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Python indexing with simplicity

I have a numpy array a,
import numpy as np
a = np.array([[[3, 2, 2], [3, 4, 2]],
[[1, 2, 2], [3, 4, 2]],
[[1, 2, 2], [3, 4, 2]]
])
print(a)
[[[3 2 2]
[3 4 2]]
[[1 2 2]
[3 4 2]]
[[1 2 2]
[3 4 2]]]
and I want to slice part of it, right now with this way:
b = []
for i in range(a.shape[0]):
if (a[i, 0, 0] > 2 and a[i, 1, 0] > 2):
b.append(a[i])
print(np.array(b))
[[[3 2 2]
[3 4 2]]]
I tried method 1
a[np.where(a[:,:,0] > 2)]
and method 2
a[a[:,:,0]> 2]
They both result in:
array([[3, 2, 2],
[3, 4, 2],
[3, 4, 2],
[3, 4, 2]])
Is there any way to deal with the index like method 1 or 2?

You can use np.all(..., axis=1):
a[np.all(a[:, :, 0] > 2, axis=1)]

You could also use a list comprehension which allow you to fit your custom conditions.
b = [i for i in a if i[0,0]>2 and i[1,0]>2]

how to append a 1d numpy array to a 2d numpy array python

I would like to append an array [3, 3, 3] to an array [[1, 1, 1], [2, 2, 2]], so that it becomes [[1, 1, 1], [2, 2, 2], [3, 3, 3]]
Here is my code:
import numpy as np
arr1 = np.array([[1, 1, 1],
[2, 2, 2]])
arr2 = np.append(arr1, [3, 3, 3])
print (arr2)
instead of printing [[1, 1, 1], [2, 2, 2], [3, 3, 3]],
it prints [1, 1, 1, 2, 2, 2, 3, 3, 3].
I am quite new to numpy and I do not understand why the 2d array suddenly becomes 1d.

You can use the useful numpy's standard method of vstack.
Here is my code.
Initialize 2-dimensional numpy array
initial_array = np.array([
[1, 1, 1],
[2, 2, 2]
])
define the array to append to initiali array
new_array = np.array([3, 3, 3])
append the new array to initial array as row
result = np.vstack((initial_array, new_array))
this is the result
print(result)
array([[1, 1, 1],
[2, 2, 2],
[3, 3, 3]])
You can read more details at here.
Hope this answer to be helpful for you. Thanks.

import numpy as np
arr1 = np.array([[1, 1, 1],
[2, 2, 2]])
arr2 = np.append(arr1, [[3, 3, 3]], axis=0)
print (arr2)
Output:
[[1 1 1]
[2 2 2]
[3 3 3]]

Use numpy.vstack:
arr2 = np.vstack((arr1, [3,3,3]))
>>> arr2
[[1 1 1]
[2 2 2]
[3 3 3]]

Swapping columns in a numpy array by a given indexs

I am trying to change column position of a matrix by a given indexs of array
import numpy as np
t = np.array([[0, 1, 2, 3, 4],
[0, 1, 2, 3, 4],
[0, 1, 2, 3, 4],
[0, 1, 2, 3, 4]])
indexs = np.array([3, 4, 2, 1, 0])
check = [False for i in range(len(indexs))]
for i in range(len(indexs)):
check[i] = True
if (i != indexs[i] and check[indexs[i]] == False):
check[indexs[i]] = True
t[:, [i, indexs[i]]] = t[:, [indexs[i], i]]
print(t)
The result I want:
[[3 4 2 1 0]
[3 4 2 1 0]
[3 4 2 1 0]
[3 4 2 1 0]]
I want to return an array whose column positions is the same as indexs but I can't.
How can I achieve that?

Just index the array along the dimension you want:
t[:, indexs]

if you transpose the matrix it's easy
transposed = t.T
result = np.array([transposed[i] for i in indexs])
result = result.T
array([[3, 4, 2, 1, 0],
[3, 4, 2, 1, 0],
[3, 4, 2, 1, 0],
[3, 4, 2, 1, 0]])

Split up numpy array

I've been looking all over but I'm not really sure how to even describe what it is I want. Essentially I need to turn
np.array(
[[0,0, 1,1, 2,2],
[0,0, 1,1, 2,2],
[3,3, 4,4, 5,5],
[3,3, 4,4, 5,5]]
)
into
np.array(
[[[0,1,2,3,4,5], [0,1,2,3,4,5]],
[[0,1,2,3,4,5], [0,1,2,3,4,5]]
)
I think I can accomplish that using np.reshape and maybe some other stuff but if I try reshape with arguments (2,2,6) I get back
[[[0 0 1 1 2 2]
[0 0 1 1 2 2]]
[[3 3 4 4 5 5]
[3 3 4 4 5 5]]]
which is not quite what I want.

Make your array with a couple of repeats:
In [208]: arr = np.arange(0,6).reshape(2,3)
In [209]: arr
Out[209]:
array([[0, 1, 2],
[3, 4, 5]])
In [210]: arr = arr.repeat(2,0).repeat(2,1)
In [211]: arr
Out[211]:
array([[0, 0, 1, 1, 2, 2],
[0, 0, 1, 1, 2, 2],
[3, 3, 4, 4, 5, 5],
[3, 3, 4, 4, 5, 5]])
Now break it into blocks which we can transpose:
In [215]: arr1 = arr.reshape(2,2,3,2)
In [216]: arr1
Out[216]:
array([[[[0, 0],
[1, 1],
[2, 2]],
[[0, 0],
[1, 1],
[2, 2]]],
[[[3, 3],
[4, 4],
[5, 5]],
[[3, 3],
[4, 4],
[5, 5]]]])
In [217]: arr1.shape
Out[217]: (2, 2, 3, 2)
In [218]: arr1.transpose(1,0,2,3)
Out[218]:
array([[[[0, 0],
[1, 1],
[2, 2]],
[[3, 3],
[4, 4],
[5, 5]]],
[[[0, 0],
[1, 1],
[2, 2]],
[[3, 3],
[4, 4],
[5, 5]]]])
Let's consolidate the middle 2 axes:
In [220]: arr1.transpose(1,0,2,3).reshape(2,6,2)
Out[220]:
array([[[0, 0],
[1, 1],
[2, 2],
[3, 3],
[4, 4],
[5, 5]],
[[0, 0],
[1, 1],
[2, 2],
[3, 3],
[4, 4],
[5, 5]]])
Almost there; just need another transpose:
In [221]: arr1.transpose(1,0,2,3).reshape(2,6,2).transpose(0,2,1)
Out[221]:
array([[[0, 1, 2, 3, 4, 5],
[0, 1, 2, 3, 4, 5]],
[[0, 1, 2, 3, 4, 5],
[0, 1, 2, 3, 4, 5]]])
The basic idea is to reshape the array into blocks, do a transpose, and reshape again. Here I needed another transpose, but if I choose the right one to start with I might not have needed that.
I don't know of a systematic way of doing this; there may be one, but so far I've just used a bit of trial and error when answering this kind of question. Everyone wants a different final arrangement.

This should work:
>>> import numpy as np
>>> A = np.array(
... [[0,0, 1,1, 2,2],
... [0,0, 1,1, 2,2],
... [3,3, 4,4, 5,5],
... [3,3, 4,4, 5,5]]
... )
>>> B = a[::2,::2].flatten()
>>> B
array([0, 1, 2, 3, 4, 5])
>>> C = np.tile(b, (2,2,1))
>>> C
array([[[0, 1, 2, 3, 4, 5],
[0, 1, 2, 3, 4, 5]],
[[0, 1, 2, 3, 4, 5],
[0, 1, 2, 3, 4, 5]]])
We can generalize this for a given n * m matrix, that contain blocks sized n/y * m/x of identical values (so there are y rows and x columns of blocks)
def transform(A, y, x):
dy = A.shape[0]/y
dx = A.shape[1]/x
B = A[::dy, ::dx].flatten()
return np.tile(B, (y,x,1))

I think this is what you are looking for:
import numpy
b=numpy.array([[0,0,1,1,2,2],[0,0,1,1,2,2],[3,3,4,4,5,5],[3,3,4,4,5,5]])
c1=(b[::2,::2].flatten(),b[::2,1::2].flatten())
c2=(b[1::2,::2].flatten(),b[1::2,1::2].flatten())
c=numpy.vstack((c1,c2)).reshape((2,2,6))
print(c)
which outputs:
[[[0 1 2 3 4 5]
[0 1 2 3 4 5]]
[[0 1 2 3 4 5]
[0 1 2 3 4 5]]]
and for general size target array and general size input array this is the algorithm with an example of 3*3 input array:
import numpy
b=numpy.array([[0,0,1,1,2,2],[0,0,1,1,2,2],[0,0,1,1,2,2],[3,3,4,4,5,5],[3,3,4,4,5,5],[3,3,4,4,5,5]])
(m,n)=b.shape
C=b[::int(m/2),::2].flatten(),b[::int(m/2),1::2].flatten()
for i in range(1,int(m/2)):
C=numpy.vstack((C,(b[i::int(m/2),::2].flatten(),b[i::int(m/2),1::2].flatten())))
print(C)
which outputs:
[[0 1 2 3 4 5]
[0 1 2 3 4 5]
[0 1 2 3 4 5]
[0 1 2 3 4 5]
[0 1 2 3 4 5]
[0 1 2 3 4 5]]

Applying a mask to an multidimensional array

I want to do this in a proper way:
data = np.array(data)
data =[
[1, 1, 2, 1],
[0, 1, 3, 2],
[0, 2, 3, 2],
[2, 4, 3, 1],
[0, 2, 1, 4],
[3, 1, 4, 1]]
this should become (delete the lines that start with 0):
[1, 1, 2, 1]
[2, 4, 3, 1]
[3, 1, 4, 1]
So far I did it like this:
lines = []
for i in range(0, len(data[0])):
if data[0,i] != 0:
lines.append(data[:,i])
lines = np.array(lines)
Then I found this fine method:
mask = 1 <= data[0,:]
and now I want to apply that mask to that array. This Mask reads: [True, False, False, True, False, True]. How do I do that?

Why not just:
[ar for ar in data if ar[0] != 0]
This assumes that arrays are not empty.

I presume you have a numpy array based on the data[0,:] and data[0,i] you have in your question and you mean data[:, 0] :
import numpy as np
data = np.array([
[1, 1, 2, 1],
[0, 1, 3, 2],
[0, 2, 3, 2],
[2, 4, 3, 1],
[0, 2, 1, 4],
[3, 1, 4, 1]])
data = data[data[:,0] != 0]
print(data)
Output:
[[1 1 2 1]
[2 4 3 1]
[3 1 4 1]]
data[0,:] is the first row [1 1 2 1] not the first column

Using List comprehension
In [56]: [elem for elem in data if elem[0] !=0]
Out[56]: [[1, 1, 2, 1], [2, 4, 3, 1], [3, 1, 4, 1]]