You could use np.array(list(result.items()), dtype=dtype):
import numpy as np
result = {0: 1.1181753789488595, 1: 0.5566080288678394, 2: 0.4718269778030734, 3: 0.48716683119447185, 4: 1.0, 5: 0.1395076201641266, 6: 0.20941558441558442}
names = ['id','data']
formats = ['f8','f8']
dtype = dict(names = names, formats=formats)
array = np.array(list(result.items()), dtype=dtype)
print(repr(array))
yields
array([(0.0, 1.1181753789488595), (1.0, 0.5566080288678394),
(2.0, 0.4718269778030734), (3.0, 0.48716683119447185), (4.0, 1.0),
(5.0, 0.1395076201641266), (6.0, 0.20941558441558442)],
dtype=[('id', '<f8'), ('data', '<f8')])
If you don’t want to create the intermediate list of tuples, list(result.items()), then you could instead use np.fromiter:
In Python2:
array = np.fromiter(result.iteritems(), dtype=dtype, count=len(result))
In Python3:
array = np.fromiter(result.items(), dtype=dtype, count=len(result))
Why using the list [key,val] does not work:
By the way, your attempt,
numpy.array([[key,val] for (key,val) in result.iteritems()],dtype)
was very close to working. If you change the list [key, val] to the tuple (key, val), then it would have worked. Of course,
numpy.array([(key,val) for (key,val) in result.iteritems()], dtype)
is the same thing as
numpy.array(result.items(), dtype)
in Python2, or
numpy.array(list(result.items()), dtype)
in Python3.
np.array treats lists differently than tuples: Robert Kern explains:
As a rule, tuples are considered “scalar” records and lists are
recursed upon. This rule helps numpy.array() figure out which
sequences are records and which are other sequences to be recursed
upon; i.e. which sequences create another dimension and which are the
atomic elements.
Since (0.0, 1.1181753789488595) is considered one of those atomic elements, it should be a tuple, not a list.