長,使用numba
可以更快一些。 第一次運行calc
通常會產生一些開銷,我們可以通過運行 function 來消除一些開銷。 下面的代碼在我的筆記本電腦上實現了大約兩倍於純 numpy 的加速。
import numba as nb
@nb.njit(parallel=True)
def calc(result, z_distances):
n = z_distances.shape[0]
for ii in nb.prange(n):
pos = pos_part[:, 2] < z_distances[ii]
result[ii] = np.sum(prop_part[pos])
return result
result4 = np.zeros_like(result)
# _t = time.time()
# calc(result4, z_distances[:10])
# print(time.time()-_t)
t3 = time.time()
result4 = calc(result4, z_distances)
print("v3 :", time.time()-t3)
plt.plot(z_distances, result4)
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