As a disclaimer I'm very new to python and numpy arrays. Reading some of the answers to similar questions and trying their solutions for my own data hasn't been very helpful so I thought I'd just post my own question. For example, Reshaping 3D Numpy Array to a 2D array. Its entirely believable though that I've just implemented those other solutions wrong.
I have a 3D numpy array "C"
C = np.reshape(np.arange(3*3*4),(3,3,4))
print(C)
[[[ 0 1 2 3]
[ 4 5 6 7]
[ 8 9 10 11]]
[[12 13 14 15]
[16 17 18 19]
[20 21 22 23]]
[[24 25 26 27]
[28 29 30 31]
[32 33 34 35]]]
I would like to reshape it into something like:
[0 12 14], [1,13,25], [2,24,26] ..... etc
where the first elements of each of the 3 arrays gets put into its own array, then the second elements of each array get put into a new array, and so on.
It seems trivial, but I'm stumped. I've tried different types combinations of .reshape, just for example,
output=C.reshape(12,3)
I've tried changing the order from "C" to "F", playing around with different .reshape() parameters, but can't seem to actually get the final result in the desired structure
Any tips would be much appreciated.
I think this is what you want:
C = np.reshape(np.arange(3*3*4),(3,3,4))
C.reshape(3,12).T
array([[ 0, 12, 24],
[ 1, 13, 25],
[ 2, 14, 26],
[ 3, 15, 27],
[ 4, 16, 28],
[ 5, 17, 29],
[ 6, 18, 30],
[ 7, 19, 31],
[ 8, 20, 32],
[ 9, 21, 33],
[10, 22, 34],
[11, 23, 35]])
Related
I made some numpy array np3
np1 = np.array(range(2*3*5))
np3 = np1.reshape(2,3,5)
and np3 has shape like this:
[[[ 0 1 2 3 4]
[ 5 6 7 8 9]
[10 11 12 13 14]]
[[15 16 17 18 19]
[20 21 22 23 24]
[25 26 27 28 29]]]
then, I made new numpy array np_55
np_55 = np.full((3,1),55)
and np_55 has shape like this:
[[55]
[55]
[55]]
I want make numpy array like below using both numpy arrays np3 and np_55 (I'll call that 'ANSWER'):
[[[ 0 1 2 3 4 55]
[ 5 6 7 8 9 55]
[10 11 12 13 14 55]]
[[15 16 17 18 19 55]
[20 21 22 23 24 55]
[25 26 27 28 29 55]]]
but I can't make it using both numpy arrays np3 and np_55. Of course I can make hard code like this:
a = np.append((np3[0]), np3_55, axis=1)
b = np.append((np3[1]), np3_55, axis=1)
a = a.reshape(1,3,6)
b = b.reshape(1,3,6)
np.append(a, b, axis=0)
but I don't know how can I solve ANSWER simply.
You can try the following:
import numpy as np
a = np.arange(2*3*5).reshape(2, 3, 5)
b = np.full((3,1),55)
np.c_[a, np.broadcast_to(b, (a.shape[0], *b.shape))]
It gives:
array([[[ 0, 1, 2, 3, 4, 55],
[ 5, 6, 7, 8, 9, 55],
[10, 11, 12, 13, 14, 55]],
[[15, 16, 17, 18, 19, 55],
[20, 21, 22, 23, 24, 55],
[25, 26, 27, 28, 29, 55]]])
I have a 2d array, and a list of <start_y, height, start_x, width>.
What I need is to select squares according to the list,
So, for example if this is my 2d array:
[[ 0 1 2 3 4 5 6 7 8 9]
[10 11 12 13 14 15 16 17 18 19]
[20 21 22 23 24 25 26 27 28 29]
[30 31 32 33 34 35 36 37 38 39]
[40 41 42 43 44 45 46 47 48 49]]
and the list is:
[[1,3,2,5],
[2,2,0,3]]
I need the output to be:
[[12,13,14,15,16,
22,23,24,25,26,
32,33,34,35,36],
[20,21,22,
30,31,32]]
i.e. - the first square starts from index 1 in the y axis, with height of 3, and in index 2 in the x axis with width of 5 - and the same logic for the second element in the list.
I obviously tried things like arr[l[:,0]:l[:,0]+l[:,1],l[:,2]:l[:,2]+l[:,3]] where arr is the array and l is the list, but it all returned an invalid syntax error.
I guess the solution involves advanced broadcasting, but I couldn't figure it out by my own.
Any help will be appreciated!
EDIT:
I'm looking for a solution without for loop (it is currently implemented with a loop, and I'm looking to make my code more efficient).
Here's a difficulty: your squares are of different sizes. Most broadcasting or useful functions will result in one array. If your squares were the same size, we could probably figure out how to get a stacked version of them into a 3d array. But if they're different sizes, how would we stack them? Nothing wrong with a for-loop here.
Read this: https://numpy.org/doc/stable/reference/arrays.indexing.html
No "advanced broadcasting" needed.
import numpy as np
arr = np.array([
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,],
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19,],
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29,],
[30, 31, 32, 33, 34, 35, 36, 37, 38, 39,],
[40, 41, 42, 43, 44, 45, 46, 47, 48, 49,],
])
coords = np.array([
[1,3,2,5],
[2,2,0,3],
])
for coord in coords:
y, h, x, w = coord
sq = arr[y:y+h, x:x+w]
print(sq)
Might be easier in your code to make a function
def get_square(arr, coord):
y, h, x, w = coord
return arr[y:y+h, x:x+w]
I have a multiband raster (84 bands). I am reading the raster using GDAL and converting it to numpy array. In numpy when I am checking the array shape it is showing as 84 = bands, 3 = row and 5 = col. I want to compute the ratio between the band(0)/band(n+1) for n in 1 to 84. Thus, I am able to get 83 arrays, each array represents pixel-by-pixel ratio. For example, I have:
Band 1
[[1, 2, 3, 4, 5],
[6, 7, 8, 9, 10],
[11, 12, 13, 14, 15]]
Band 2
[[21, 22, 23, 24, 25],
[26, 27, 28, 29, 30],
[31, 32, 33, 34, 35]]
Band 3
[[31, 32, 33, 34, 35],
[36, 37, 38, 39, 40],
[41, 42, 43, 44, 45]]
...
...
Band84
I need to loop through all the bands in such a way that I get these: Band2/Band1; Band3/Band1; ... ; Band84/Band1
Band2/Band1
[[1/21, 2/22, 3/23, 4/24, 5/25],
[6/26, 7/27, 8/28, 9/29, 10/30],
[11,31, 12/32, 13/33, 14/34, 15/35]]
And so on...
There is any way to vectorize this calculation?
I really appreciate your advice.
If I understand you need
Band2/Band1; Band3/Band1 ... Band84/Band1
Band3/Band2; Band4/Band2 ... Band84/Band2
...
Band84/Band83
It should be something like this
for a in range(0, len(all_bands)-1):
for b in range(a+1, len(all_bands)):
print( all_bands[b]/all_bands[a] )
I have a code in Matlab which I need to translate in Python. A point here that shapes and indexes are really important since it works with tensors. I'm a little bit confused since it seems that it's enough to use order='F' in python reshape(). But when I work with 3D data I noticed that it does not work. For example, if A is an array from 1 to 27 in python
array([[[ 1, 2, 3],
[ 4, 5, 6],
[ 7, 8, 9]],
[[10, 11, 12],
[13, 14, 15],
[16, 17, 18]],
[[19, 20, 21],
[22, 23, 24],
[25, 26, 27]]])
if I perform A.reshape(3, 9, order='F') I get
[[ 1 4 7 2 5 8 3 6 9]
[10 13 16 11 14 17 12 15 18]
[19 22 25 20 23 26 21 24 27]]
In Matlab for A = 1:27 reshaped to [3, 3, 3] and then to [3, 9] it seems that I get another array:
1 4 7 10 13 16 19 22 25
2 5 8 11 14 17 20 23 26
3 6 9 12 15 18 21 24 27
And SVD in Matlab and Python gives different results. So, is there a way to fix this?
And maybe you know the correct way of operating with multidimensional arrays in Matlab -> python, like should I get the same SVD for arrays like arange(1, 13).reshape(3, 4) and in Matlab 1:12 -> reshape(_, [3, 4]) or what is the correct way to work with that? Maybe I can swap axes somehow in python to get the same results as in Matlab? Or change the order of axes in reshape(x1, x2, x3,...) in Python?
I was having the same issues, until I found this wikipedia article: row- and column-major order
Python (and C) organizes the data arrays in row major order. As you can see in your first example code, the elements first increases with the columns:
array([[[ 1, 2, 3],
- - - -> increasing
Then in the rows
array([[[ 1, 2, 3],
[ 4, <--- new element
When all columns and rows are full, it moves to the next page.
array([[[ 1, 2, 3],
[ 4, 5, 6],
[ 7, 8, 9]],
[[10, <-- new element in next page
In matlab (as fortran) increases first the rows, then the columns, and so on.
For N-dimensionals arrays it looks like:
Python (row major -> last dimension is contiguous): [dim1,dim2,...,dimN]
Matlab (column major -> first dimension is contiguous): the same tensor in memory would look the other way around .. [dimN,...,dim2,dim1]
If you want to export n-dim. arrays from python to matlab, the easiest way is to permute the dimensions first:
(in python)
import numpy as np
import scipy.io as sio
A=np.reshape(range(1,28),[3,3,3])
sio.savemat('A',{'A':A})
(in matlab)
load('A.mat')
A=permute(A,[3 2 1]);%dimensions in reverse ordering
reshape(A,9,3)' %gives the same result as A.reshape([3,9]) in python
Just notice that the (9,3) an the (3,9) are intentionally putted in reverse order.
In Matlab
A = 1:27;
A = reshape(A,3,3,3);
B = reshape(A,9,3)'
B =
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18
19 20 21 22 23 24 25 26 27
size(B)
ans =
3 9
In Python
A = np.array(range(1,28))
A = A.reshape(3,3,3)
B = A.reshape(3,9)
B
array([[ 1, 2, 3, 4, 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14, 15, 16, 17, 18],
[19, 20, 21, 22, 23, 24, 25, 26, 27]])
np.shape(B)
(3, 9)
I am trying to cummatively add a value to the previous value and each time, store the value in an array.
This code is just part of a larger project. For simplicity i am going to define my variables as follows:
ele_ini = [12]
smb = [2, 5, 7, 8, 9, 10]
val = ele_ini
for i in range(len(smb)):
val += smb[i]
print(val)
elevation_smb.append(val)
Problem
Each time, the previous value stored in elevation_smb is replaced by the current value such that the result i obtain is:
elevation_smb = [22, 22, 22, 22, 22, 22]
The result i am expecting however is
elevation_smb = [14, 19, 26, 34, 43, 53]
NOTE:
ele_ini is a vector with n elements. I am only using 1 element just for simplicity.
Don use loops, because slow. Better is fast vectorized solution below.
I think need numpy.cumsum and add vector ele_ini for 2d numpy array:
ele_ini = [12, 10, 1, 0]
smb = [2, 5, 7, 8, 9, 10]
elevation_smb = np.cumsum(np.array(smb)) + np.array(ele_ini)[:, None]
print (elevation_smb)
[[14 19 26 34 43 53]
[12 17 24 32 41 51]
[ 3 8 15 23 32 42]
[ 2 7 14 22 31 41]]
It seems vector in your case is using pointers. That's why it is not creating new values. Try adding copy() which copies the value.
elevation_smb.append(val.copy())
Do with reduce,
In [6]: reduce(lambda c, x: c + [c[-1] + x], smb, ele_ini)
Out[6]: [12, 14, 19, 26, 34, 43, 53]