user2789194 - 1 year ago 79

Python Question

Okay, so after going through the tutorials on numpy's structured arrays I am able to create some simple examples:

`from numpy import array, ones`

names=['scalar', '1d-array', '2d-array']

formats=['float64', '(3,)float64', '(2,2)float64']

my_dtype = dict(names=names, formats=formats)

struct_array1 = ones(1, dtype=my_dtype)

struct_array2 = array([(42., [0., 1., 2.], [[5., 6.],[4., 3.]])], dtype=my_dtype)

(My intended use case would have more than three entries and would use very long 1d-arrays.) So, all goes well until we try to perform some basic math. I get errors for all of the following:

`struct_array1 + struct_array2`

struct_array1 * struct_array2

1.0 + struct_array1

2.0 * struct_array2

Apparently, simple operators (+, -, *, /) are not supported for even the simplest structured arrays. Or am I missing something? Should I be looking at some other package (and don't say Pandas, because it is total overkill for this)? This seems like an obvious capability, so I'm a little dumbfounded. But it's difficult to find any chatter about this on the net. Doesn't this severely limit the usefulness of structured arrays? Why would anyone use a structure array rather than arrays packed into a dict? Is there a technical reason why this might be intractable? Or, if the correct solution is to perform the arduous work of overloading, then how is that done while keeping the operations fast?

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Answer Source

Another way to operate on the whole array is to use the 'union' dtype described in the documentation. In your example, you could expand your dtype by adding a 'union' field, and specifying overlapping 'offsets':

```
from numpy import array, ones, zeros
names=['scalar', '1d-array', '2d-array', 'union']
formats=['float64', '(3,)float64', '(2,2)float64', '(8,)float64']
offsets=[0, 8, 32, 0]
my_dtype = dict(names=names, formats=formats, offsets=offsets)
struct_array3=zeros((4,), dtype=my_dtype)
```

`['union']`

now gives access to all the data as a `(n,8)`

array

```
struct_array3['union'] # == struct_array3.view('(8,)f8')
struct_array3['union'].shape # (4,8)
```

You can operate on 'union' or any other fields:

```
struct_array3['union'] += 2
struct_array3['scalar']= 1
```

The 'union' field could another compatible shape, such as `'(2,4)float64'`

. A 'row' of such an array might look like:

```
array([ (3.0, [0.0, 0.0, 0.0], [[2.0, 2.0], [0.0, 0.0]],
[[3.0, 0.0, 0.0, 0.0], [2.0, 2.0, 0.0, 0.0]])],
dtype={'names':['scalar','1d-array','2d-array','union'],
'formats':['<f8',('<f8', (3,)),('<f8', (2, 2)),('<f8', (2, 4))],
'offsets':[0,8,32,0],
'itemsize':64})
```

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