If by "array" you actually mean a Python list, you can use
Use with AIa = [0] * 10
or
Use with AIa = [None] * 10
Answer from Sven Marnach on Stack Overflowpython - How do I check if a list is empty? - Stack Overflow
python - How do I create an empty array and then append to it in NumPy? - Stack Overflow
python - How to create a list of empty lists - Stack Overflow
Create an empty list with certain size in Python - Stack Overflow
How do I create an empty array in Python?
How do I declare an empty array of size n in Python?
Is a Python list the same as an array?
If by "array" you actually mean a Python list, you can use
Use with AIa = [0] * 10
or
Use with AIa = [None] * 10
You can't do exactly what you want in Python (if I read you correctly). You need to put values in for each element of the list (or as you called it, array).
But, try this:
Use with AIa = [0 for x in range(N)] # N = size of list you want
a[i] = 5 # as long as i < N, you're okay
For lists of other types, use something besides 0. None is often a good choice as well.
if not a:
print("List is empty")
Using the implicit booleanness of the empty list is quite Pythonic.
The Pythonic way to do it is from the PEP 8 style guide.
For sequences, (strings, lists, tuples), use the fact that empty sequences are false:
# Correct: if not seq: if seq: # Wrong: if len(seq): if not len(seq):
That is the wrong mental model for using NumPy efficiently. NumPy arrays are stored in contiguous blocks of memory. To append rows or columns to an existing array, the entire array needs to be copied to a new block of memory, creating gaps for the new elements to be stored. This is very inefficient if done repeatedly.
Instead of appending rows, allocate a suitably sized array, and then assign to it row-by-row:
>>> import numpy as np
>>> a = np.zeros(shape=(3, 2))
>>> a
array([[ 0., 0.],
[ 0., 0.],
[ 0., 0.]])
>>> a[0] = [1, 2]
>>> a[1] = [3, 4]
>>> a[2] = [5, 6]
>>> a
array([[ 1., 2.],
[ 3., 4.],
[ 5., 6.]])
A NumPy array is a very different data structure from a list and is designed to be used in different ways. Your use of hstack is potentially very inefficient... every time you call it, all the data in the existing array is copied into a new one. (The append function will have the same issue.) If you want to build up your matrix one column at a time, you might be best off to keep it in a list until it is finished, and only then convert it into an array.
e.g.
mylist = []
for item in data:
mylist.append(item)
mat = numpy.array(mylist)
item can be a list, an array or any iterable, as long
as each item has the same number of elements.
In this particular case (data is some iterable holding the matrix columns) you can simply use
mat = numpy.array(data)
(Also note that using list as a variable name is probably not good practice since it masks the built-in type by that name, which can lead to bugs.)
EDIT:
If for some reason you really do want to create an empty array, you can just use numpy.array([]), but this is rarely useful!
For arbitrary length lists, you can use [ [] for _ in range(N) ]
Do not use [ [] ] * N, as that will result in the list containing the same list object N times
For manually creating a specified number of lists, this would be good:
empty_list = [ [], [], ..... ]
In case, you want to generate a bigger number of lists, then putting it inside a for loop would be good:
empty_lists = [ [] for _ in range(n) ]
You cannot assign to a list like xs[i] = value, unless the list already is initialized with at least i+1 elements (because the first index is 0). Instead, use xs.append(value) to add elements to the end of the list. (Though you could use the assignment notation if you were using a dictionary instead of a list.)
Creating an empty list:
>>> xs = [None] * 10
>>> xs
[None, None, None, None, None, None, None, None, None, None]
Assigning a value to an existing element of the above list:
>>> xs[1] = 5
>>> xs
[None, 5, None, None, None, None, None, None, None, None]
Keep in mind that something like xs[15] = 5 would still fail, as our list has only 10 elements.
range(x) creates a list from [0, 1, 2, ... x-1]
# 2.X only. Use list(range(10)) in 3.X.
>>> xs = range(10)
>>> xs
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Using a function to create a list:
>>> def display():
... xs = []
... for i in range(9): # This is just to tell you how to create a list.
... xs.append(i)
... return xs
...
>>> print display()
[0, 1, 2, 3, 4, 5, 6, 7, 8]
List comprehension (Using the squares because for range you don't need to do all this, you can just return range(0,9) ):
>>> def display():
... return [x**2 for x in range(9)]
...
>>> print display()
[0, 1, 4, 9, 16, 25, 36, 49, 64]
Try this instead:
lst = [None] * 10
The above will create a list of size 10, where each position is initialized to None. After that, you can add elements to it:
lst = [None] * 10
for i in range(10):
lst[i] = i
Admittedly, that's not the Pythonic way to do things. Better do this:
lst = []
for i in range(10):
lst.append(i)
Or even simpler, in Python 2.x you can do this to initialize a list with values from 0 to 9:
lst = range(10)
And in Python 3.x:
lst = list(range(10))