import numpy as np
a=np.array([1,2,3,4,5,6,7,8,9])
b=np.array([\"a\",\"b\",\"c\",\"d\",\"e\",\"f\",\"g\",\"h\",\"i\"])
c=np.array([9,8,7,6,5,4,3,2,1])
datatype=np.d
You might as well try numpy.rec.fromarrays
.
import numpy as np
a=np.array([1,2,3,4,5,6,7,8,9])
b=np.array(["a","b","c","d","e","f","g","h","i"])
c=np.array([9,8,7,6,5,4,3,2,1])
d = np.rec.fromarrays([a,b,c], formats=['i','S32','i'], names=['num','char','len'])
Although timings are not as good as using itertools
.
In [2]: %timeit d = np.rec.fromarrays([a,b,c], formats=['i','S32','i'], names=['num','char','len'])
10000 loops, best of 3: 86.5 us per loop
In [6]: import itertools
In [7]: %timeit np.fromiter(itertools.izip(a,b,c),dtype=datatype)
100000 loops, best of 3: 11.5 us per loop
zip
does create a list of tuples, which could be memory-intensive if the arrays are big. You could use itertools.izip
to be more memory-efficient:
import itertools
d=np.fromiter(itertools.izip(a,b,c),dtype=datatype)
For small arrays of length ~10:
In [68]: %timeit np.fromiter(itertools.izip(a,b,c),dtype=datatype)
100000 loops, best of 3: 15.8 us per loop
In [69]: %timeit np.array(zip(a,b,c),dtype=datatype)
10000 loops, best of 3: 20.8 us per loop
For arrays of length ~10000:
In [72]: A=np.tile(a,1000)
In [74]: B=np.tile(b,1000)
In [75]: C=np.tile(c,1000)
In [83]: %timeit np.fromiter(itertools.izip(A,B,C),dtype=datatype)
100 loops, best of 3: 10.7 ms per loop
In [84]: %timeit np.array(zip(A,B,C),dtype=datatype)
100 loops, best of 3: 12.7 ms per loop
So np.fromiter
appears to be slightly faster than np.array
.