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I use Python and NumPy and have some problems with "transpose":
import numpy as np
a = np.array([5,4])
print(a)
print(a.T)
Invoking a.T is not transposing the array. If a is for example [[],[]] then it transposes correctly, but I need the transpose of [...,...,...].
It's working exactly as it's supposed to. The transpose of a 1D array is still a 1D array! (If you're used to matlab, it fundamentally doesn't have a concept of a 1D array. Matlab's "1D" arrays are 2D.)
If you want to turn your 1D vector into a 2D array and then transpose it, just slice it with np.newaxis (or None, they're the same, newaxis is just more readable).
import numpy as np
a = np.array([5,4])[np.newaxis]
print(a)
print(a.T)
Generally speaking though, you don't ever need to worry about this. Adding the extra dimension is usually not what you want, if you're just doing it out of habit. Numpy will automatically broadcast a 1D array when doing various calculations. There's usually no need to distinguish between a row vector and a column vector (neither of which are vectors. They're both 2D!) when you just want a vector.
Use two bracket pairs instead of one. This creates a 2D array, which can be transposed, unlike the 1D array you create if you use one bracket pair.
import numpy as np
a = np.array([[5, 4]])
a.T
More thorough example:
>>> a = [3,6,9]
>>> b = np.array(a)
>>> b.T
array([3, 6, 9]) #Here it didn't transpose because 'a' is 1 dimensional
>>> b = np.array([a])
>>> b.T
array([[3], #Here it did transpose because a is 2 dimensional
[6],
[9]])
Use numpy's shape method to see what is going on here:
>>> b = np.array([10,20,30])
>>> b.shape
(3,)
>>> b = np.array([[10,20,30]])
>>> b.shape
(1, 3)
For 1D arrays:
a = np.array([1, 2, 3, 4])
a = a.reshape((-1, 1)) # <--- THIS IS IT
print a
array([[1],
[2],
[3],
[4]])
Once you understand that -1 here means "as many rows as needed", I find this to be the most readable way of "transposing" an array. If your array is of higher dimensionality simply use a.T.
You can convert an existing vector into a matrix by wrapping it in an extra set of square brackets...
from numpy import *
v=array([5,4]) ## create a numpy vector
array([v]).T ## transpose a vector into a matrix
numpy also has a matrix class (see array vs. matrix)...
matrix(v).T ## transpose a vector into a matrix
numpy 1D array --> column/row matrix:
>>> a=np.array([1,2,4])
>>> a[:, None] # col
array([[1],
[2],
[4]])
>>> a[None, :] # row, or faster `a[None]`
array([[1, 2, 4]])
And as #joe-kington said, you can replace None with np.newaxis for readability.
To 'transpose' a 1d array to a 2d column, you can use numpy.vstack:
>>> numpy.vstack(numpy.array([1,2,3]))
array([[1],
[2],
[3]])
It also works for vanilla lists:
>>> numpy.vstack([1,2,3])
array([[1],
[2],
[3]])
instead use arr[:,None] to create column vector
You can only transpose a 2D array. You can use numpy.matrix to create a 2D array. This is three years late, but I am just adding to the possible set of solutions:
import numpy as np
m = np.matrix([2, 3])
m.T
Basically what the transpose function does is to swap the shape and strides of the array:
>>> a = np.ones((1,2,3))
>>> a.shape
(1, 2, 3)
>>> a.T.shape
(3, 2, 1)
>>> a.strides
(48, 24, 8)
>>> a.T.strides
(8, 24, 48)
In case of 1D numpy array (rank-1 array) the shape and strides are 1-element tuples and cannot be swapped, and the transpose of such an 1D array returns it unchanged. Instead, you can transpose a "row-vector" (numpy array of shape (1, n)) into a "column-vector" (numpy array of shape (n, 1)). To achieve this you have to first convert your 1D numpy array into row-vector and then swap the shape and strides (transpose it). Below is a function that does it:
from numpy.lib.stride_tricks import as_strided
def transpose(a):
a = np.atleast_2d(a)
return as_strided(a, shape=a.shape[::-1], strides=a.strides[::-1])
Example:
>>> a = np.arange(3)
>>> a
array([0, 1, 2])
>>> transpose(a)
array([[0],
[1],
[2]])
>>> a = np.arange(1, 7).reshape(2,3)
>>> a
array([[1, 2, 3],
[4, 5, 6]])
>>> transpose(a)
array([[1, 4],
[2, 5],
[3, 6]])
Of course you don't have to do it this way since you have a 1D array and you can directly reshape it into (n, 1) array by a.reshape((-1, 1)) or a[:, None]. I just wanted to demonstrate how transposing an array works.
Another solution.... :-)
import numpy as np
a = [1,2,4]
[1, 2, 4]
b = np.array([a]).T
array([[1],
[2],
[4]])
The name of the function in numpy is column_stack.
>>>a=np.array([5,4])
>>>np.column_stack(a)
array([[5, 4]])
I am just consolidating the above post, hope it will help others to save some time:
The below array has (2, )dimension, it's a 1-D array,
b_new = np.array([2j, 3j])
There are two ways to transpose a 1-D array:
slice it with "np.newaxis" or none.!
print(b_new[np.newaxis].T.shape)
print(b_new[None].T.shape)
other way of writing, the above without T operation.!
print(b_new[:, np.newaxis].shape)
print(b_new[:, None].shape)
Wrapping [ ] or using np.matrix, means adding a new dimension.!
print(np.array([b_new]).T.shape)
print(np.matrix(b_new).T.shape)
There is a method not described in the answers but described in the documentation for the numpy.ndarray.transpose method:
For a 1-D array this has no effect, as a transposed vector is simply the same vector. To convert a 1-D array into a 2D column vector, an additional dimension must be added. np.atleast2d(a).T achieves this, as does a[:, np.newaxis].
One can do:
import numpy as np
a = np.array([5,4])
print(a)
print(np.atleast_2d(a).T)
Which (imo) is nicer than using newaxis.
As some of the comments above mentioned, the transpose of 1D arrays are 1D arrays, so one way to transpose a 1D array would be to convert the array to a matrix like so:
np.transpose(a.reshape(len(a), 1))
To transpose a 1-D array (flat array) as you have in your example, you can use the np.expand_dims() function:
>>> a = np.expand_dims(np.array([5, 4]), axis=1)
array([[5],
[4]])
np.expand_dims() will add a dimension to the chosen axis. In this case, we use axis=1, which adds a column dimension, effectively transposing your original flat array.
I have a numpy array of a shape (400, 3, 3, 3) and I want to split it into two parts, so I would get arrays like (100, 3, 3, 3) and (300, 3, 3, 3).
I was playing with numpy split methods, e.g.:
subsets = np.array_split(arr, 2)
which gives me what I want, but it divides the original array into two halves the same size and I don't know how to specify these sizes. It'd be probably easy with some indexing (I guess) but I'm not sure how to do it.
As mentioned in my comment, you can use the Ellipsis notation to specify all axes:
x, y = arr[:100, ...], arr[100:, ...]
I want to create a 3d numpy array form 2d numpy array using for loop.I tried in many different methods to create 3d arrays from 2d but each time its giving me errors. This is what I have done, the end array should have a dimension of (10,3,3).
#this is a sample code
arr=[]
for i in range(10):
a=np.random.rand(3,3)
arr=np.stack(a,arr)
#arr=np.append(arr,a)
#arr=np.array([arr,a])
#arr[i]=a
You can append the 2d arrays to the arr list using list.append method, and after you are done with the for loop, convert arr to 3d array by wrapping it with np.array:
arr = []
for i in range(10):
a = np.random.rand(3,3)
arr.append(a)
np.array(arr).shape
# (10, 3, 3)
Or numpy.stack:
np.stack(arr).shape
# (10, 3, 3)
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'm currently trying to append multiple Numpy arrays together. Basically, what I want to do is to start from a (1 x m) matrix (technically a vector), and end up with a (n x m) matrix. So going from n (1 x m) matrices (vectors) to one (n x m) matrix (If that makes any sense). The ultimate goal with this is to write the matrix into a csv-file with the numpy.savetxt() function so I'll end up with a csv-file with n columns of m length.
The problem with this is that numpy.append() appends the vectors together into a (1 x 2m) vector. So let's say a1 and a2 are Numpy arrays with 10000 elements each. I'll append a2 into a1 by using the append function and simultaneously creating a new array called a, which contains both a1 and a2.
a=np.append(a1, a2, axis=0)
a.shape
>>(20000,)
What I want instead is for the shape to be of the form
>>(2, 10000)
or more generally
>>(n, m)
What should I do? Please note, that I want to continue adding the vectors into the array. Thanks for your time!
you can use the transpose of numpy.column_stack
For example:
import numpy as np
a=np.array([1,2,3,4,5])
b=np.array([9,8,7,6,5])
c=np.column_stack((a,b)).T
print c
>>> array([[1, 2, 3, 4, 5],
[9, 8, 7, 6, 5]])
print a.shape,b.shape,c.shape
>>> (5,) (5,) (2, 5)
EDIT:
you can keep adding columns like so:
d=np.array([2,2,2,2,2])
c=np.column_stack((c.T,d)).T
print c
>>> array([[1, 2, 3, 4, 5],
[9, 8, 7, 6, 5],
[2, 2, 2, 2, 2]])
print c.shape
>>> (3, 5)
This should work
a=np.append(a1, a2, axis=0).reshape(2,10000)
a.shape
>>(2,10000)
In order to merge arrays vertically I would use np.vstack
import numpy as np
np.vstack((a1,a2))
However, from my point of view, numpy.array shouldn't be created using for loops and appending the new array to the old one. Instead, either you create first the whole numpy.array (nxm) and you write the data from the for loop into that array,
data = np.zeros((n,m))
for i in range(n):
data[i] = ...
or you first create your array as an ordinary python list using append which you can transform at the end into an numpy.array.
data = []
for i in range(n):
data.append(...)
data = np.asarray(data)