组合两个不同维的numpy数组

Question

As I read through numpy tutorials, I give myself challenges to build my understanding. I was reading through tutorialpoint.com's numpy resource when I saw in the last example that their end-product modified array was not really an array.

Bottom of page, Broadcasting Iteration example: https://www.tutorialspoint.com/numpy/numpy_iterating_over_array.htm

So I decided it would be a nice challenge to try and create the same endproduct as an array. I succeeded but I was not able to use np.nditer nor was I able to utilize broadcasting although I'm sure there must be a way to utilize either/both.

Here is my code:

a = np.arange(0,60,5) 
a = a.reshape(12,1)
b = np.arange(1,5)
arr = np.zeros((12,2))

counter = 0
for i in range(arr.shape[0]):
    if counter < 4:
        arr[i,:] = np.array([a[i],b[counter]])
        counter += 1
    else:
        counter = 0
        arr[i,:] = np.array([a[i],b[counter]])

print arr

How can I do this more efficiently?

Answer 1

I haven't seen that particular nditer tutorial before.
https://docs.scipy.org/doc/numpy-1.13.0/reference/arrays.nditer.html

is the one I've used. And I keep telling people, nditer , with this Python interface, is not efficient. This page is most useful as stepping stone to using nditer in C code, as illustrated in the last cython example.

There aren't many numpy functions that use np.nditer (in Python code). np.ndindex is one of the few. It's worth reading its code. np.einsum uses this iterator, but in compiled code.

I'll take time later to read and comment on the example in question. Learning to use broadcasting well is more important than using nditer .

In [212]: a=np.arange(0,60,5).reshape(3,4)
In [213]: a
Out[213]: 
array([[ 0,  5, 10, 15],
       [20, 25, 30, 35],
       [40, 45, 50, 55]])
In [214]: b=np.arange(1,5)
In [215]: b
Out[215]: array([1, 2, 3, 4])

In [225]: for x,y in np.nditer([a,b]):
     ...:     print("%d:%d"%(x,y), end=' ')
     ...: print()
0:1 5:2 10:3 15:4 20:1 25:2 30:3 35:4 40:1 45:2 50:3 55:4 

Equivalent plain Python iteration:

In [231]: for row in a:
     ...:     for x,y in zip(row,b):
     ...:         print("%d:%d"%(x,y), end=' ')
     ...: print()
     ...: 
0:1 5:2 10:3 15:4 20:1 25:2 30:3 35:4 40:1 45:2 50:3 55:4 

np.broadcast will broadcast the (3,4) array with (4,):

In [234]: np.broadcast(a,b)
Out[234]: <numpy.broadcast at 0x9c2a7f8>
In [235]: list(_)
Out[235]: 
[(0, 1),
 (5, 2),
 (10, 3),
 (15, 4),
 (20, 1),
 (25, 2),
 (30, 3),
 (35, 4),
 (40, 1),
 (45, 2),
 (50, 3),
 (55, 4)]

Use np.array(list(np.broadcast(a,b))) to make a (12,2) array.

Or with the same print:

In [237]: for x,y in np.broadcast(a,b):
     ...:     print("%d:%d"%(x,y), end=' ')
     ...: print()
     ...: 
0:1 5:2 10:3 15:4 20:1 25:2 30:3 35:4 40:1 45:2 50:3 55:4

---

Your iteration:

In [251]: arr = np.zeros((12,2),dtype=int)
     ...: counter = 0
     ...: for i in range(arr.shape[0]):
     ...:     if counter < 4:
     ...:         arr[i,:] = np.array([a.flat[i],b[counter]])
     ...:         counter += 1
     ...:     else:
     ...:         counter = 0
     ...:         arr[i,:] = np.array([a.flat[i],b[counter]])
     ...:         
In [252]: arr
Out[252]: 
array([[ 0,  1],
       [ 5,  2],
       [10,  3],
       [15,  4],
       [20,  1],
       [25,  1],
       [30,  2],
       [35,  3],
       [40,  4],
       [45,  1],
       [50,  1],
       [55,  2]])

Oops, looks like something off, if you expect the second column to be a repeated b .

There are many ways of combining a and b into this kind of array.

This turns the 2d a into a 1d; replicates b with tile , and joins them with stack ( column_stack would have worked as well):

In [264]: np.stack((a.flat, np.tile(b,3)),1)
Out[264]: 
array([[ 0,  1],
       [ 5,  2],
       [10,  3],
       [15,  4],
       [20,  1],
       [25,  2],
       [30,  3],
       [35,  4],
       [40,  1],
       [45,  2],
       [50,  3],
       [55,  4]])