The way to "start" the array that you want is:
arr = np.empty((0,3), int)
Which is an empty array but it has the proper dimensionality.
>>> arr
array([], shape=(0, 3), dtype=int64)
Then be sure to append along axis 0:
arr = np.append(arr, np.array([[1,2,3]]), axis=0)
arr = np.append(arr, np.array([[4,5,6]]), axis=0)
But, @jonrsharpe is right. In fact, if you're going to be appending in a loop, it would be much faster to append to a list as in your first example, then convert to a numpy array at the end, since you're really not using numpy as intended during the loop:
In [210]: %%timeit
.....: l = []
.....: for i in xrange(1000):
.....: l.append([3*i+1,3*i+2,3*i+3])
.....: l = np.asarray(l)
.....:
1000 loops, best of 3: 1.18 ms per loop
In [211]: %%timeit
.....: a = np.empty((0,3), int)
.....: for i in xrange(1000):
.....: a = np.append(a, 3*i+np.array([[1,2,3]]), 0)
.....:
100 loops, best of 3: 18.5 ms per loop
In [214]: np.allclose(a, l)
Out[214]: True
The numpythonic way to do it depends on your application, but it would be more like:
In [220]: timeit n = np.arange(1,3001).reshape(1000,3)
100000 loops, best of 3: 5.93 µs per loop
In [221]: np.allclose(a, n)
Out[221]: True
The way to "start" the array that you want is:
arr = np.empty((0,3), int)
Which is an empty array but it has the proper dimensionality.
>>> arr
array([], shape=(0, 3), dtype=int64)
Then be sure to append along axis 0:
arr = np.append(arr, np.array([[1,2,3]]), axis=0)
arr = np.append(arr, np.array([[4,5,6]]), axis=0)
But, @jonrsharpe is right. In fact, if you're going to be appending in a loop, it would be much faster to append to a list as in your first example, then convert to a numpy array at the end, since you're really not using numpy as intended during the loop:
In [210]: %%timeit
.....: l = []
.....: for i in xrange(1000):
.....: l.append([3*i+1,3*i+2,3*i+3])
.....: l = np.asarray(l)
.....:
1000 loops, best of 3: 1.18 ms per loop
In [211]: %%timeit
.....: a = np.empty((0,3), int)
.....: for i in xrange(1000):
.....: a = np.append(a, 3*i+np.array([[1,2,3]]), 0)
.....:
100 loops, best of 3: 18.5 ms per loop
In [214]: np.allclose(a, l)
Out[214]: True
The numpythonic way to do it depends on your application, but it would be more like:
In [220]: timeit n = np.arange(1,3001).reshape(1000,3)
100000 loops, best of 3: 5.93 µs per loop
In [221]: np.allclose(a, n)
Out[221]: True
Here is my solution:
arr = []
arr.append([1,2,3])
arr.append([4,5,6])
np_arr = np.array(arr)
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!
1. A Hack
You can create an empty array using the np.empty() function and specify the dimensions to be (0, 0) and then use np.append() to append items later.
>>> a = np.empty((0, 0))
>>> a
array([], shape=(0, 0), dtype=float64)
>>> b = np.append(a, [1, 2])
>>> b
array([1., 2.])
2. However...
The hack above is not advisable, use it with caution. Appending to lists has O(N) complexity, while appending to arrays has O(N^2) (besides different memory use). The proper way should be, then, to append to lists. Note that using list() on numpy arrays to transform them into lists is not correct, as you will get a list of numpy arrays. Instead, use the .tolist() method.
>>> a = np.array([[1, 2], [3, 4]])
>>>
>>> list(a)
[array([1, 2]), array([3, 4])]
>>>
>>> a.tolist()
[[1, 2], [3, 4]]
It is advised to use normal python lists and then at the end convert the list to a numpy array.
x = np.asarray(x)
I suspect you are trying to replicate this working list code:
In [56]: x = []
In [57]: x.append([1,2])
In [58]: x
Out[58]: [[1, 2]]
In [59]: np.array(x)
Out[59]: array([[1, 2]])
But with arrays:
In [53]: x = np.empty((2,2),int)
In [54]: x
Out[54]:
array([[73096208, 10273248],
[ 2, -1]])
Despite the name, the np.empty array is NOT a close of the empty list. It has 4 elements, the shape that you specified.
In [55]: np.append(x, np.array([1,2]), axis=0)
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-55-64dd8e7900e3> in <module>
----> 1 np.append(x, np.array([1,2]), axis=0)
<__array_function__ internals> in append(*args, **kwargs)
/usr/local/lib/python3.6/dist-packages/numpy/lib/function_base.py in append(arr, values, axis)
4691 values = ravel(values)
4692 axis = arr.ndim-1
-> 4693 return concatenate((arr, values), axis=axis)
4694
4695
<__array_function__ internals> in concatenate(*args, **kwargs)
ValueError: all the input arrays must have same number of dimensions, but the array at index 0 has 2 dimension(s) and the array at index 1 has 1 dimension(s)
Note that np.append has passed the task on to np.concatenate. With the axis parameter, that's all this append does. It is NOT a list append clone.
np.concatenate demands consistency in the dimensions of its inputs. One is (2,2), the other (2,). Mismatched dimensions.
np.append is a dangerous function, and not that useful even when used correctly. np.concatenate (and the various stack) functions are useful. But you need to pay attention to shapes. And don't use them iteratively. List append is more efficient for that.
When you got this error, did you look up the np.append, np.empty (and np.concatenate) functions? Read and understand the docs? In the long run SO questions aren't a substitute for reading the documentation.
You can create empty list by []. In order to add new item use append. For add other list use extend.
x = [1, 2, 3]
x.append(4)
x.extend([5, 6])
print(x)
# [1, 2, 3, 4, 5, 6]
I might understand the question incorrectly, but if you want to declare an array of a certain shape but with nothing inside, the following might be helpful:
Initialise empty array:
>>> a = np.zeros((0,3)) #or np.empty((0,3)) or np.array([]).reshape(0,3)
>>> a
array([], shape=(0, 3), dtype=float64)
Now you can use this array to append rows of similar shape to it. Remember that a numpy array is immutable, so a new array is created for each iteration:
>>> for i in range(3):
... a = np.vstack([a, [i,i,i]])
...
>>> a
array([[ 0., 0., 0.],
[ 1., 1., 1.],
[ 2., 2., 2.]])
np.vstack and np.hstack is the most common method for combining numpy arrays, but coming from Matlab I prefer np.r_ and np.c_:
Concatenate 1d:
>>> a = np.zeros(0)
>>> for i in range(3):
... a = np.r_[a, [i, i, i]]
...
>>> a
array([ 0., 0., 0., 1., 1., 1., 2., 2., 2.])
Concatenate rows:
>>> a = np.zeros((0,3))
>>> for i in range(3):
... a = np.r_[a, [[i,i,i]]]
...
>>> a
array([[ 0., 0., 0.],
[ 1., 1., 1.],
[ 2., 2., 2.]])
Concatenate columns:
>>> a = np.zeros((3,0))
>>> for i in range(3):
... a = np.c_[a, [[i],[i],[i]]]
...
>>> a
array([[ 0., 1., 2.],
[ 0., 1., 2.],
[ 0., 1., 2.]])
numpy.append is pretty different from list.append in python. I know that's thrown off a few programers new to numpy. numpy.append is more like concatenate, it makes a new array and fills it with the values from the old array and the new value(s) to be appended. For example:
import numpy
old = numpy.array([1, 2, 3, 4])
new = numpy.append(old, 5)
print old
# [1, 2, 3, 4]
print new
# [1, 2, 3, 4, 5]
new = numpy.append(new, [6, 7])
print new
# [1, 2, 3, 4, 5, 6, 7]
I think you might be able to achieve your goal by doing something like:
result = numpy.zeros((10,))
result[0:2] = [1, 2]
# Or
result = numpy.zeros((10, 2))
result[0, :] = [1, 2]
Update:
If you need to create a numpy array using loop, and you don't know ahead of time what the final size of the array will be, you can do something like:
import numpy as np
a = np.array([0., 1.])
b = np.array([2., 3.])
temp = []
while True:
rnd = random.randint(0, 100)
if rnd > 50:
temp.append(a)
else:
temp.append(b)
if rnd == 0:
break
result = np.array(temp)
In my example result will be an (N, 2) array, where N is the number of times the loop ran, but obviously you can adjust it to your needs.
new update
The error you're seeing has nothing to do with types, it has to do with the shape of the numpy arrays you're trying to concatenate. If you do np.append(a, b) the shapes of a and b need to match. If you append an (2, n) and (n,) you'll get a (3, n) array. Your code is trying to append a (1, 0) to a (2,). Those shapes don't match so you get an error.
if you know the number of columns before hand:
>>> xs = np.array([[1,2,3,4,5],[10,20,30,40,50]])
>>> ys = np.array([], dtype=np.int64).reshape(0,5)
>>> ys
array([], shape=(0, 5), dtype=int64)
>>> np.vstack([ys, xs])
array([[ 1., 2., 3., 4., 5.],
[ 10., 20., 30., 40., 50.]])
if not:
>>> ys = np.array([])
>>> ys = np.vstack([ys, xs]) if ys.size else xs
array([[ 1, 2, 3, 4, 5],
[10, 20, 30, 40, 50]])
In Python, if possible to work with the individual vectors, to append you should use list.append()
>>> E = []
>>> B = np.array([1,2,3,4,5])
>>> C = np.array([10,20,30,40,50])
>>> E = E.append(B)
>>> E = E.append(C)
[array([1, 2, 3, 4, 5]), array([10, 20, 30, 40, 50])]
and then after all append operations are done, return to np.array thusly
>>> E = np.array(E)
array([[ 1, 2, 3, 4, 5],
[10, 20, 30, 40, 50]])