In numpy, I have two "arrays", X is (m,n) and y is a vector (n,1)
using
X*y
I am getting the error
ValueError: operands could not be broadcast together with shapes (97,2) (2,1)
When (97,2)x(2,1) is clearly a legal matrix operation and should give me a (97,1) vector
EDIT:
I have corrected this using X.dot(y) but the original question still remains.
dot is matrix multiplication, but * does something else.
We have two arrays:
X, shape (97,2)
y, shape (2,1)
With Numpy arrays, the operation
X * y
is done element-wise, but one or both of the values can be expanded in one or more dimensions to make them compatible. This operation is called broadcasting. Dimensions, where size is 1 or which are missing, can be used in broadcasting.
In the example above the dimensions are incompatible, because:
97 2
2 1
Here there are conflicting numbers in the first dimension (97 and 2). That is what the ValueError above is complaining about. The second dimension would be ok, as number 1 does not conflict with anything.
For more information on broadcasting rules: http://docs.scipy.org/doc/numpy/user/basics.broadcasting.html
(Please note that if X and y are of type numpy.matrix, then asterisk can be used as matrix multiplication. My recommendation is to keep away from numpy.matrix, it tends to complicate more than simplifying things.)
Your arrays should be fine with numpy.dot; if you get an error on numpy.dot, you must have some other bug. If the shapes are wrong for numpy.dot, you get a different exception:
ValueError: matrices are not aligned
If you still get this error, please post a minimal example of the problem. An example multiplication with arrays shaped like yours succeeds:
In [1]: import numpy
In [2]: numpy.dot(numpy.ones([97, 2]), numpy.ones([2, 1])).shape
Out[2]: (97, 1)
Per numpy docs:
When operating on two arrays, NumPy compares their shapes element-wise. It starts with the trailing dimensions, and works its way forward. Two dimensions are compatible when:
they are equal, or
one of them is 1
In other words, if you are trying to multiply two matrices (in the linear algebra sense) then you want X.dot(y) but if you are trying to broadcast scalars from matrix y onto X then you need to perform X * y.T.
Example:
>>> import numpy as np
>>>
>>> X = np.arange(8).reshape(4, 2)
>>> y = np.arange(2).reshape(1, 2) # create a 1x2 matrix
>>> X * y
array([[0,1],
[0,3],
[0,5],
[0,7]])
You are looking for np.matmul(X, y). In Python 3.5+ you can use X # y.
It's possible that the error didn't occur in the dot product, but after.
For example try this
a = np.random.randn(12,1)
b = np.random.randn(1,5)
c = np.random.randn(5,12)
d = np.dot(a,b) * c
np.dot(a,b) will be fine; however np.dot(a, b) * c is clearly wrong (12x1 X 1x5 = 12x5 which cannot element-wise multiply 5x12) but numpy will give you
ValueError: operands could not be broadcast together with shapes (12,1) (1,5)
The error is misleading; however there is an issue on that line.
Use np.mat(x) * np.mat(y), that'll work.
We might confuse ourselves that a * b is a dot product.
But in fact, it is broadcast.
Dot Product :
a.dot(b)
Broadcast:
The term broadcasting refers to how numpy treats arrays with different
dimensions during arithmetic operations which lead to certain
constraints, the smaller array is broadcast across the larger array so
that they have compatible shapes.
(m,n) +-/* (1,n) → (m,n) : the operation will be applied to m rows
Convert the arrays to matrices, and then perform the multiplication.
X = np.matrix(X)
y = np.matrix(y)
X*y
we should consider two points about broadcasting.
first: what is possible.
second: how much of the possible things is done by numpy.
I know it might look a bit confusing, but I will make it clear by some example.
lets start from the zero level.
suppose we have two matrices. first matrix has three dimensions (named A) and the second has five (named B).
numpy tries to match last/trailing dimensions. so numpy does not care about the first two dimensions of B.
then numpy compares those trailing dimensions with each other. and if and only if they be equal or one of them be 1, numpy says "O.K. you two match". and if it these conditions don't satisfy, numpy would "sorry...its not my job!".
But I know that you may say comparison was better to be done in way that can handle when they are devisable(4 and 2 / 9 and 3). you might say it could be replicated/broadcasted by a whole number(2/3 in out example). and i am agree with you. and this is the reason I started my discussion with a distinction between what is possible and what is the capability of numpy.
This is because X and y are not the same types. for example X is a numpy matrix and y is a numpy array!
Error: operands could not be broadcast together with shapes (2,3) (2,3,3)
This kind of error occur when the two array does not have the same shape.
to correct this you need reshape one array to match the other.
see example below
a1 = array([1, 2, 3]), shape = (2,3)
a3 =array([[[1., 2., 3.],
[2., 3., 2.],
[2., 4., 5.]],
[[1., 0., 3.],
[2., 3., 7.],
[2., 4., 6.]]])
with shape = (2,3,3)
IF i try to run np.multiply(a2,a3) it will return the error below
Error: operands could not be broadcast together with shapes (2,3) (2,3,3)
to solve this check out the broadcating rules
which state hat Two dimensions are compatible when:
#1.they are equal, or
#2.one of them is 1`
Therefore lets reshape a2.
reshaped = a2.reshape(2,3,1)
Now try to run np.multiply(reshaped,a3)
the multiplication will run SUCCESSFUL!!
ValueError: operands could not be broadcast together with shapes (x ,y) (a ,b)
where x ,y are variables
Basically this error occurred when value of y (no. of columns) doesn't equal to the number of elements in another multidimensional array.
Now let's go through by ex=>
coding apart
import numpy as np
arr1= np.arange(12).reshape(3,
output of arr1
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
arr2= np.arange(4).reshape(1,4)
or (both are same 1 rows and 4 columns)
arr2= np.arange(4)
ouput of arr2=>
array([0, 1, 2, 3])
no of elements in arr2 is equal no of no. of the columns in arr1 it will be excute.
for x,y in np.nditer([a,b]):
print(x,y)
output =>
0 0
1 1
2 2
3 3
4 0
5 1
6 2
7 3
8 0
9 1
10 2
11 3
Related
EDIT
So I just understand now that Python and Numpy just don't do polymorphism very well. Given the same data, it has to be put into the right form before most functions can use it. So expecting Python to be able to 'one-line' something is beyond its capabilities.
Just for comparison, MATLAB doesn't do polymorphism very well either but it's much less noticeable as by default it keeps all numeric data as a 2D array so the majority of functions will work with the majority of data - making it soooo much easier
EDIT
I'm pretty new to Python and struggling to create new arrays out of existing arrays:
In MATLAB the following works to create a column vector from other column vectors:
a = [b(5:6); c(N:M); d(1:P); e(Q)]
with a lot of computational flexibility (Q could be a vector for example).
In Python, I can't find a nice command to add multiple 1D NumPy arrays together and it seems to have lots of issues with single values as it changes them from NumPy arrays to some other format, WHY?!
Can anyone give me a single line of code to carry out the above? It'd be great to see - so far all I've got is lines and lines of checking for the indexing variables (N, M, P, Q) and soooo many np.array(..)'s everywhere to try and keep things the same data type.
I've tried np.append but that doesn't work for multiple vectors (I could nest them but that seems very ugly, esp if I need to add many arrays) and np.concatenate complains that something is 0-dimensional, I don't understand that at all.
concatenate has no problems with a bunch of 1d array:
In [52]: np.concatenate([np.array([1,2,3]), np.ones((2,)), np.array([1])])
Out[52]: array([1., 2., 3., 1., 1., 1.])
If one argument is scalar:
In [53]: np.concatenate([np.array([1,2,3]), np.ones((2,)), 1])
Traceback (most recent call last):
File "<ipython-input-53-51b00c09f677>", line 1, in <module>
np.concatenate([np.array([1,2,3]), np.ones((2,)), 1])
File "<__array_function__ internals>", line 5, in concatenate
ValueError: all the input arrays must have same number of dimensions, but the array at index 0 has 1 dimension(s) and the array at index 2 has 0 dimension(s)
It make a array from the last
In [57]: np.array(1)
Out[57]: array(1)
but that's a 0d array. In MATLAB that would be 2d - everything is 2d, there's not true scalars. Remember, numpy works in python, which has scalars and lists. MATLAB is matrices all the way down ...
Also numpy arrays can be 0d or 1d. There's no artificial 2d lower bound. It's a general array language, not just matrices. In MATLAB even 3d is a tweak on the original 2d.
hstack adds a tweak to make sure all arguments are at least 1d:
In [54]: np.hstack([np.array([1,2,3]), np.ones((2,)), 1])
Out[54]: array([1., 2., 3., 1., 1., 1.])
Even in MATLAB/Octave mismatched dimensions give problems:
>> a = [3:5; [1,2,3]]
a =
3 4 5
1 2 3
>> a = [3:5; [1,2,3]; 4]
error: vertical dimensions mismatch (2x3 vs 1x1)
>> a = [3:5; [1,2,3,4]]
error: vertical dimensions mismatch (1x3 vs 1x4)
>> a = [3:5, [1,2,3,4]]
a =
3 4 5 1 2 3 4
>> a = [3:5, [1,2,3,4],5]
a =
3 4 5 1 2 3 4 5
you would want to look on
np.concatenate
which take sequence of arrays and axis to concatnate them on
What is the best way to perform normal vector addition, where one of the operands is an n x 1 matrix?
Why do I care? Sometimes, a function that should return a vector returns an n x 1 matrix (because the function would equivalently work element-wise on a matrix). When I want to further work with the returned "vector", I always have to reshape - there must be a better way.
For example:
v = np.zeros(shape=(2,1))
w = np.array([1,1])
print('{}, {}'.format(v.shape,w.shape))
Prints: (2, 1), (2,)
print(v+w)
[[1. 1.]
[1. 1.]]
print(v+w.reshape((2,1)))
[[1.]
[1.]] (the desired output!)
Sounds a bit like you are coming from MATLAB where everything is 2d (scalars size is (1,1)) and the trailing dimension is outermost. Or a linear algebra that treats 'vectors' as single column matrices.
In numpy, 0 and 1d arrays are just a normal as 2d. A shape like (n,) is common. By the rules of broadcasting adding a leading dimension is automatic (1,n), but adding a trailing dimension requires user action. That a[:,None] is most idiomatic, though not the only option.
The v+w broadcasting logic is
(2,1) + (2,) => (2,1) + (1,2) => (2,2)
The auto-leading logic avoids ambiguity (what should happen if you try to add a (2,) to a (3,)?). And since leading dimensions are 'outer-most' it makes most sense to expand in that direction. MATLAB on the other hand 'naturally' expends and contracts the trailing dimensions.
So to some degree or other a (n,1) shape is more awkward in numpy, though it is still relatively easy to create.
Another example of an auto leading dimension:
In [129]: np.atleast_2d(np.arange(3)).shape
Out[129]: (1, 3)
On the other hand expand_dims lets us add dimensions all over the place
In [132]: np.expand_dims(np.arange(3),(0,2,3)).shape
Out[132]: (1, 3, 1, 1)
If w has the desired shape of the result (e.g. (2,)), and v has the same size (e.g. (2,1), or (2,)), this is safe and easy:
w + v.reshape(w.shape)
Less generally, if all you want is to get rid of the last dimension, knowing it is of length 1, you can do:
w + v[..., 0]
I don't understand broadcasting. The documentation explains the rules of broadcasting but doesn't seem to define it in English. My guess is that broadcasting is when NumPy fills a smaller dimensional array with dummy data in order to perform an operation. But this doesn't work:
>>> x = np.array([1,3,5])
>>> y = np.array([2,4])
>>> x+y
*** ValueError: operands could not be broadcast together with shapes (3,) (2,)
The error message hints that I'm on the right track, though. Can someone define broadcasting and then provide some simple examples of when it works and when it doesn't?
The term broadcasting describes how numpy treats arrays with different shapes during arithmetic operations.
It's basically a way numpy can expand the domain of operations over arrays.
The only requirement for broadcasting is a way aligning array dimensions such that either:
Aligned dimensions are equal.
One of the aligned dimensions is 1.
So, for example if:
x = np.ndarray(shape=(4,1,3))
y = np.ndarray(shape=(3,3))
You could not align x and y like so:
4 x 1 x 3
3 x 3
But you could like so:
4 x 1 x 3
3 x 3
How would an operation like this result?
Suppose we have:
x = np.ndarray(shape=(1,3), buffer=np.array([1,2,3]),dtype='int')
array([[1, 2, 3]])
y = np.ndarray(shape=(3,3), buffer=np.array([1,1,1,1,1,1,1,1,1]),dtype='int')
array([[1, 1, 1],
[1, 1, 1],
[1, 1, 1]])
The operation x + y would result in:
array([[2, 3, 4],
[2, 3, 4],
[2, 3, 4]])
I hope you caught the drift. If you did not, you can always check the official documentation here.
Cheers!
1.What is Broadcasting?
Broadcasting is a Tensor operation. Helpful in Neural Network (ML, AI)
2.What is the use of Broadcasting?
Without Broadcasting addition of only identical Dimension(shape) Tensors is supported.
Broadcasting Provide us the Flexibility to add two Tensors of Different Dimension.
for Example: adding a 2D Tensor with a 1D Tensor is not possible without broadcasting see the image explaining Broadcasting pictorially
Run the Python example code understand the concept
x = np.array([1,3,5,6,7,8])
y = np.array([2,4,5])
X=x.reshape(2,3)
x is reshaped to get a 2D Tensor X of shape (2,3), and adding this 2D Tensor X with 1D Tensor y of shape(1,3) to get a 2D Tensor z of shape(2,3)
print("X =",X)
print("\n y =",y)
z=X+y
print("X + y =",z)
You are almost correct about smaller Tensor, no ambiguity, the smaller tensor will be broadcasted to match the shape of the larger tensor.(Small vector is repeated but not filled with Dummy Data or Zeros to Match the Shape of larger).
3. How broadcasting happens?
Broadcasting consists of two steps:
1 Broadcast axes are added to the smaller tensor to match the ndim of
the larger tensor.
2 The smaller tensor is repeated alongside these new axes to match the full shape
of the larger tensor.
4. Why Broadcasting not happening in your code?
your code is working but Broadcasting can not happen here because both Tensors are different in shape but Identical in Dimensional(1D).
Broadcasting occurs when dimensions are nonidentical.
what you need to do is change Dimension of one of the Tensor, you will experience Broadcasting.
5. Going in Depth.
Broadcasting(repetition of smaller Tensor) occurs along broadcast axes but since both the Tensors are 1 Dimensional there is no broadcast Axis.
Don't Confuse Tensor Dimension with the shape of tensor,
Tensor Dimensions are not same as Matrices Dimension.
Broadcasting is numpy trying to be smart when you tell it to perform an operation on arrays that aren't the same dimension. For example:
2 + np.array([1,3,5]) == np.array([3, 5, 7])
Here it decided you wanted to apply the operation using the lower dimensional array (0-D) on each item in the higher-dimensional array (1-D).
You can also add a 0-D array (scalar) or 1-D array to a 2-D array. In the first case, you just add the scalar to all items in the 2-D array, as before. In the second case, numpy will add row-wise:
In [34]: np.array([1,2]) + np.array([[3,4],[5,6]])
Out[34]:
array([[4, 6],
[6, 8]])
There are ways to tell numpy to apply the operation along a different axis as well. This can be taken even further with applying an operation between a 3-D array and a 1-D, 2-D, or 0-D array.
>>> x = np.array([1,3,5])
>>> y = np.array([2,4])
>>> x+y
*** ValueError: operands could not be broadcast together with shapes (3,) (2,)
Broadcasting is how numpy do math operations with array of different shapes. Shapes are the format the array has, for example the array you used, x , has 3 elements of 1 dimension; y has 2 elements and 1 dimension.
To perform broadcasting there are 2 rules:
1) Array have the same dimensions(shape) or
2)The dimension that doesn't match equals one.
for example x has shape(2,3) [or 2 lines and 3 columns];
y has shape(2,1) [or 2 lines and 1 column]
Can you add them? x + y?
Answer: Yes, because the mismatched dimension is equal to 1 (the column in y). If y had shape(2,4) broadcasting would not be possible, because the mismatched dimension is not 1.
In the case you posted:
operands could not be broadcast together with shapes (3,) (2,);
it is because 3 and 2 mismatched altough both have 1 line.
I would like to suggest to try the np.broadcast_arrays, run some demos may give intuitive ideas. Official Document is also helpful. From my current understanding, numpy will compare the dimension from tail to head. If one dim is 1, it will broadcast in the dimension, if one array has more axes, such (256*256*3) multiply (1,), you can view (1) as (1,1,1). And broadcast will make (256,256,3).
I don't understand broadcasting. The documentation explains the rules of broadcasting but doesn't seem to define it in English. My guess is that broadcasting is when NumPy fills a smaller dimensional array with dummy data in order to perform an operation. But this doesn't work:
>>> x = np.array([1,3,5])
>>> y = np.array([2,4])
>>> x+y
*** ValueError: operands could not be broadcast together with shapes (3,) (2,)
The error message hints that I'm on the right track, though. Can someone define broadcasting and then provide some simple examples of when it works and when it doesn't?
The term broadcasting describes how numpy treats arrays with different shapes during arithmetic operations.
It's basically a way numpy can expand the domain of operations over arrays.
The only requirement for broadcasting is a way aligning array dimensions such that either:
Aligned dimensions are equal.
One of the aligned dimensions is 1.
So, for example if:
x = np.ndarray(shape=(4,1,3))
y = np.ndarray(shape=(3,3))
You could not align x and y like so:
4 x 1 x 3
3 x 3
But you could like so:
4 x 1 x 3
3 x 3
How would an operation like this result?
Suppose we have:
x = np.ndarray(shape=(1,3), buffer=np.array([1,2,3]),dtype='int')
array([[1, 2, 3]])
y = np.ndarray(shape=(3,3), buffer=np.array([1,1,1,1,1,1,1,1,1]),dtype='int')
array([[1, 1, 1],
[1, 1, 1],
[1, 1, 1]])
The operation x + y would result in:
array([[2, 3, 4],
[2, 3, 4],
[2, 3, 4]])
I hope you caught the drift. If you did not, you can always check the official documentation here.
Cheers!
1.What is Broadcasting?
Broadcasting is a Tensor operation. Helpful in Neural Network (ML, AI)
2.What is the use of Broadcasting?
Without Broadcasting addition of only identical Dimension(shape) Tensors is supported.
Broadcasting Provide us the Flexibility to add two Tensors of Different Dimension.
for Example: adding a 2D Tensor with a 1D Tensor is not possible without broadcasting see the image explaining Broadcasting pictorially
Run the Python example code understand the concept
x = np.array([1,3,5,6,7,8])
y = np.array([2,4,5])
X=x.reshape(2,3)
x is reshaped to get a 2D Tensor X of shape (2,3), and adding this 2D Tensor X with 1D Tensor y of shape(1,3) to get a 2D Tensor z of shape(2,3)
print("X =",X)
print("\n y =",y)
z=X+y
print("X + y =",z)
You are almost correct about smaller Tensor, no ambiguity, the smaller tensor will be broadcasted to match the shape of the larger tensor.(Small vector is repeated but not filled with Dummy Data or Zeros to Match the Shape of larger).
3. How broadcasting happens?
Broadcasting consists of two steps:
1 Broadcast axes are added to the smaller tensor to match the ndim of
the larger tensor.
2 The smaller tensor is repeated alongside these new axes to match the full shape
of the larger tensor.
4. Why Broadcasting not happening in your code?
your code is working but Broadcasting can not happen here because both Tensors are different in shape but Identical in Dimensional(1D).
Broadcasting occurs when dimensions are nonidentical.
what you need to do is change Dimension of one of the Tensor, you will experience Broadcasting.
5. Going in Depth.
Broadcasting(repetition of smaller Tensor) occurs along broadcast axes but since both the Tensors are 1 Dimensional there is no broadcast Axis.
Don't Confuse Tensor Dimension with the shape of tensor,
Tensor Dimensions are not same as Matrices Dimension.
Broadcasting is numpy trying to be smart when you tell it to perform an operation on arrays that aren't the same dimension. For example:
2 + np.array([1,3,5]) == np.array([3, 5, 7])
Here it decided you wanted to apply the operation using the lower dimensional array (0-D) on each item in the higher-dimensional array (1-D).
You can also add a 0-D array (scalar) or 1-D array to a 2-D array. In the first case, you just add the scalar to all items in the 2-D array, as before. In the second case, numpy will add row-wise:
In [34]: np.array([1,2]) + np.array([[3,4],[5,6]])
Out[34]:
array([[4, 6],
[6, 8]])
There are ways to tell numpy to apply the operation along a different axis as well. This can be taken even further with applying an operation between a 3-D array and a 1-D, 2-D, or 0-D array.
>>> x = np.array([1,3,5])
>>> y = np.array([2,4])
>>> x+y
*** ValueError: operands could not be broadcast together with shapes (3,) (2,)
Broadcasting is how numpy do math operations with array of different shapes. Shapes are the format the array has, for example the array you used, x , has 3 elements of 1 dimension; y has 2 elements and 1 dimension.
To perform broadcasting there are 2 rules:
1) Array have the same dimensions(shape) or
2)The dimension that doesn't match equals one.
for example x has shape(2,3) [or 2 lines and 3 columns];
y has shape(2,1) [or 2 lines and 1 column]
Can you add them? x + y?
Answer: Yes, because the mismatched dimension is equal to 1 (the column in y). If y had shape(2,4) broadcasting would not be possible, because the mismatched dimension is not 1.
In the case you posted:
operands could not be broadcast together with shapes (3,) (2,);
it is because 3 and 2 mismatched altough both have 1 line.
I would like to suggest to try the np.broadcast_arrays, run some demos may give intuitive ideas. Official Document is also helpful. From my current understanding, numpy will compare the dimension from tail to head. If one dim is 1, it will broadcast in the dimension, if one array has more axes, such (256*256*3) multiply (1,), you can view (1) as (1,1,1). And broadcast will make (256,256,3).
I'm currently wondering how the numpy array behaves. I feel like the dimensions are not consistent from vectors (Nx1 dimensional) to 'real arrays' (NxN dimensional).
I dont get, why this isn't working:
a = array(([1,2],[3,4],[5,6]))
concatenate((a[:,0],a[:,1:]), axis = 1)
# ValueError: all the input arrays must have same number of dimensions
It seems like the : (at 1:]) makes the difference, but (:0 is not working)
Thanks in advance!
Detailled Version: So I would expect that shape(b)[0] references the vertical direction in (Nx1 arrays), like in an 2D (NxN) array. But it seems like dimension [0] is the horizontal direction in arrays (Nx1 arrays)?
from numpy import *
a = array(([1,2],[3,4],[5,6]))
b = a[:,0]
print shape(a) # (3L, 2L), [0] is vertical
print a # [1,2],[3,4],[5,6]
print shape(b) # (3L, ), [0] is horizontal
print b # [1 3 5]
c = b * ones((shape(b)[0],1))
print shape(c) # (3L, 3L), I'd expect (3L, 1L)
print c # [[ 1. 3. 5.], [ 1. 3. 5.], [ 1. 3. 5.]]
What did I get wrong? Is there a nicer way than
d = b * ones((1, shape(b)[0]))
d = transpose(d)
print shape(d) # (3L, 1L)
print d # [[ 1.], [ 3.], [ 5.]]
to get the (Nx1) vector that I expect or want?
There are two overall issues here. First, b is not an (N, 1) shaped array, it is an (N,) shaped array. In numpy, 1D and 2D arrays are different things. 1D arrays simply have no direction. Vertical vs. horizontal, rows vs. columns, these are 2D concepts.
The second has to do with something called "broadcasting". In numpy arrays, you are able to broadcast lower-dimensional arrays to higher-dimensional ones, and the lower-dimensional part is applied elementwise to the higher-dimensional one.
The broadcasting rules are pretty simple:
When operating on two arrays, NumPy compares their shapes element-wise. It starts with the trailing dimensions, and works its way forward. Two dimensions are compatible when
they are equal, or
one of them is 1
In your case, it starts with the last dimension of ones((shape(b)[0],1)), which is 1. This meets the second criteria. So it multiplies the array b elementwise for each element of ones((shape(b)[0],1)), resulting in a 3D array.
So it is roughly equivalent to:
c = np.array([x*b for x in ones(shape(b))])
Edit:
To answer your original question, what you want to do is to keep both the first and second arrays as 2D arrays.
numpy has a very simple rule for this: indexing reduces the number of dimensions, slicing doesn't. So all you need is to have a length-1 slice. So in your example, just change a[:,0] to a[:,:1]. This means 'get every column up to the second one'. Of course that only includes the first column, but it is still considered a slice operation rather than getting an element, so it still preservers the number of dimensions:
>>> print(a[:, 0])
[1 3 5]
>>> print(a[:, 0].shape)
(3,)
>>> print(a[:, :1])
[[1]
[3]
[5]]
>>> print(a[:, :1].shape)
(3, 1)
>>> print(concatenate((a[:,:1],a[:,1:]), axis = 1))
[[1 2]
[3 4]
[5 6]]