How do I get the dimensions of an array? For instance, this is 2x2:
a = np.array([[1,2],[3,4]])
Use .shape to obtain a tuple of array dimensions:
>>> a.shape
(2, 2)
First:
By convention, in Python world, the shortcut for numpy is np, so:
In [1]: import numpy as np
In [2]: a = np.array([[1,2],[3,4]])
Second：
In Numpy, dimension, axis/axes, shape are related and sometimes similar concepts:
dimension
In Mathematics/Physics, dimension or dimensionality is informally defined as the minimum number of coordinates needed to specify any point within a space. But in Numpy, according to the numpy doc, it's the same as axis/axes:
In Numpy dimensions are called axes. The number of axes is rank.
In [3]: a.ndim # num of dimensions/axes, *Mathematics definition of dimension*
Out[3]: 2
axis/axes
the nth coordinate to index an array in Numpy. And multidimensional arrays can have one index per axis.
In [4]: a[1,0] # to index `a`, we specific 1 at the first axis and 0 at the second axis.
Out[4]: 3 # which results in 3 (locate at the row 1 and column 0, 0-based index)
shape
describes how many data (or the range) along each available axis.
In [5]: a.shape
Out[5]: (2, 2) # both the first and second axis have 2 (columns/rows/pages/blocks/...) data
import numpy as np
>>> np.shape(a)
(2,2)
Also works if the input is not a numpy array but a list of lists
>>> a = [[1,2],[1,2]]
>>> np.shape(a)
(2,2)
Or a tuple of tuples
>>> a = ((1,2),(1,2))
>>> np.shape(a)
(2,2)
Use .shape:
In: a = np.array([[1,2,3],[4,5,6]])
In: a.shape
Out: (2, 3)
In: a.shape[0] # x axis
Out: 2
In: a.shape[1] # y axis
Out: 3
You can use .ndim for dimension and .shape to know the exact dimension:
>>> var = np.array([[1,2,3,4,5,6], [1,2,3,4,5,6]])
>>> var.ndim
2
>>> varshape
(2, 6)
You can change the dimension using .reshape function:
>>> var_ = var.reshape(3, 4)
>>> var_.ndim
2
>>> var_.shape
(3, 4)
The shape method requires that a be a Numpy ndarray. But Numpy can also calculate the shape of iterables of pure python objects:
np.shape([[1,2],[1,2]])
a.shape is just a limited version of np.info(). Check this out:
import numpy as np
a = np.array([[1,2],[1,2]])
np.info(a)
Out
class: ndarray
shape: (2, 2)
strides: (8, 4)
itemsize: 4
aligned: True
contiguous: True
fortran: False
data pointer: 0x27509cf0560
byteorder: little
byteswap: False
type: int32
rows = a.shape[0] # 2
cols = a.shape[1] # 2
a.shape #(2,2)
a.size # rows * cols = 4
Execute below code block in python notebook.
import numpy as np
a = np.array([[1,2],[1,2]])
print(a.shape)
print(type(a.shape))
print(a.shape[0])
output
(2, 2)
<class 'tuple'>
2
then you realized that a.shape is a tuple.
so you can get any dimension's size by a.shape[index of dimention]
Related
import numpy as np
a = np.array([1,2,3,4])
print a.shape[0]
Why it will output 4?
The array [1,2,3,4], it's rows should be 1, I think , so who can explain the reason for me?
because
print(a.shape) # -> (4,)
what you think (or want?) to have is
a = np.array([[1],[2],[3],[4]])
print(a.shape) # -> (4, 1)
or rather (?)
a = np.array([[1, 2 , 3 , 4]])
print(a.shape) # -> (1, 4)
If you'll print a.ndim you'll get 1. That means that a is a one-dimensional array (has rank 1 in numpy terminology), with axis length = 4. It's different from 2D matrix with a single row or column (rank 2).
More on ranks
Related questions:
numpy: 1D array with various shape
Python: Differentiating between row and column vectors
The shape attribute for numpy arrays returns the dimensions of the array. If a has n rows and m columns, then a.shape is (n,m). So a.shape[0] is n and a.shape[1] is m.
numpy arrays returns the dimensions of the array. So, when you create an array using,
a = np.array([1,2,3,4])
you get an array with 4 dimensions. You can check it by printing the shape,
print(a.shape) #(4,)
So, what you get is NOT a 1x4 matrix. If you want that do,
a = numpy.array([1,2,3,4]).reshape((1,4))
print(a.shape)
Or even better,
a = numpy.array([[1,2,3,4]])
a = np.array([1, 2, 3, 4])
by doing this, you get a a as a ndarray, and it is a one-dimension array. Here, the shape (4,) means the array is indexed by a single index which runs from 0 to 3. You can access the elements by the index 0~3. It is different from multi-dimensional arrays.
You can refer to more help from this link Difference between numpy.array shape (R, 1) and (R,).
I have a Numpy array of shape (5,5,3,2). I want to take the element (1,4) of that matrix, which is also a matrix of shape (3,2), and add an element to it -so it becomes a (4,2) array.
The code I'm using is the following:
import numpy as np
a = np.random.rand(5,5,3,2)
a = np.array(a, dtype = object) #So I can have different size sub-matrices
a[2][3] = np.append(a[2][3],[[1.0,1.0]],axis=0) #a[2][3] shape = (3,2)
I'm always obtaining the error:
ValueError: could not broadcast input array from shape (4,2) into shape (3,2)
I understand that the shape returned by the np.append function is not the same as the a[2][3] sub-array, but I thought that the dtype=object would solve my problem. However, I need to do this. Is there any way to go around this limitation?
I also tried to use the insert function but I don't know how could I add the element in the place I want.
Make sure you understand what you have produced. That requires checking the shape and dtype, and possibly looking at the values
In [29]: a = np.random.rand(5,5,3,2)
In [30]: b=np.array(a, dtype=object)
In [31]: a.shape
Out[31]: (5, 5, 3, 2) # a is a 4d array
In [32]: a.dtype
Out[32]: dtype('float64')
In [33]: b.shape
Out[33]: (5, 5, 3, 2) # so is b
In [34]: b.dtype
Out[34]: dtype('O')
In [35]: b[2,3].shape
Out[35]: (3, 2)
In [36]: c=np.append(b[2,3],[[1,1]],axis=0)
In [37]: c.shape
Out[37]: (4, 2)
In [38]: c.dtype
Out[38]: dtype('O')
b[2][3] is also an array. b[2,3] is the proper numpy way of indexing 2 dimensions.
I suspect you wanted b to be a (5,5) array containing arrays (as objects), and you think that you you can simply replace one of those with a (4,2) array. But the b constructor simply changes the floats of a to objects, without changing the shape (or 4d nature) of b.
I could construct a (5,5) object array, and fill it with values from a. And then replace one of those values with a (4,2) array:
In [39]: B=np.empty((5,5),dtype=object)
In [40]: for i in range(5):
...: for j in range(5):
...: B[i,j]=a[i,j,:,:]
...:
In [41]: B.shape
Out[41]: (5, 5)
In [42]: B.dtype
Out[42]: dtype('O')
In [43]: B[2,3]
Out[43]:
array([[ 0.03827568, 0.63411023],
[ 0.28938383, 0.7951006 ],
[ 0.12217603, 0.304537 ]])
In [44]: B[2,3]=c
In [46]: B[2,3].shape
Out[46]: (4, 2)
This constructor for B is a bit crude. I've answered other questions about creating/filling object arrays, but I'm not going to take the time here to streamline this case. It's for illustration purposes only.
In an array of object, any element can be indeed an array (or any kind of object).
import numpy as np
a = np.random.rand(5,5,3,2)
a = np.array(a, dtype=object)
# Assign an 1D array to the array element ``a[2][3][0][0]``:
a[2][3][0][0] = np.arange(10)
a[2][3][0][0][9] # 9
However a[2][3] is not an array element, it is a whole array.
a[2][3].ndim # 2
Therefore when you do a[2][3] = (something) you are using broadcasting instead of assigning an element: numpy tries to replace the content of the subarray a[2][3] and fails because of shape mismatch. The memory layout of numpy arrays does not allow to change the shape of subarrays.
Edit: Instead of using numpy arrays you could use nested lists. These nested lists can have arbitrary sizes. Note that the memory is higher and that the access time is higher compared to numpy array.
import numpy as np
a = np.random.rand(5,5,3,2)
a = np.array(a, dtype=object)
b = np.append(a[2][3], [[1.0,1.0]],axis=0)
a_list = a.tolist()
a_list[2][3] = b.tolist()
The problem here, is that you try to assign to a[2][3]
Make a new array instead.
new_array = np.append(a[2][3],np.array([[1.0,1.0]]),axis=0)
With numpy and python3 I have to following problem:
I have a function which returns a 2 dimensional array of integers of fixed size (2x3 in this case). What is the most idiomatic way to run this function n times and stack these together to a 3 dimensional 2x3xn array? What about performance? Something which only does the minimum number of allocations would be nice.
You are probably looking for np.dstack:
>>> import numpy as np
>>> arrs = [np.random.rand(2, 3) for x in range(5)]
>>> np.dstack(arrs).shape
(2, 3, 5)
If you know the final shape you can do something like the following:
>>> out = np.empty((2, 3, 5))
>>> out[..., 0] = np.random.rand(2, 3)
If I create an array X = np.random.rand(D, 1) it has shape (3,1):
[[ 0.31215124]
[ 0.84270715]
[ 0.41846041]]
If I create my own array A = np.array([0,1,2]) then it has shape (1,3) and looks like
[0 1 2]
How can I force the shape (3, 1) on my array A?
You can assign a shape tuple directly to numpy.ndarray.shape.
A.shape = (3,1)
As of 2022, the docs state:
Setting arr.shape is discouraged and may be deprecated in the future.
Using ndarray.reshape is the preferred approach.
The current best solution would be
A = np.reshape(A, (3,1))
A=np.array([0,1,2])
A.shape=(3,1)
or
A=np.array([0,1,2]).reshape((3,1)) #reshape takes the tuple shape as input
The numpy module has a reshape function and the ndarray has a reshape method, either of these should work to create an array with the shape you want:
import numpy as np
A = np.reshape([1, 2, 3, 4], (4, 1))
# Now change the shape to (2, 2)
A = A.reshape(2, 2)
Numpy will check that the size of the array does not change, ie prod(old_shape) == prod(new_shape). Because of this relation, you're allowed to replace one of the values in shape with -1 and numpy will figure it out for you:
A = A.reshape([1, 2, 3, 4], (-1, 1))
You can set the shape directy i.e.
A.shape = (3L, 1L)
or you can use the resize function:
A.resize((3L, 1L))
or during creation with reshape
A = np.array([0,1,2]).reshape((3L, 1L))
Your 1-D array has the shape (3,):
>>>A = np.array([0,1,2]) # create 1-D array
>>>print(A.shape) # print array shape
(3,)
If you create an array with shape (1,3), you can use the numpy.reshape mentioned in other answers or numpy.swapaxes:
>>>A = np.array([[0,1,2]]) # create 2-D array
>>>print(A.shape) # print array shape
>>>A = np.swapaxes(A,0,1) # swap 0th and 1st axes
>>>A # display array with swapped axes
(1, 3)
array([[0],
[1],
[2]])
Suppose I have 3 numpy arrays a, b, c, of the same shape, say
a.shape == b.shape == c.shape == (7,9)
Now I'd like to create a 3-dimensional array of size (7,9,3), say x, such that
x[:,:,0] == a
x[:,:,1] == b
x[:,:,2] == c
What is the "pythonic" way of doing it (perhaps in one line)?
Thanks in advance!
There's a function that does exactly that: numpy.dstack ("d" for "depth"). For example:
In [10]: import numpy as np
In [11]: a = np.ones((7, 9))
In [12]: b = a * 2
In [13]: c = a * 3
In [15]: x = np.dstack((a, b, c))
In [16]: x.shape
Out[16]: (7, 9, 3)
In [17]: (x[:, :, 0] == a).all()
Out[17]: True
In [18]: (x[:, :, 1] == b).all()
Out[18]: True
In [19]: (x[:, :, 2] == c).all()
Out[19]: True
TL;DR:
Use numpy.stack (docs), which joins a sequence of arrays along a new axis of your choice.
Although #NPE answer is very good and cover many cases, there are some scenarios in which numpy.dstack isn't the right choice (I've just found that out while trying to use it). That's because numpy.dstack, according to the docs:
Stacks arrays in sequence depth wise (along third axis).
This is equivalent to concatenation along the third axis after 2-D
arrays of shape (M,N) have been reshaped to (M,N,1) and 1-D arrays of
shape (N,) have been reshaped to (1,N,1).
Let's walk through an example in which this function isn't desirable. Suppose you have a list with 512 numpy arrays of shape (3, 3, 3) and want to stack them in order to get a new array of shape (3, 3, 3, 512). In my case, those 512 arrays were filters of a 2D-convolutional layer. If you use numpy.dstack:
>>> len(arrays_list)
512
>>> arrays_list[0].shape
(3, 3, 3)
>>> numpy.dstack(arrays_list).shape
(3, 3, 1536)
That's because numpy.dstack always stacks the arrays along the third axis! Alternatively, you should use numpy.stack (docs), which joins a sequence of arrays along a new axis of your choice:
>>> numpy.stack(arrays_list, axis=-1).shape
(3, 3, 3, 512)
In my case, I passed -1 to the axis parameter because I wanted the arrays stacked along the last axis.