Develop Reference

Numpy append multiple arrays

In R it is very easy to combine multiple vectors, for instance:
a<-c(1,2,3)
b<-c(4,5,6,7)
c<-c(8,9,10)
#combine to
d<-c(a,b,c)
This is what I want to recreate by using NumPy.
I tried to achieve this using np.append, and it works fine as long as all arrays have the same length, for instance:
a = np.array([1,2,3])
b = np.array([4,5,6])
c = np.array([7,8,9])
d = np.append(a,(b,c)) #works fine
However
a = np.array([1,2,3])
b = np.array([4,5,6,7])
c = np.array([8,9,10])
d = np.append(a,(b,c)) #does not work fine
The result is: [1,2 3 array([4,5,6,7]) array([8,9,10])]. How do I turn this into a classic NumPy array [1 2 3 4 5 6 7 8 9 10]?

I think you need this function:
np.concatenate([a,b,c])
am I right?
np.concatenate is used to concatenate arrays with same number of dimension, but different size, along one specific axis (0 axis by default). In your case, since you have just one dimension and you want to concatenate on the unique dimension you have, you can also use np.hstack([a,b,c]), as mentioned by #Sembei Norimaki in the comments.
EDIT to answer you question in the comments:
in the source code of numpy:
if axis is None:
if arr.ndim != 1:
arr = arr.ravel()
values = ravel(values)
axis = arr.ndim-1
return concatenate((arr, values), axis=axis)
as you can see, the values to append are forced to be a numpy array before appending them (this happens in the ravel function). Since your arrays have different shape, it is not possible to cast them in an integers numpy array and so a numpy array of numpy arrays is created (try np.array((b,c)) and see what happens). For this reason your are appending a numpy array of numpy arrays to an integer numpy array and this causes the problem.

Iterating through rows in python

I have a (68x2) matrix named shape and I am trying to iterate through all the 68 rows by placing column 0 and column 1 of shape in array B. This is then multiplied by a (3x3) transformation matrix A. Then my intent was to create a single array (which is why I used np.append) but actually all I am getting are 68 singular 2 dimensional matrices and I do not know why.
Here is my code:
import numpy as np
for row in shape:
B = np.array([[row[0]],[row[1]],[1]])
result = np.matmul(A,B)
result = np.append(result[0], result[1], axis = 0)
print(result)
Anyone know how I can fix my problem?

You can concatenate a new column onto your shape array and then multiply all your rows by the transform matrix at once using a single matrix multiplication.
result = (np.concatenate((shape, np.ones((68, 1))), axis=1) # A)[:,:2]
It's possible you need to multiply by the transpose of the transformation matrix, A.T, rather than by A itself.

add column Numpy array python

I am very new to python and am very familiar with R, but my question is very simple using Numpy Arrays:
Observe:
I have one array X of dimension (100,2) of floating point type and I want to add a 3rd column, preferably into a new Numpy array of dimension (100,3) such that the 3rd column = col(1)^2 for every row in array of X.
My understanding is Numpy arrays are generally of fixed dimension so I'm OK with creating a new array of dim 100x3, I just don't know how to do so using Numpy arrays.
Thanks!

One way to do this is by creating a new array and then concatenating it. For instance, say that M is currently your array.
You can compute col(1)^2 as C = M[:,0] ** 2 (which I'm interpreting as column 1 squared, not column 1 to the power of the values in column two). C will now be an array with shape (100, ), so we can reshape it using C = np.expand_dims(C, 1) which will create a new axis of length 1, so our new column now has shape (100, 1). This is important because we want all both of our arrays to have the same number of dimensions when concatenating them.
The last step here is to concatenate them using np.concatenate. In total, our result looks like this
C = M[:, 0] ** 2
C = np.expand_dims(C, 1)
M = np.concatenate([M, C], axis=1) #third row will now be col(1) ^ 2
If you're the kind of person who likes to do things in one line, you have:
M = np.concatenate([M, np.expand_dims(M[:, 0] ** 2, 0)], axis=1)
That being said, I would recommend looking at Pandas, it supports these actions more naturally, in my opinion. In Pandas, it would be
M["your_col_3_name"] = M["your_col_1_name"] ** 2
where M is a pandas dataframe.

Append with axis=1 should work.
a = np.zeros((5,2))
b = np.ones((5,1))
print(np.append(a,b,axis=1))
This should return:
[[0,0,1],
[0,0,1],
[0,0,1],
[0,0,1],
[0,0,1]]

# generate an array with shape (100,2), fill with 2.
a = np.full((100,2),2)
# calcuate the square to first column, this will be a 1-d array.
squared=a[:,0]**2
# concatenate the 1-d array to a,
# first need to convert it to 2-d arry with shape (100,1) by reshape(-1,1)
c = np.concatenate((a,squared.reshape(-1,1)),axis=1)

way to create a 3d matrix of 2 vectors and 1 matrix

Hello i have a question regarding a problem I am facing in python. I was studying about tensors and I saw that each row/column of a tensor must have the same size. Is it possible to create a tensor of perhaps a 3d object or matrix where lets say we have 3 axis : x,y,z
In the x axis I want to create a vector to work as an index. So let x be from 0 to N
Then on the y axis I want to have N random integer vectors of size m (where mm
Is it possible?
My first approach was to create a big vector of Nm and a big matrix of (Nm,Nm) dimensions where i would store all my random vectors and matrices and then if I wanted to change for example the my second vector then i would have to play with the indexes. However is there another way to approach this problem with tensors or numpy that I m unaware of?
Thank you in advance for your advices

First vector, N = 3, [1,2, 3]
Second N vectors with length m, m = 2
[[4,5], [6,7], [7,8]]
So, N matrices of size (m,m)
[[[1,1], [2,2]], [[1,1], [2,2]], [[1,1], [2,2]] ]
Lets create numpy arrays from them.
import numpy as np
N = 3
m = 2
a = np.array([1,2,3])
b = np.random.randn(N, m)
c = np.random.randn(N, m, m)
You see the problem here? The last matrix c has already 3 dimensions according to your definitions.
Your argument can be simplified.
Let's say our final matrix is -
a = np.zeros((3,2,2)) # 3 dimensions, x,y,z
1) For first dimension -
a[0,:,:] = 0 # first axis, first index = 0
a[1,:,:] = 1 # first axis, 2nd index = 1
a[2,:,:] = 2 # first axis, 3rd index = 2
2) Now, we need to fill up the rest of the positions, but dimensions don't match up.
So, it's better to create separate tensors for them.

Replicating a matrix in pandas or numpy to a certain size

I have a matrix A which is (41, 41) which is a dataframe.
B is a matrix of size (7154, 8240), ndarray.
I want replicate A (keeping the whole 41x41 matrix intact) to the size of B. It will not fit exactly, but then it should just clip the rows that does not fit.
This is to be able to multiply A*B.
I tried this code, but I cannot multiply with a float.
repeat = pd.concat([A]*(B.shape[0]/A.shape[0]), axis=0, ignore_index=True)
filter_large = pd.concat([repeat]*(B.shape[1]/A.shape[1]), axis=1, ignore_index=True)
filter_l = filter_large.values # change from a dataframe to a numpy array
AB = A*filter_l
I should mention that I've tried numpy.resize but it does not keep the matrix intact, mixing up all rows which is not what I want.

This code will do what you ask for:
shapeMultiples = (np.ceil(B.shape[0]/A.shape[0]).astype(int), np.ceil(B.shape[1]/A.shape[1]).astype(int))
res = np.tile(A, shapeMultiples)[:B.shape[0], :B.shape[1]]
Explanation:
np.tile(A, reps) repeats the matrix A multiple times along each axis. How often it is repeated is specified for each axis in reps.
For your example it should be repeated b.shape[0]/a.shape[0] times along axis 0 and b.shape[1]/a.shape[1] times along axis 1. However you have to round these values up, to make sure it extends the size of matrix B, which is what np.ceil does. Since reps is expected to be a shape of integers but ceil returns floats, we have to cast the type to int.
In the final step we cut of the result to make it fit the size of B with [:B.shape[0], :B.shape[1]].