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name Punditsdkoslkdosdkoskdo

Convert map object to numpy array in python 3

In Python 2 I could do the following:

import numpy as np    
f = lambda x: x**2
seq = map(f, xrange(5))
seq = np.array(seq)
print seq
# prints: [ 0  1  4  9 16]

In Python 3 it does not work anymore:

import numpy as np    
f = lambda x: x**2
seq = map(f, range(5))
seq = np.array(seq)
# prints: <map object at 0x10341e310>

How do I get the old behaviour (converting the map results to numpy array)?

Edit: As @jonrsharpe pointed out in his answer this could be fixed if I converted seq to a list first:

seq = np.array(list(seq))

but I would prefer to avoid the extra call to list.

Although you refer to it as seq, the map object in Python 3 is not a sequence (it's an iterator, see what's new in Python 3). numpy.array needs a sequence so the len can be determined and the appropriate amount of memory reserved; it won't consume an iterator. For example, the range object, which does support most sequence operations, can be passed directly;

seq = np.array(range(5))
# prints: [0 1 2 3 4]

To restore the previous behaviour, as you're aware, you can explicitly convert the map object back to a sequence (e.g. list or tuple):

seq = np.array(list(seq))  # should probably change the name!

However, as the documentation puts it:

a quick fix is to wrap map() in list(), e.g. list(map(...)), but a better fix is often to use a list comprehension (especially when the original code uses lambda)

So another option would be:

seq = [f(x) for x in range(5)]

or just:

seq = [x**2 for x in range(5)]

Alternatively, actually use numpy from the start:

import numpy as np    
arr = np.arange(5)
arr **= 2
# prints [ 0  1  4  9 16] in 2.x and 3.x
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    • @btel no, np.array won't consume the iterator. You need the full list up-front so the appropriate memory can be reserved.

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