在python中如何定义这样的空矩阵？

[lp_grace]

def checkMat(aMat)
  bMat = []    #这是matlab中的定义方式
  for i in range(aMat.shape[1]):
    if aMat[i,] == ...  #满足一条件
      bMat = concatenate((bMat,aMat[i,]))
  return bMat

python 中有类似的bMat的空矩阵的定义方式吗，empty()函数因为随机填充了值，所以不适合

[panghuhu250]

python用append.

In [279]: def checkMat(aMat):
     ...:     bMat = []
     ...:     for row in aMat:
     ...:         if sum(row) > 0:
     ...:           bMat.append(row)
     ...:     return bMat

In [280]: aMat = [[0, 1, 0], [1, -2, 0], [0, 0, 0]]

In [281]: checkMat(aMat)
Out[281]: [[0, 1, 0]]

更方便的是用list comprehension:

In [282]: [row for row in aMat if sum(row) > 0]
Out[282]: [[0, 1, 0]]

如果你希望按列取的话, numpy应该支持按列取或增加一列. 如果你是做数值计算的话,建议使用numpy. 如果是简单的应用,不想依赖于numpy,可以先把矩阵转置,对行处理,再转置回去.

In [283]: def checkColumn(aMat):
     ...:     tranpose = zip(*aMat)
     ...:     bMat = [row for row in tranpose if sum(row)>0]
     ...:     return zip(*bMat)

In [284]: checkColumn(aMat)
Out[284]: [(0,), (1,), (0,)]

写在一行也不算长(可读性如何就见仁见智了):

In [285]: zip(*(row for row in zip(*aMat) if sum(row) > 0))
Out[285]: [(0,), (1,), (0,)]