returning subarrays with the same dimensions when looping over numpy array

Question

Consider the following convenient looping idiom.

import numpy

print "shape of"
x = numpy.array([['a', 'b'], ['c', 'd']])
print x
print "is", x.shape
for row in x:
    print "shape of", row, "is", row.shape

This gives

shape of
[['a' 'b']
 ['c' 'd']]
is (2, 2)
shape of ['a' 'b'] is (2,)
shape of ['c' 'd'] is (2,)

My question is, can one preserve the convenient for row in x idiom while returning arrays which have shape (2,1), in this case? Thanks. A function which converts the shape of the subarray from (2,) to (2,0) would be fine. E.g.

for row in x:
    print "shape of", somefunc(row), "is", row.shape

returning

shape of ['a' 'b'] is (2,1)

talonmies · Accepted Answer · 2011-11-09 13:39:37Z

2

You can use numpy.expand_dim to increase the rank of any numpy array:

In [7]: x=np.array([1,2,3,4])

In [8]: x.shape
Out[8]: (4,)

In [9]: np.expand_dims(x,axis=-1).shape
Out[9]: (4, 1)

answered Nov 9, 2011 at 13:39

talonmies

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5 Comments

Faheem Mitha Over a year ago

That's a neat solution. How does it compare to reshape in generality?

Woltan Over a year ago

@FaheemMitha Since neither expand_dims nor reshape copies the array to a new location in memory, both approaches are really equivalent. So it comes down to a matter of taste which one you prefer.

Faheem Mitha Over a year ago

@Woltan: Isn't reshape more general?

Woltan Over a year ago

@FaheemMitha What do you mean with general? Both are valid functions of the numpy library. It is a matter of taste, and what sort of parameters you have at hand, which one you want to use. I for myself use reshape a lot more often. In your case I would suggest using expand_dims since you don`t need to know what the size of the array is...

Faheem Mitha Over a year ago

@Woltan: By general I mean that one applies more generally than the other. The usual meaning of general. As for using reshape, one can always just use .size as you do, so I don't see the problem.

Woltan · Accepted Answer · 2011-11-09 09:37:18Z

1

I don't see why you would want that but you could give this a try:

for row in x:
    print "shape of", row, "is", numpy.reshape(row, (1, row.size)).shape

In my optinion, a 1d-array is easier to handle. So it does not make much sense to me to reshape it to a "1d-matrix".

answered Nov 9, 2011 at 9:37

Woltan

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3 Comments

Faheem Mitha Over a year ago

Thanks, but I think you meant to write (row.size, 1).

Woltan Over a year ago

@FaheemMitha Since ['a' 'b'] is a row vector it should be (1, number_of_elements).

Faheem Mitha Over a year ago

You're right. BTW, I have a function that requires a 2d array, so I have no choice but to change the dimension.

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