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.

bootstrap numpy 2D array

I am trying to sample with replacement a base 2D numpy array with shape of (4,2) by rows, say 10 times. The final output should be a 3D numpy array.
Have tried the code below, it works. But is there a way to do it without the for loop?
base=np.array([[20,30],[50,60],[70,80],[10,30]])
print(np.shape(base))
nsample=10
tmp=np.zeros((np.shape(base)[0],np.shape(base)[1],10))
for i in range(nsample):
id_pick = np.random.choice(np.shape(base)[0], size=(np.shape(base)[0]))
print(id_pick)
boot1=base[id_pick,:]
tmp[:,:,i]=boot1
print(tmp)

Here's one vectorized approach -
m,n = base.shape
idx = np.random.randint(0,m,(m,nsample))
out = base[idx].swapaxes(1,2)
Basic idea is that we generate all the possible indices with np.random.randint as idx. That would an array of shape (m,nsample). We use this array to index into the input array along the first axis. Thus, it selects random rows off base. To get the final output with a shape (m,n,nsample), we need to swap last two axes.

You can use the stack function from numpy. Your code would then look like:
base=np.array([[20,30],[50,60],[70,80],[10,30]])
print(np.shape(base))
nsample=10
tmp = []
for i in range(nsample):
id_pick = np.random.choice(np.shape(base)[0], size=(np.shape(base)[0]))
print(id_pick)
boot1=base[id_pick,:]
tmp.append(boot1)
tmp = np.stack(tmp, axis=-1)
print(tmp)

Based on #Divakar 's answer, if you already know the shape of this 2D-array, you can treat it as an (8,) 1D array while bootstrapping, and then reshape it:
m, n = base.shape
flatbase = np.reshape(base, (m*n,))
idxs = np.random.choice(range(8), (numReps, m*n))
bootflats = flatbase[idx]
boots = np.reshape(flatbase, (numReps, m, n))

Append numpy one dimensional arrays does not lead to a matrix

I am trying to get a 2d array, by randomly generating its rows and appending
import numpy as np
my_nums = np.array([])
for i in range(100):
x = np.random.rand(2, 1)
my_nums = np.append(my_nums, np.array(x))
But I do not get what I want but instead get a 1d array.
What is wrong?
Transposing x did not help either.

You could do this by using np.append(axis=0) or np.vstack. This however requires the rows appended to have the same length as the rows already in the array.
You cannot use the same code to append a row with two values to an empty array, and to append a row to an already existing 2D array: numpy will throw a
ValueError: all the input arrays must have same number of dimensions.
You could initialize my_nums to work around this:
my_nums = np.random.rand(1, 2)
for i in range(99):
x = np.random.rand(1, 2)
my_nums = np.append(my_nums, x, axis=0)
Note the decrease in the range by one due to the initialization row. Also note that I changed the dimensions to (1, 2) to get actual row vectors.
Much easier than appending row-wise will of course be to create the array in the wanted final shape:
my_nums = np.random.rand(100, 2)

Reshaping array of matrices in Python

I have a Numpy array X of n 2x2 matrices, arranged so that X.shape = (2,2,n), that is, to get the first matrix I call X[:,:,0]. I would like to reshape X into an array Y such that I can get the first matrix by calling Y[0] etc., but performing X.reshape(n,2,2) messes up the matrices. How can I get it to preserve the matrices while reshaping the array?
I am essentially trying to do this:
import numpy as np
Y = np.zeros([n,2,2])
for i in range(n):
Y[i] = X[:,:,i]
but without using the for loop. How can I do this with reshape or a similar function?
(To get an example array X, try X = np.concatenate([np.identity(2)[:,:,None]] * n, axis=2) for some n.)

numpy.moveaxis can be used to take a view of an array with one axis moved to a different position in the shape:
numpy.moveaxis(X, 2, 0)
numpy.moveaxis(a, source, destination) takes a view of array a where the axis originally at position source ends up at position destination, so numpy.moveaxis(X, 2, 0) makes the original axis 2 the new axis 0 in the view.
There's also numpy.transpose, which can be used to perform arbitrary rearrangements of an array's axes in one go if you pass it the optional second argument, and numpy.rollaxis, an older version of moveaxis with a more confusing calling convention.

Use swapaxis:
Y = X.swapaxes(0,2)

Convert a 1D array to a 2D array in numpy

I want to convert a 1-dimensional array into a 2-dimensional array by specifying the number of columns in the 2D array. Something that would work like this:
> import numpy as np
> A = np.array([1,2,3,4,5,6])
> B = vec2matrix(A,ncol=2)
> B
array([[1, 2],
[3, 4],
[5, 6]])
Does numpy have a function that works like my made-up function "vec2matrix"? (I understand that you can index a 1D array like a 2D array, but that isn't an option in the code I have - I need to make this conversion.)

You want to reshape the array.
B = np.reshape(A, (-1, 2))
where -1 infers the size of the new dimension from the size of the input array.

You have two options:
If you no longer want the original shape, the easiest is just to assign a new shape to the array
a.shape = (a.size//ncols, ncols)
You can switch the a.size//ncols by -1 to compute the proper shape automatically. Make sure that a.shape[0]*a.shape[1]=a.size, else you'll run into some problem.
You can get a new array with the np.reshape function, that works mostly like the version presented above
new = np.reshape(a, (-1, ncols))
When it's possible, new will be just a view of the initial array a, meaning that the data are shared. In some cases, though, new array will be acopy instead. Note that np.reshape also accepts an optional keyword order that lets you switch from row-major C order to column-major Fortran order. np.reshape is the function version of the a.reshape method.
If you can't respect the requirement a.shape[0]*a.shape[1]=a.size, you're stuck with having to create a new array. You can use the np.resize function and mixing it with np.reshape, such as
>>> a =np.arange(9)
>>> np.resize(a, 10).reshape(5,2)

Try something like:
B = np.reshape(A,(-1,ncols))
You'll need to make sure that you can divide the number of elements in your array by ncols though. You can also play with the order in which the numbers are pulled into B using the order keyword.

If your sole purpose is to convert a 1d array X to a 2d array just do:
X = np.reshape(X,(1, X.size))

convert a 1-dimensional array into a 2-dimensional array by adding new axis.
a=np.array([10,20,30,40,50,60])
b=a[:,np.newaxis]--it will convert it to two dimension.

There is a simple way as well, we can use the reshape function in a different way:
A_reshape = A.reshape(No_of_rows, No_of_columns)

You can useflatten() from the numpy package.
import numpy as np
a = np.array([[1, 2],
[3, 4],
[5, 6]])
a_flat = a.flatten()
print(f"original array: {a} \nflattened array = {a_flat}")
Output:
original array: [[1 2]
[3 4]
[5 6]]
flattened array = [1 2 3 4 5 6]

some_array.shape = (1,)+some_array.shape
or get a new one
another_array = numpy.reshape(some_array, (1,)+some_array.shape)
This will make dimensions +1, equals to adding a bracket on the outermost

Change 1D array into 2D array without using Numpy.
l = [i for i in range(1,21)]
part = 3
new = []
start, end = 0, part
while end <= len(l):
temp = []
for i in range(start, end):
temp.append(l[i])
new.append(temp)
start += part
end += part
print("new values: ", new)
# for uneven cases
temp = []
while start < len(l):
temp.append(l[start])
start += 1
new.append(temp)
print("new values for uneven cases: ", new)

import numpy as np
array = np.arange(8)
print("Original array : \n", array)
array = np.arange(8).reshape(2, 4)
print("New array : \n", array)