CODEWITHSUNDEEP
pythonnumpy
Lucas
Lucas

Reputation: 980

Error after zip in python

I tried to run the following python code.

import numpy as np

recarr = np.zeros((2,), dtype=('i4,f4,a10'))

col1 = np.arange(1,3)
col2 = np.arange(4,6, dtype=np.float32)
col3 = ['Man', 'Woman']

tmp = zip(col1, col2, col3)
recarr[:] = tmp

But I had the following error message.

File "<ipython-input-55-0c1735078108>", line 1, in <module>
recarr[:] = tmp
ValueError: setting an array element with a sequence.

Could you please help me to solve this problem? Thank you.

Upvotes: 0

Views: 153

Answers (1)

juanpa.arrivillaga
juanpa.arrivillaga

Reputation: 95993

The zip function returns an iterator. To properly assign it to a slice you must materialize the iterator: 

>>> recarr[:] = list(tmp)
>>> recarr
array([(1,  4., b'Man'), (2,  5., b'Woman')],
      dtype=[('f0', '<i4'), ('f1', '<f4'), ('f2', 'S10')])

Note, if you don't want to materialize the iterator into a wasteful intermediate data-structure, you can use np.fromiter:

>>> recarr = np.fromiter(tmp, dtype=('i4,f4,a10'), count=2)
>>> recarr
array([(1,  4., b'Man'), (2,  5., b'Woman')],
      dtype=[('f0', '<i4'), ('f1', '<f4'), ('f2', 'S10')])
>>>

Note, I passed the count argument, which is optional, but if you can provide a count argument, it will make this constructor much more efficient.

Upvotes: 2

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