dtype parameter in numpy.array()

Question

I am trying to understand the logic of dtypes in Numpy.

numpy.min_scalar_type(10)
-> uint8

and:

a = numpy.array([10])
print(a.dtype)
-> int32 (on my machine)

I was expecting uint8 instead of int32, because the (1.9) doc says:

numpy.array(object, dtype=None, ...)

dtype : data-type, optional. The desired data-type for the array. If not given, then the type will be determined as the minimum type required to hold the objects in the sequence

and:

numpy.min_scalar_type(a)

For scalar a, returns the data type with the smallest size and smallest scalar kind which can hold its value.

...

Returns: out : dtype. The minimal data type.

Am I misinterpreting something ?

This is somewhat vague, but I think the key thing here to understand is that minimum type and minimum type with the smallest size are usually different. np.array gives you the minimum type. The minimum type is integer, it does not give you the smallest sized integer type (which is totally reasonable). — cel
– cel, Commented Aug 28, 2015 at 8:01
@cel. I see. Minimum type is with regard with type hierarchy, not with type size. But then, minimum type is not unique since signedinteger and unsignedinteger are not in the same branch of the dtype hierarchy (docs.scipy.org/doc/numpy/reference/arrays.scalars.html). So why does it return int32 (or int64) more than let's say uint32 (or uint64) ? — M. Page
– M. Page, Commented Aug 28, 2015 at 8:04

Paul · Accepted Answer · 2015-08-28 08:54:25Z

4

A python list can contain disparately typed objects, e.g. X = ['apples', 'oranges',10]. If you do type([10]) you'll see the Python type for the container is technically called a list, not an array.

In contrast, in a numpy array all objects are of the same type, the dtype.

The docs are telling you that on creation of a numpy array, the dtype is set to the type that will hold all of the existing objects.

See, look:

the type will be determined as the minimum type required to hold the objects in the sequence

The writers perhaps should have added "and not their values"

We can make a uint8 ten easily enough:

ten = np.uint8(10)

If that is put into a Python list, it retains its type because Python lists preserve types. If that list is sent to numpy.array() to make a numpy array, then the numpy array will use dtype np.uint8 because it is big enough to hold all (1) of the pre-existing Python list objects.

In [49]: np.array([ten]).dtype
Out[49]: dtype('uint8')

But if we use a literal 10, python will create an int object for it instead of an np.uint8 because np.uint8 is peculiar to numpy and all 10 does is invoke python to create that number.

If we make a Python list containing a literal 10, we duplicate your result (with machine-architecture ints):

In [50]: np.array([10]).dtype
Out[50]: dtype('int64')

And if we put the two types together in a python list, and send that list to np.array for creation of a numpy array, then the dtype must be big enough to hold both objects, in this case int64.

In [51]: np.array([ten, 10]).dtype
Out[51]: dtype('int64')

edited Aug 28, 2015 at 8:54

answered Aug 28, 2015 at 8:07

Paul

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3 Comments

M. Page Over a year ago

That's a very nice explanation !

Paul Over a year ago

Thanks. I'm glad you get it.

hpaulj Over a year ago

I'd write Python list instead of Python array. There is a less commonly used Array module.

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