Question

我正在使用pyDOE包生成DOE，然后使用此函数将其转换为正态分布：

def Transform_DOE(means,std,DOE):
    import numpy as np
    import sys
    from scipy.stats.distributions import norm

    p=np.shape(DOE)[1]#dimension of the DOE
    if np.shape(means)[0]==1:
        means=means*p
    elif p!=np.shape(means)[0]:
        print 'the dimensions of the DOE is not consistent with that of "means"!'
        sys.exit(1)

    if np.shape(std)[0]==1:
        std=std*p
    elif p!=np.shape(std)[0]:
        print 'the dimensions of the DOE is not consistent with that of "std"!'
        sys.exit(1)  

    if np.max(DOE)>1 or np.min(DOE)<0:
        print 'The input DOE should be in [0,1]'
        sys.exit(1)

    DOEtrf=np.mat(DOE)
    for i in xrange(p):
        DOEtrf[:,i]=norm(loc=means[i], scale=std[i]).ppf(DOEtrf[:, i])

    return DOEtrf

示例运行如下： DOEtrf=Transform_DOE([1,2,3],[0.1,0.3,0.2],DOE)

当我运行此代码时，DOE将会改变！我使用指针还是我不知道的东西？

我使用Anaconda for python 2.7，这就是我生成DOE的方法：

def DOE_LHS(n,p,Min,Max):
import numpy as np
import pyDOE
import sys

if type(Min)!=list or type(Max)!=list :
    print "Min and Max should be lists!"
    sys.exit(1)
elif np.size(Min)!=p or np.size(Max)!=p:
    print "The dimension of Min and Max is not consistent with p!"
    sys.exit(1)
Min=map(float,Min)  #or Min=[float(i) for i in Min]
Max=map(float,Max)
Min=np.array(Min)
Max=np.array(Max)
Range=Max-Min
DOE=pyDOE.lhs(p,samples=n,criterion="centermaximin",iterations=1000)

DOE=DOE*np.tile(Range,(n,1))+Min
return DOE

提前感谢。

为什么我的功能改变了它的输入？ python 2.7

0 个答案: