我想要:
我玩过subprocess.Popen,os.exec,os.spawn,os.system ......但没有成功.
另一种解释问题的方法:如果有人杀死了myexe.exe(arg0)的"进程树",如何保护myexe.exe(arg1)?
编辑:同样的问题(没有答案)在这里
编辑:以下命令不保证子进程的独立性
subprocess.Popen(["myexe.exe",arg[1]],creationflags = DETACHED_PROCESS | CREATE_NEW_PROCESS_GROUP,close_fds = True)
Run Code Online (Sandbox Code Playgroud) 我需要从特定的面/顶点列表计算最小值,最大值和平均值.我尝试使用Numpy来优化这种计算,但没有成功.
这是我的测试用例:
#!/usr/bin/python
# -*- coding: iso-8859-15 -*-
'''
Module Started 22 févr. 2013
@note: test case comparaison numpy vs python
@author: Python4D/damien
'''
import numpy as np
import time
def Fnumpy(vertices):
np_vertices=np.array(vertices)
_x=np_vertices[:,:,0]
_y=np_vertices[:,:,1]
_z=np_vertices[:,:,2]
_min=[np.min(_x),np.min(_y),np.min(_z)]
_max=[np.max(_x),np.max(_y),np.max(_z)]
_mean=[np.mean(_x),np.mean(_y),np.mean(_z)]
return _mean,_max,_min
def Fpython(vertices):
list_x=[item[0] for sublist in vertices for item in sublist]
list_y=[item[1] for sublist in vertices for item in sublist]
list_z=[item[2] for sublist in vertices for item in sublist]
taille=len(list_x)
_mean=[sum(list_x)/taille,sum(list_y)/taille,sum(list_z)/taille]
_max=[max(list_x),max(list_y),max(list_z)]
_min=[min(list_x),min(list_y),min(list_z)]
return _mean,_max,_min
if __name__=="__main__":
vertices=[[[1.1,2.2,3.3,4.4]]*4]*1000000
_t=time.clock() …Run Code Online (Sandbox Code Playgroud) A级和B级有什么区别?
自我有什么问题?
为什么我需要逐行声明自己?
class A(dict):
def __init__(self):
self={1:"you", 2:"and me"}
print "inside of class A",self
class B(dict):
def __init__(self):
self[1]="you"
self[2]="and me"
print "inside of class B",self
a=A()
print "outside of class A",a
b=B()
print "outside of class B",b
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结果:
inside of class A {1: 'you', 2: 'and me'}
outside of class A {}
inside of class B {1: 'you', 2: 'and me'}
outside of class B {1: 'you', 2: 'and me'}
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