This is my first time handling multidimensional arrays and I'm having problems accessing elements. I'm trying to get the red pixels of a picture but just the first 8 elements within the array. Here's the code
import Image
import numpy as np
im = Image.open("C:\Users\Jones\Pictures\1.jpg")
pix = im.load()
r, g, b = np.array(im).T
print r[0:8]
Since you're dealing with images, r is a 2-D array. To get the first 8 pixels in the image, try
r.flatten()[:8]
This will wrap around automatically if the first row has less than 8 pixels.
do you want all rows too? Try this r[:,:8]
only want the first row? Try this r[0,:8]
You can do it like this:
r[0][:8]
Note, however, that this will not work if the first row has less than 8 pixels. To fix that, do this:
from itertools import chain
r = list(chain.from_iterable(r))
r[:8]
or (if you don't want to import an entire module):
r = [val for element in r for val in element]
r[:8]
I think it could be more simple. This example uses a random matrix (this will be your r matrix):
In [7]: from pylab import * # convention
In [8]: r = randint(0,10,(10,10)) # this is your image
In [9]: r
array([[7, 9, 5, 5, 6, 8, 1, 4, 3, 4],
[5, 4, 4, 4, 2, 6, 2, 6, 4, 2],
[1, 4, 9, 9, 2, 6, 1, 9, 0, 6],
[5, 9, 0, 7, 9, 9, 5, 2, 0, 7],
[8, 3, 3, 9, 0, 0, 5, 9, 2, 2],
[5, 3, 7, 8, 8, 1, 6, 3, 2, 0],
[0, 2, 5, 7, 0, 1, 0, 2, 1, 2],
[4, 0, 4, 5, 9, 9, 3, 8, 3, 7],
[4, 6, 9, 9, 5, 9, 3, 0, 5, 1],
[6, 9, 9, 0, 3, 4, 9, 7, 9, 6]])
Then, extract first 8 columns and do something
In [17]: r_8 = r[:,:8] # extract columns
In [18]: r_8
Out[18]:
array([[7, 9, 5, 5, 6, 8, 1, 4],
[5, 4, 4, 4, 2, 6, 2, 6],
[1, 4, 9, 9, 2, 6, 1, 9],
[5, 9, 0, 7, 9, 9, 5, 2],
[8, 3, 3, 9, 0, 0, 5, 9],
[5, 3, 7, 8, 8, 1, 6, 3],
[0, 2, 5, 7, 0, 1, 0, 2],
[4, 0, 4, 5, 9, 9, 3, 8],
[4, 6, 9, 9, 5, 9, 3, 0],
[6, 9, 9, 0, 3, 4, 9, 7]])
In [19]: r_8 = r_8 * 2 # do something
In [20]: r_8
Out[20]:
array([[14, 18, 10, 10, 12, 16, 2, 8],
[10, 8, 8, 8, 4, 12, 4, 12],
[ 2, 8, 18, 18, 4, 12, 2, 18],
[10, 18, 0, 14, 18, 18, 10, 4],
[16, 6, 6, 18, 0, 0, 10, 18],
[10, 6, 14, 16, 16, 2, 12, 6],
[ 0, 4, 10, 14, 0, 2, 0, 4],
[ 8, 0, 8, 10, 18, 18, 6, 16],
[ 8, 12, 18, 18, 10, 18, 6, 0],
[12, 18, 18, 0, 6, 8, 18, 14]])
Now, this is the trick. Replace the first 8 columns in r using hstack:
In [21]: r = hstack((r_8, r[:,8:])) # it replaces the FISRT 8 columns, note the indexing notation
In [22]: r
Out[22]:
array([[14, 18, 10, 10, 12, 16, 2, 8, 3, 4], # it does not touch the last 2 columns
[10, 8, 8, 8, 4, 12, 4, 12, 4, 2],
[ 2, 8, 18, 18, 4, 12, 2, 18, 0, 6],
[10, 18, 0, 14, 18, 18, 10, 4, 0, 7],
[16, 6, 6, 18, 0, 0, 10, 18, 2, 2],
[10, 6, 14, 16, 16, 2, 12, 6, 2, 0],
[ 0, 4, 10, 14, 0, 2, 0, 4, 1, 2],
[ 8, 0, 8, 10, 18, 18, 6, 16, 3, 7],
[ 8, 12, 18, 18, 10, 18, 6, 0, 5, 1],
[12, 18, 18, 0, 6, 8, 18, 14, 9, 6]])
EDIT: as to what DSM pointed out, OP is infact using a numpy array.
i retract my answer as nneonneo's correct
Related
I have a matrix like the following:
A = array([[12, 6, 14, 8, 4, 1],
[18, 13, 8, 10, 9, 19],
[ 8, 15, 6, 5, 6, 18],
[ 3, 0, 2, 14, 13, 12],
[ 4, 4, 5, 19, 0, 14],
[16, 8, 7, 7, 11, 0],
[ 3, 11, 2, 19, 11, 5],
[ 4, 2, 1, 9, 12, 12]])
For each cell I want to select the values in a radius of k=2 closest cells.
For instance if I select the A[3,4] I would like a submatrix like the following
array([[18, 13, 8, 10, 9],
[ 8, 15, 6, 5, 6],
[ 3, 0, 2, 14, 13],
[ 4, 4, 5, 19, 0],
[16, 8, 7, 7, 11]])
I defined the following function
def queen_neighbourhood(Adj, in_row, in_col, k):
j=k
k+=1
neighbourhood = Adj[in_row-j:in_row+k, in_col-j:in_col+k]
return neighbourhood
such as queen_neighbourhood(A, 3, 2, 2) returns
array([[18, 13, 8, 10, 9],
[ 8, 15, 6, 5, 6],
[ 3, 0, 2, 14, 13],
[ 4, 4, 5, 19, 0],
[16, 8, 7, 7, 11]])
However it does not work in borders.
For instance, for the cell [0,0] I would like to have
array([[12, 6, 14],
[18, 13, 8],
[ 8, 15, 16])
but it returns queen_neighbourhood(A, 0, 0, 2)
array([], shape=(0, 0), dtype=int64)
You could avoid negative indices:
neighbourhood = Adj[max(in_row-j, 0) : in_row+k,
max(in_col-j, 0) : in_col+k]
Adding to the previous answer; taking into consideration the extreme values
def queen_neighbourhood(Adj, in_row, in_col, k):
j=k
k+=1
neighbourhood = Adj[max(in_row-j, 0) : min(in_row+k,Adj.shape[0]),
max(in_col-j, 0) : min(in_col+k,Adj.shape[1])]
return(neighbourhood)
You can use numpy roll to ensure you are always dealing with the middle value,
import numpy as np
def queen_neighbourhood(Adj, in_row, in_col, k):
j=k
k+=1
midrow = int(Adj.shape[0]/2.)+1
midcol = int(Adj.shape[1]/2.)+1
Ashift = np.roll(Adj,(in_row-midrow,in_col-midcol),(0,1))
neighbourhood = Ashift[1:k+1, 1:k+1]
return neighbourhood
A = np.array([[18, 13, 8, 10, 9],
[ 8, 15, 6, 5, 6],
[ 3, 0, 2, 14, 13],
[ 4, 4, 5, 19, 0],
[16, 8, 7, 7, 11]])
print(A)
An = queen_neighbourhood(A, 0, 0, 2)
print(An)
which gives,
[[11 16 8]
[ 9 18 13]
[ 6 8 15]]
I need to create 2D np array by the known shape and parameters in this way:
import numpy as np
rows = 9
cols = 8
xrows = 3
xcols = 2
the wanted results:
([[ 1, 1, 4, 4, 7, 7, 10, 10],
[ 1, 1, 4, 4, 7, 7, 10, 10],
[ 1, 1, 4, 4, 7, 7, 10, 10],
[ 2, 2, 5, 5, 8, 8, 11, 11],
[ 2, 2, 5, 5, 8, 8, 11, 11],
[ 2, 2, 5, 5, 8, 8, 11, 11],
[ 3, 3, 6, 6, 9, 9, 12, 12],
[ 3, 3, 6, 6, 9, 9, 12, 12],
[ 3, 3, 6, 6, 9, 9, 12, 12]])
rows%xrows = 0
cols%xcols = 0
rows is the number of rows of the array, cols is the number of columns of the array, xrows is the number of rows in each slice, xcols is the number of column in each slice.
the answer should be general not for this parameters
IIUC, you can use:
((np.arange(rows//xrows*cols//xcols, dtype=int)+1)
.reshape((rows//xrows,cols//xcols), order='F')
.repeat(xrows,0)
.repeat(xcols,1)
)
Output:
array([[ 1, 1, 4, 4, 7, 7, 10, 10],
[ 1, 1, 4, 4, 7, 7, 10, 10],
[ 1, 1, 4, 4, 7, 7, 10, 10],
[ 2, 2, 5, 5, 8, 8, 11, 11],
[ 2, 2, 5, 5, 8, 8, 11, 11],
[ 2, 2, 5, 5, 8, 8, 11, 11],
[ 3, 3, 6, 6, 9, 9, 12, 12],
[ 3, 3, 6, 6, 9, 9, 12, 12],
[ 3, 3, 6, 6, 9, 9, 12, 12]])
I have a list(T) of 6500 images(arrays) that I am using for image classification, and I would like to see how increasing the data affects the accuracy.
So, starting from n=2000 images, I am thinking of having a loop that will add 500(n+=500) images at each iteration till it reaches 6500 and therefore compare the accuracy between 2000, 2500, 3000, ... 6500. I have simplified the problem below by having a list of 20 elements.
lst = [1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9,0]
My second list (slist) contains the first 9 elements of the first list (lst).
I am trying to add 2 values to slist at each iteration, starting from lst[9:]. I know rather than using append, extend should be used to add multiple values at once. However, I couldn't find a way to do it.
In the following code, one element is added to slist (from lst) at each loop.
lst = [1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9,0]
slist = lst[:9]
for i in lst[9:]:
slist.append(i)
How can I add 2 or 3 elements simultaneously at each loop? An example output would be:
[1,2,3,4,5,6,7,8,9,0,1]
[1,2,3,4,5,6,7,8,9,0,1,2,3]
[1,2,3,4,5,6,7,8,9,0,1,2,3,4,5]
You could try using extend:
l = [1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9,0]
slist = l[:9]
for i in l[9:][::2]:
if i == l[9]:
slist.extend(l[9+i: 9+i+1])
else:
slist.extend(l[9+i-1: 9+i+1])
print(slist)
Output:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 0]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0]
lst=[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,
16,17,18,19,20]
iter = 9
while True:
print(lst[:iter])
iter+=2
if len(lst) <= iter:
print(lst[:iter])
break
This code does the job
lst=[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,
16,17,18,19,21]
slist= lst[:9]
s,f=0,2
while True:
slist.extend(lst[9:][s:f])
print(slist)
s+=2
f+=2
if len(slist) >= len(lst):
break
It prints out:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21]
I have created the array from a csv using pandas and numpy.
This is my code that convert 2D csv to 3D array:
>>> import pandas as pd
>>> import numpy as npp
>>> df = pd.read_csv("test.csv")
>>> df_mat = df.values
>>> seq_len = 3
>>> data=[]
>>> for index in range(len(df_mat) - seq_len):
... data.append(df_mat[index: index + seq_len])
...
>>> data = np.array(data)
>>> data.shape
(4, 3, 9)
The csv is used is:
input1,input2,input3,input4,input5,input6,input7,input8,output
1,2,3,4,5,6,7,8,1
2,3,4,5,6,7,8,9,0
3,4,5,6,7,8,9,10,-1
4,5,6,7,8,9,10,11,-1
5,6,7,8,9,10,11,12,1
6,7,8,9,10,11,12,13,0
7,8,9,10,11,12,13,14,1
Now I want to get the 3D array back to 2D array format.
Kindly, let me know how I can I do that. Not getting any clue.
Slice on the 0th rows along each each block until the last block and stack with the last one -
np.vstack((data[np.arange(data.shape[0]-1),0],data[-1]))
Output with given sample data -
In [24]: np.vstack((data[np.arange(data.shape[0]-1),0],data[-1]))
Out[24]:
array([[ 1, 2, 3, 4, 5, 6, 7, 8, 1],
[ 2, 3, 4, 5, 6, 7, 8, 9, 0],
[ 3, 4, 5, 6, 7, 8, 9, 10, -1],
[ 4, 5, 6, 7, 8, 9, 10, 11, -1],
[ 5, 6, 7, 8, 9, 10, 11, 12, 1],
[ 6, 7, 8, 9, 10, 11, 12, 13, 0],
[ 7, 8, 9, 10, 11, 12, 13, 14, 1]], dtype=int64)
Or slice 0th rows across all blocks and stack with the last block skipping the first row -
In [28]: np.vstack((data[np.arange(data.shape[0]),0],data[-1,1:]))
Out[28]:
array([[ 1, 2, 3, 4, 5, 6, 7, 8, 1],
[ 2, 3, 4, 5, 6, 7, 8, 9, 0],
[ 3, 4, 5, 6, 7, 8, 9, 10, -1],
[ 4, 5, 6, 7, 8, 9, 10, 11, -1],
[ 5, 6, 7, 8, 9, 10, 11, 12, 1],
[ 6, 7, 8, 9, 10, 11, 12, 13, 0],
[ 7, 8, 9, 10, 11, 12, 13, 14, 1]], dtype=int64)
I am trying to write a function that would create a regular grid of 5 pixels by 5 pixels inside a 2d array. I was hoping some combination of numpy.arange and numpy.repeat might do it, but so far I haven't had any luck because numpy.repeat will just repeat along the same row.
Here is an example:
Let's say I want a 5x5 grid inside a 2d array of shape (20, 15). It should look like:
array([[ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
[ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
[ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
[ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
[ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
[ 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
[ 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
[ 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
[ 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
[ 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
[ 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8],
[ 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8],
[ 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8],
[ 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8],
[ 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8],
[ 9, 9, 9, 9, 9,10,10,10,10,10,11,11,11,11,11],
[ 9, 9, 9, 9, 9,10,10,10,10,10,11,11,11,11,11],
[ 9, 9, 9, 9, 9,10,10,10,10,10,11,11,11,11,11],
[ 9, 9, 9, 9, 9,10,10,10,10,10,11,11,11,11,11],
[ 9, 9, 9, 9, 9,10,10,10,10,10,11,11,11,11,11]])
I realize I could simply use a loop and slicing to accomplish this, but I could be applying this to very large arrays and I worry that the performance of that would be too slow or impractical.
Can anyone recommend a method to accomplish this?
Thanks in advance.
UPDATE:
All the answers provided seem to work well. Can anyone tell me which will be the most efficient to use for large arrays? By large array I mean it could be 100000 x 100000 or more with 15 x 15 grid cell sizes.
Broadcasting is the answer here:
m, n, d = 20, 15, 5
arr = np.empty((m, n), dtype=np.int)
arr_view = arr.reshape(m // d, d, n // d, d)
vals = np.arange(m // d * n // d).reshape(m // d, 1, n // d, 1)
arr_view[:] = vals
>>> arr
array([[ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
[ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
[ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
[ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
[ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
[ 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
[ 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
[ 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
[ 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
[ 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
[ 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8],
[ 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8],
[ 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8],
[ 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8],
[ 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8],
[ 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11],
[ 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11],
[ 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11],
[ 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11],
[ 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11]])
Similar to Jaime's answer:
np.repeat(np.arange(0, 10, 3), 4)[..., None] + np.repeat(np.arange(3), 5)[None, ...]
kron will do this expansion (as Brionius also suggested in the comments):
xi, xj, ni, nj = 5, 5, 4, 3
r = np.kron(np.arange(ni*nj).reshape((ni,nj)), np.ones((xi, xj)))
Although I haven't tested it, I assume it's less efficient than the broadcasting approach, but a bit more concise and easier to understand (I hope). It's likely less efficient because: 1) it requires the array of ones, 2) it does xi*xj multiplications by 1, and 3) it does a bunch of concats.