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我有一个列表列表。我想将数组的元素添加到子列表之一，但我添加的元素取决于数组的长度。

import numpy as np
import numpy.linalg
from numpy import matrix
from scipy.linalg import inv,det,eig
import random
import matplotlib.pyplot as plt
import pylab
from numpy import vstack
from scipy.optimize import curve_fit
from array import array
import copy

def getstabmat(orign, small, specsize, alln):
    #FIRST MAKE THE MATRIX
    matfound=0
    while (matfound==0):
        n=orign
        A=np.empty((n,n))
        I=np.ones((n))
        for i in range(0,n):
            for j in range(i+1,n):
                A[j,i]=random.random()
        for i in range(0,n):
            for j in range(i+1,n):
                    A[i,j] = A[j,i]
        for i in range(n):
            A[i,i]=1
        #NOW REMOVE NEGATIVE ELEMENTS AND KEEP SOLVING
        allpos=0
        while (allpos !=1): #loop for dealing just with getting it positive
            x = numpy.linalg.solve(A,I)
            if any(tl<small for tl in x): #if any of the solutions in x are negative or small
                p=np.where(x==min(x)) # find the most negative solution, p is the position
                x=np.delete(x, p, 0)
                A=np.delete(A, p, 0)
                A=np.delete(A, p, 1)
                I=np.delete(I, p, 0)
                n=n-1
            else:
                allpos=1
        #now test for stability, only do it once and remove one element before returning to check positiveness.
        J=np.empty((n,n)) # make empty jacobian
        for i in range (0,n):
            for j in range (0,n):
                if (i==j): # if we are looking at the diagonal of the matrix, there is a special formula for species dealing with itself
                    tsum = 0
                    for k in range (0,n): #for the summation part
                        tsum = tsum + A[i][k]*x[k] # x is vector of fixed points obtained before
                    J[i][j] = 1 - A[i][j]*x[i] - tsum
                else:
                    J[i][j] = -A[i][j]*x[i]
                    #now jacobian at fixed point has been constructed
        Jeig, Jvec =  eig(J) # get the eigenvalues and eigenvectors
        #run through eigenvalues and find out if any of them are positive
        if any(tl>0 for tl in Jeig.real): #if any eigenvalues are positive
            matfound=0
        else:
            if ((alln==0 and len(A)==specsize) or alln==1): # if the matrix found has five species
                matfound=1
            else:
                matfound=0
    return A, x

def main():

    mats=3 #number of matrices to find
    orign=15
    alln=1 #if alln=1, that means that all sizes of stable matrix should be returned
    n=5 # the number of different species wanted in each matrix
    small=0.0001 #the fractional size that a species is when it is considered to be extinct
    a=0
    sortedspec=[[]]*10
    specad=[]
    while (a<mats): #while all the mats have not been found
        print a
        A, specfp = getstabmat(orign, small, n, alln) #15 is the original size of matrix to pass to fnc.n is the size that will be returned
        a=a+1
        print specfp
        print len(specfp)
        for i in range (0,len(specfp)):
            (sortedspec[len(specfp)]).append(specfp[i])
    print sortedspec

if __name__ == '__main__':
  main()

因此，如果：

specfp = [ 0.78076862  0.79608003  0.50719552]

然后我希望将每个元素添加到列表 sortedspec[3] 中。但是，我最终将数组的每个元素都添加到每个列表元素中。为什么会这样，我可以修复它吗？谢谢你。

4

1 回答 1

3 
sortedspec=[[]]*10

这将使您的列表包含对相同（单个）子列表的 10 个引用，因此更新其中一个也将反映在其他子列表中：

>>> sortedspec[0].append(1)
>>> sortedspec
[[1], [1], [1], [1], [1], [1], [1], [1], [1], [1]]

您需要做的是创建 10 个不同的列表，例如使用列表推导：

sortedspec=[[] for i in range(10)]
于 2012-11-08T13:03:50.493 回答