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Numpy append: automatically overlay an array of the wrong dimension

Is there a way to do the following without an if clause?

I am reading a set of netcdf files with pupynere and want to build an array with numpy append. Sometimes the input data is multi-dimensional (see below “a”), sometimes it is one-dimensional (“b”), but the number of elements in the first dimension is always the same (“9” in the example below).

> import numpy as np
> a = np.arange(27).reshape(3,9)
> b = np.arange(9)
> a.shape
(3, 9)
> b.shape
(9,)

this works as expected:

> np.append(a,a, axis=0)
array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8],
   [ 9, 10, 11, 12, 13, 14, 15, 16, 17],
   [18, 19, 20, 21, 22, 23, 24, 25, 26],
   [ 0,  1,  2,  3,  4,  5,  6,  7,  8],
   [ 9, 10, 11, 12, 13, 14, 15, 16, 17],
   [18, 19, 20, 21, 22, 23, 24, 25, 26]])

but adding b doesn't work so elegantly:

> np.append(a,b, axis=0)
ValueError: arrays must have same number of dimensions

Problem with append (from numpy manual)

When an axis is specified, the values ​​must be in the correct shape.

I need to do the first to get the correct result.

> np.append(a,b.reshape(1,9), axis=0)
array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8],
   [ 9, 10, 11, 12, 13, 14, 15, 16, 17],
   [18, 19, 20, 21, 22, 23, 24, 25, 26],
   [ 0,  1,  2,  3,  4,  5,  6,  7,  8]])

So, in my file reading loop, I am currently using an if clause similar to this:

for i in [a, b]:
    if np.size(i.shape) == 2:
        result = np.append(result, i, axis=0)
    else:
        result = np.append(result, i.reshape(1,9), axis=0)

Is there a way to add "a" and "b" without an if statement?

: @Sven ( np.atleast_2d()), ( ) , . . . .

+5
4

numpy.atleast_2d():

result = np.append(result, np.atleast_2d(i), axis=0)

, , numpy.append() - NumPy - . , , .

+3

, np.vstack(), . .

|1> a = np.arange(27).reshape(3,9)

|2> b = np.arange(9)

|3> np.vstack([a,b])
array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8],
       [ 9, 10, 11, 12, 13, 14, 15, 16, 17],
       [18, 19, 20, 21, 22, 23, 24, 25, 26],
       [ 0,  1,  2,  3,  4,  5,  6,  7,  8]])
+2

@Sven, @Henry @Robert. @Sven , , , , , .

python, O (1), numpy.append() O ( ** 2). numpy:

, i a b:

> a = np.arange(27).reshape(3,9)
> b = np.arange(9)
> a.shape
(3, 9)
> b.shape
(9,)

, . "aaba".

> mList = []
> for i in [a,a,b,a]:
     mList.append(i)

mList :

> mList
[array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8],
   [ 9, 10, 11, 12, 13, 14, 15, 16, 17],
   [18, 19, 20, 21, 22, 23, 24, 25, 26]]),
 array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8],
   [ 9, 10, 11, 12, 13, 14, 15, 16, 17],
   [18, 19, 20, 21, 22, 23, 24, 25, 26]]),
 array([0, 1, 2, 3, 4, 5, 6, 7, 8]),
 array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8],
   [ 9, 10, 11, 12, 13, 14, 15, 16, 17],
   [18, 19, 20, 21, 22, 23, 24, 25, 26]])]

, vstack numpy:

> result = np.vstack(mList[:])
> result.shape
(10, 9)

.

+1

, append numpy. , , :

total_size = 0
for i in [a,b]:
    total_size += i.size

result = numpy.empty(total_size, dtype=a.dtype)
offset = 0
for i in [a,b]:
    # copy in the array
    result[offset:offset+i.size] = i.ravel()
    offset += i.size

# if you know its always divisible by 9:
result = result.reshape(result.size//9, 9)

, , , , . :

result = result[0:known_final_size]
0

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