选择除numpy数组范围内的所有行

Question

I have a matrix and I want to iterate over chunks of it. Each iteration I want to grab that chunk and everything NOT in that chunk. For example

#grab all rows between index1 and index2
chunk = arr[index1:index2, :]

#want to grab the rest of the matrix that wasn't already grabbed in chunk
#this is the wrong syntax but demonstrates the behavior I'm looking for
rest = arr[ not(index1:index2), :] 

Any good way to do this?

Answer 1

One approach is to vstack the 2 pieces:

In [68]: arr
Out[68]: 
array([[ 0,  5, 10, 15],
       [ 1,  6, 11, 16],
       [ 2,  7, 12, 17],
       [ 3,  8, 13, 18],
       [ 4,  9, 14, 19]])

In [69]: arr[2:4,:]
Out[69]: 
array([[ 2,  7, 12, 17],
       [ 3,  8, 13, 18]])

In [70]: np.vstack([arr[:2,:],arr[4:,:]])
Out[70]: 
array([[ 0,  5, 10, 15],
       [ 1,  6, 11, 16],
       [ 4,  9, 14, 19]])

Or you could construct a discontinuous list of indices, eg [0,1,4]

arr[[0,1,4],:]

In np.lib.index_tricks there is a function (class actually) that streamlines constructing such a list (from lists or slices):

np.r_[:2,4:5]  # == np.array([0,1,4])

In fact r_ can be used like vstack :

np.r_[arr[:2,:],arr[4:,:]]

Often you want to divide an array based on some condition, in which case a boolean index is useful:

I=np.ones(5,dtype=bool)
I[2:4]=False   # array True, with False for the rows to omit
arr[I,:]
arr[~I,:] # the original slice

Answer 2

如果您知道不想要index1：index2 ..为什么不这样做呢？

rest=arr[index3:]