Python 2.7 在列表列表上运行计算的最快方法
我有一个这样的列表:Python 2.7 在列表列表上运行计算的最快方法,python-2.7,numpy,numba,Python 2.7,Numpy,Numba,我有一个这样的列表: import numpy as np import random import time import itertools N = 1000 x =np.random.random((N,N)) y = np.zeros((N,N)) z = np.random.random((N,N)) list_of_lists = [[x, y], [y,z], [z,x]] 对于每个子列表,我想计算非零的数量,平均值和标准偏差 我是这样做的: distribution = [
import numpy as np
import random
import time
import itertools
N = 1000
x =np.random.random((N,N))
y = np.zeros((N,N))
z = np.random.random((N,N))
list_of_lists = [[x, y], [y,z], [z,x]]
对于每个子列表,我想计算非零的数量,平均值和标准偏差
我是这样做的:
distribution = []
alb_mean = []
alb_std = []
start = time.time()
for i in range(len(list_of_lists)):
one_mean = []
non_zero_l = []
one_list = list_of_lists[i]
for n in one_list:
#count non_zeros
non_zero_count = np.count_nonzero(n)
non_zero_l.append(non_zero_count)
#assign nans
n = n.astype(float)
n[n == 0.0] = np.nan
#flatten the matrix
n = np.array(n.flatten())
one_mean.append(n)
#append means and stds
distribution.append(sum(non_zero_l))
alb_mean.append(np.nanmean(one_mean))
alb_std.append(np.nanstd(one_mean))
end = time.time()
print "Loop took {} seconds".format((end - start))
这需要0.23秒
我尝试通过第二个选项加快速度:
distribution = []
alb_mean = []
alb_std = []
start = time.time()
for i in range(len(list_of_lists)):
for_mean = []
#get one list
one_list = list_of_lists[i]
#flatten the list
chain = itertools.chain(*one_list)
flat = list(chain)
#count non_zeros
non_zero_count = np.count_nonzero(flat)
distribution.append(non_zero_count)
#remove zeros
remove_zero = np.setdiff1d(flat ,[0.0])
alb_mean.append(np.nanmean(remove_zero))
alb_std.append(np.nanstd(remove_zero))
end = time.time()
print "Loop took {} seconds".format((end - start))
这实际上比较慢,需要0.88秒
大量的循环让我想到有更好的方法来实现这一点。我尝试过numba
,但它不喜欢在函数中添加附加。Version#1
在使用loopy解决方案的示例中,您使用两个循环进行循环-一个是3
迭代,另一个是2
迭代。所以,它已经接近矢量化了。唯一的瓶颈是append
步骤
完全矢量化,这里有一种方法-
a = np.array(list_of_lists, dtype=float)
zm = a!=0
avgs = np.einsum('ijkl,ijkl->i',zm,a)/zm.sum(axis=(1,2,3)).astype(float)
a[~zm] = np.nan
stds = np.nanstd(a, axis=(1,2,3))
使用与问题中相同的设置,以下是关于计时的内容-
Loop took 0.150925159454 seconds
Proposed solution took 0.121352910995 seconds
Loop took 0.155035018921 seconds
Proposed solution took 0.0648851394653 seconds
版本#2
我们可以使用average
计算std
,从而重复使用avgs
进一步提升:
因此,需要修改版本-
a = np.asarray(list_of_lists)
zm = a!=0
N = zm.sum(axis=(1,2,3)).astype(float)
avgs = np.einsum('ijkl,ijkl->i',zm,a)/N
diffs = ((a-avgs[:,None,None,None])**2)
stds = np.sqrt(np.einsum('ijkl,ijkl->i',zm,diffs)/N)
更新时间-
Loop took 0.150925159454 seconds
Proposed solution took 0.121352910995 seconds
Loop took 0.155035018921 seconds
Proposed solution took 0.0648851394653 seconds
为什么要在列表列表上使用numpy函数?为什么不使用
numpy
数组呢?请原谅,我对numpy的世界还不熟悉,但我现在做的事情就是这样,因为列表中的数据代表numpy 2d矩阵,所以最好使用带零的ints
输入数组。目前,对于np.random.random((N,N))
,它不太可能有任何零,因此像np.count\u nonzero(N)
这样的计算是多余的。在我的实际数据中有0,也许我应该选择一个更好的例子matrixTrynp.random.randint()
。