Create a numpy.recarray from two lists python

Question

Is there a easy way to create a numpy.recarray from two lists. For instance, give the following lists:

list1 = ["a","b","c"]
list2 = [1,2,3,4,5,6,7,8,9,10,11,12]

What I am trying to do is to get the following result:

rec_array = np.rec.array([('a', 1), ('a', 2),('a', 3),('a', 4),
                          ('b', 5), ('b', 6),('b', 7),('b', 8),
                          ('c', 9), ('c', 10),('c', 11),('c', 12)] dtype = [('string','|U5'),('int', '<i4')])

I mean I know how a rec.array works but don't really know how to create one from lists. Maybe dicts could make things easy since the key ,value option. But from lists is there a way to do this?.

I don't see a pattern in paring items of list1 with list2. There are lots of 'a' values, and only one 'b'. But you can make a structured array from a list of tuples. I'd focus on generating such a list. — hpaulj
– hpaulj, Commented Apr 21, 2018 at 20:09

hpaulj · Accepted Answer · 2018-04-21 20:27:39Z

In [73]: list1 = ["a","b","c"]
    ...: list2 = [1,2,3,4,5,6,7,8,9,10,11,12]
    ...: 
In [74]: dt = [('string','|U5'),('int', '<i4')]

A simple pairing of elements:

In [75]: [(i,j) for i, j in zip(list1,list2)]
Out[75]: [('a', 1), ('b', 2), ('c', 3)]

break list2 into 3 groups:

In [79]: list3 = [list2[i:i+4] for i in range(0,12,4)]
In [80]: list3
Out[80]: [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]

double list comprehension:

In [81]: [(i,j) for i,row in zip(list1,list3) for j in row]
Out[81]: 
[('a', 1),
 ('a', 2),
 ('a', 3),
 ('a', 4),
 ('b', 5),
 ('b', 6),
 ('b', 7),
 ('b', 8),
 ('c', 9),
 ('c', 10),
 ('c', 11),
 ('c', 12)]

make a structured array from that:

In [82]: np.array(_, dtype=dt)
Out[82]: 
array([('a',  1), ('a',  2), ('a',  3), ('a',  4), ('b',  5), ('b',  6),
       ('b',  7), ('b',  8), ('c',  9), ('c', 10), ('c', 11), ('c', 12)],
      dtype=[('string', '<U5'), ('int', '<i4')])

OR to make a (3,4) array:

In [86]: [[(i,j) for j in row] for i,row in zip(list1, list3)]
Out[86]: 
[[('a', 1), ('a', 2), ('a', 3), ('a', 4)],
 [('b', 5), ('b', 6), ('b', 7), ('b', 8)],
 [('c', 9), ('c', 10), ('c', 11), ('c', 12)]]
In [87]: np.array(_, dt)
Out[87]: 
array([[('a',  1), ('a',  2), ('a',  3), ('a',  4)],
       [('b',  5), ('b',  6), ('b',  7), ('b',  8)],
       [('c',  9), ('c', 10), ('c', 11), ('c', 12)]],
      dtype=[('string', '<U5'), ('int', '<i4')])
In [88]: _.shape
Out[88]: (3, 4)

Or replicate list1 to same size as list2:

In [97]: np.array([(i,j) for i,j in zip(np.repeat(list1,4),list2)],dt).reshape(3
    ...: ,4)
Out[97]: 
array([[('a',  1), ('a',  2), ('a',  3), ('a',  4)],
       [('b',  5), ('b',  6), ('b',  7), ('b',  8)],
       [('c',  9), ('c', 10), ('c', 11), ('c', 12)]],
      dtype=[('string', '<U5'), ('int', '<i4')])

Paul Panzer · Accepted Answer · 2018-04-21 20:31:09Z

1

In addition to @hpaulj's methods you could also allocate and then fill the array like so:

dtype = [('string','|U5'),('int', '<i4')]
>>> list1 = ["a","b","c"]
>>> list2 = [1,2,3,4,5,6,7,8,9,10,11,12]
>>> 
>>> result = np.recarray((12,), dtype=dtype)
>>> result['string'].reshape(3, 4).T[...] = list1
>>> result['int'] = list2
>>> result
rec.array([('a',  1), ('a',  2), ('a',  3), ('a',  4), ('b',  5),
           ('b',  6), ('b',  7), ('b',  8), ('c',  9), ('c', 10),
           ('c', 11), ('c', 12)],
          dtype=[('string', '<U5'), ('int', '<i4')])

The (small) advantage here is that one can use broadcasting on list1.

answered Apr 21, 2018 at 20:31

Paul Panzer

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