How to concatenate an empty array with Numpy.concatenate?

Question

I need to create an array of a specific size mxn filled with empty values so that when I concatenate to that array the initial values will be overwritten with the added values.

My current code:

a = numpy.empty([2,2])  # Make empty 2x2 matrix
b = numpy.array([[1,2],[3,4]])    # Make example 2x2 matrix
myArray = numpy.concatenate((a,b))    # Combine empty and example arrays

Unfortunately, I end up making a 4x2 matrix instead of a 2x2 matrix with the values of b.

Is there anyway to make an actually empty array of a certain size so when I concatenate to it, the values of it become my added values instead of the default + added values?

Why not just do a[:]=b? Or a=b.copy()? Do you want to keep any initial elements of a after this operation? — hpaulj
– hpaulj, Commented Feb 2, 2016 at 7:50

eli · Accepted Answer · 2016-02-02 07:13:06Z

2

Like Oniow said, concatenate does exactly what you saw.

If you want 'default values' that will differ from regular scalar elements, I would suggest you to initialize your array with NaNs (as your 'default value'). If I understand your question, you want to merge matrices so that regular scalars will override your 'default value' elements.

Anyway I suggest you to add the following:

def get_default(size_x,size_y):
    # returns a new matrix filled with 'default values'
    tmp = np.empty([size_x,size_y])
    tmp.fill(np.nan)
    return tmp

And also:

def merge(a, b):
    l = lambda x, y: y if np.isnan(x) else x
    l = np.vectorize(l)
    return map(l, a, b)

Note that if you merge 2 matrices, and both values are non 'default' then it will take the value of the left matrix.

Using NaNs as default value, will result the expected behavior from a default value, for example all math ops will result 'default' as this value indicates that you don't really care about this index in the matrix.

edited Feb 2, 2016 at 7:13

answered Feb 2, 2016 at 6:36

eli

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1 Comment

hpaulj Over a year ago

Be ware of 2 things - np.nan is a float, so this is a good thing to do with dtype=int arrays. And your vectorize operation is almost as slow as plain iteration over all elements (at most a 2x speedup). It will be much slower than tmp[:]=b, especially for large arrays.

Oniow · Accepted Answer · 2016-02-02 05:27:03Z

If I understand your question correctly - concatenate is not what you are looking for. Concatenate does as you saw: joins along an axis.

If you are trying to have an empty matrix that becomes the values of another you could do the following:

import numpy as np

a = np.zeros([2,2])
b = np.array([[1,2],[3,4]])

my_array = a + b

--or--

import numpy as np

my_array = np.zeros([2,2]) # you can use empty here instead in this case.

my_array[0,0] = float(input('Enter first value: ')) # However you get your data to put them into the arrays.

But, I am guessing that is not what you really want as you could just use my_array = b. If you edit your question with more info I may be able to help more.

If you are worried about values adding over time to your array...

import numpy as np

a = np.zeros([2,2])

my_array = b # b is some other 2x2 matrix
''' Do stuff '''
new_b # New array has appeared
my_array = new_b # update your array to these new values. values will not add. 

# Note: if you make changes to my_array those changes will carry onto new_b. To avoid this at the cost of some memory:
my_array = np.copy(new_b)

David Traparic · Accepted Answer · 2024-08-26 11:42:56Z

0

I'll add a similar case, if someone wants to do something in the same manner than:

growing_list = []
for value in values_to_add:
    growing_list.append(value)

But with a numpy empty array, I've sometimes seen a pythonic way to do it like this:

for value in values_to_add:
    try:
        growing_ndarray.concatenate((growing_ndarray, value), axis=0)
    except UnboundLocalError as e:
        growing_ndarray = np.asarray(value)

answered Aug 26, 2024 at 11:42

David Traparic

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