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I have the following back and white image
import numpy as np
thresh = np.array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1]]).astype('uint8')
I am trying to find contour in the thresh image, like this
import cv2
contours, hierarchy = cv2.findContours(
thresh,
cv2.RETR_CCOMP,
cv2.CHAIN_APPROX_SIMPLE
)
Just looking at the threshold image, it's intuitive that there is 1 big contour around the 1, that is shaped like an arrow.
However, visually inspection of the returned contour from cv2 found using
canvas = np.zeros_like(thresh)
for ct in contours:
cv2.drawContours(canvas, ct,-1, 1, 1)
yeilds the following;
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1],
[0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1]],
dtype=uint8)
In summary, how would be best get the contour from thresh image?
The result you obtained is correct.
Contours allow you to find the points along the boundary of any shape. The third parameter in cv2.findContours lets you decide how you want to store the boundary points. You have 2 ways of doing that (1) either store all the points OR (2) find a good approximation.
In your case, you are using the flag cv2.CHAIN_APPROX_SIMPLE. This method does not store all the boundary points of the shape. For every line along the boundary, it stores just 2 points (the ends of each line). This is the best way to approximate the shape of any contour and it also saves memory.
If you want to store all the boundary points you need to use cv2.CHAIN_APPROX_NONE.
Here's the documentation for more
Illustration:
Consider the following array as input:
thresh = np.array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]],
dtype=uint8)
Using the flag cv2.CHAIN_APPROX_SIMPLE you would get:
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]],
dtype=uint8)
And using the flag cv2.CHAIN_APPROX_NONE you would get:
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]],
dtype=uint8)
I have an array called predictions
array([0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0,
1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1,
1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1,
1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1,
1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1,
0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1,
1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0,
0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0,
1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1,
0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0,
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0,
0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0,
1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0,
0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0,
1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1,
0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0],
dtype=int64)
I try to convert this array to dataframe column but have an error mentioned in the title. I try to convert elements of ndarray to an integer, but none of this works.
Xtest["Survived"] = [int(x) for x in predictions]
Xtest["Survived"] = [x.astype(int) for x in predictions]
Xtest["Survived"] = pd.Series(predictions)
Also when I type
type(predictions[1])
I got
numpy.int64
As solved in the comments, the error is thrown because Xdata was an ndarray and not a DataFrame. An ndarray can't be indexed with a string ("Survived"), thus the error message about indeces. The problem is unrelated to the predictions data. If Xtest is a DataFrame, any of the lines above should work. They could be simplified to
Xtest["Survived"] = predictions
Consider a 3 x 3 x 3 cube where each of the 27 elements is connected to other elements along faces. A cube-shaped element has 6 sides, thus a maximum of 6 connections is possible per element (for example, the center-most element in a 3 x 3 x 3 cube is bounded by 6 elements, and has 6 connections).
Then, let m1, m2, and m3 be the first, second, and third layers of the cube respectively. The name of each element is xyz, where x, y, z are the row number, column number, and layer number of the element. For example, the element 213 is in the second row, first column, and 3rd layer of the cube. This element is connected to 4 other elements: three are in its layer (113, 313, 223), and one is one layer above it (212).
x = 3 # nrow
y = 3 # ncol
z = 3 # nlay
# print each layer as a 2D matrix
for(k in 1:z){
m = paste0(rep(1:x, each=x), rep(1:y, times = y), k)
print(matrix(m, nrow=x, byrow=T))
}
[,1] [,2] [,3]
[1,] "111" "121" "131"
[2,] "211" "221" "231"
[3,] "311" "321" "331"
[,1] [,2] [,3]
[1,] "112" "122" "132"
[2,] "212" "222" "232"
[3,] "312" "322" "332"
[,1] [,2] [,3]
[1,] "113" "123" "133"
[2,] "213" "223" "233"
[3,] "313" "323" "333"
Is there an out-of-the-box function in igraph or a related package for creating either an adjacency matrix OR an edge list for a network like this? I need a solution that scales to any number of rows, columns, and layers. Python solutions are welcome.
I manually created the 2D adjacency matrix, where the rows and columns are given by c(m1, m2, m3) below:
m1 = paste0(rep(1:x, each=x), rep(1:y, times = y), 1)
m2 = paste0(rep(1:x, each=x), rep(1:y, times = y), 2)
m3 = paste0(rep(1:x, each=x), rep(1:y, times = y), 3)
c(m1, m2, m3)
[1] "111" "121" "131" "211" "221" "231" "311" "321" "331" "112" "122" "132" "212" "222" "232" "312" "322" "332"
[19] "113" "123" "133" "213" "223" "233" "313" "323" "333"
For this simple example the adjacency matrix is sparse, has 0's along the diagonal, and is symmetric. It looks like this:
And here's a dput() to C&P and validate with.
dput(temp)
structure(c(0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0,
1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0,
1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0,
1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0,
1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0,
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0,
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0), .Dim = c(27L,
27L), .Dimnames = list(c("111", "121", "131", "211", "221", "231",
"311", "321", "331", "112", "122", "132", "212", "222", "232",
"312", "322", "332", "113", "123", "133", "213", "223", "233",
"313", "323", "333"), c("111", "121", "131", "211", "221", "231",
"311", "321", "331", "112", "122", "132", "212", "222", "232",
"312", "322", "332", "113", "123", "133", "213", "223", "233",
"313", "323", "333")))
There's an edge when the Manhattan distance between the nodes is 1, so you can use dist() in R to create the adjacency matrix:
cube_mat = expand.grid(
x = 1:3,
y = 1:3,
z = 1:3
)
m_dist = as.matrix(dist(cube_mat[, 1:3], method = "manhattan", diag = TRUE))
# Zero out any distances != 1
m_dist[m_dist != 1] = 0
rownames(m_dist) = paste0(cube_mat$x, cube_mat$y, cube_mat$z)
colnames(m_dist) = paste0(cube_mat$x, cube_mat$y, cube_mat$z)
# Plot of the adjacency matrix (looks reversed because 111 is in the bottom left):
image(m_dist)
If you want to just use a package funtion from igraph:
#adj <- my.adjacency.matrix
as_edgelist(graph.adjacency(adj))
In general you can use the functions in the igraph package to go between edgelists, adjacency matrices, and also produce graphs using plot.igraph. Here's the default cube:
plot.igraph(graph.adjacency(adj))
I am still coding a fingerprint image preprocessor on Python. I see in MATLAB there is a special function to remove H breaks and spurs:
bwmorph(a , 'hbreak')
bwmorph(a , 'spur')
I have searched scikit, OpenCV and others but couldn't find an equivalent for these two use of bwmorph. Can anybody point me to right direction or do i have to implement my own?
Edit October 2017
the skimage module now has at least 2 options:
skeletonize and thin
Example with comparison
from skimage.morphology import thin, skeletonize
import numpy as np
import matplotlib.pyplot as plt
square = np.zeros((7, 7), dtype=np.uint8)
square[1:-1, 2:-2] = 1
square[0, 1] = 1
thinned = thin(square)
skel = skeletonize(square)
f, ax = plt.subplots(2, 2)
ax[0,0].imshow(square)
ax[0,0].set_title('original')
ax[0,0].get_xaxis().set_visible(False)
ax[0,1].axis('off')
ax[1,0].imshow(thinned)
ax[1,0].set_title('morphology.thin')
ax[1,1].imshow(skel)
ax[1,1].set_title('morphology.skeletonize')
plt.show()
Original post
I have found this solution by joefutrelle on github.
It seems (visually) to give similar results as the Matlab version.
Hope that helps!
Edit:
As it was pointed out in the comments, I'll extend my initial post as the mentioned link might change:
Looking for a substitute in Python for bwmorph from Matlab I stumbled upon the following code from joefutrelle on Github (at the end of this post as it's very long).
I have figured out two ways to implement this into my script (I'm a beginner and I'm sure there are better ways!):
1) copy the whole code into your script and then call the function (but this makes the script harder to read)
2) copy the code it in a new python file 'foo' and save it. Now copy it in the Python\Lib (eg. C:\Program Files\Python35\Lib) folder. In your original script you can call the function by writing:
from foo import bwmorph_thin
Then you'll feed the function with your binary image:
skeleton = bwmorph_thin(foo_image, n_iter = math.inf)
import numpy as np
from scipy import ndimage as ndi
# lookup tables for bwmorph_thin
G123_LUT = np.array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0,
1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0,
0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1,
0, 0, 0], dtype=np.bool)
G123P_LUT = np.array([0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0,
1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0,
0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0,
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1,
0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0], dtype=np.bool)
def bwmorph_thin(image, n_iter=None):
"""
Perform morphological thinning of a binary image
Parameters
----------
image : binary (M, N) ndarray
The image to be thinned.
n_iter : int, number of iterations, optional
Regardless of the value of this parameter, the thinned image
is returned immediately if an iteration produces no change.
If this parameter is specified it thus sets an upper bound on
the number of iterations performed.
Returns
-------
out : ndarray of bools
Thinned image.
See also
--------
skeletonize
Notes
-----
This algorithm [1]_ works by making multiple passes over the image,
removing pixels matching a set of criteria designed to thin
connected regions while preserving eight-connected components and
2 x 2 squares [2]_. In each of the two sub-iterations the algorithm
correlates the intermediate skeleton image with a neighborhood mask,
then looks up each neighborhood in a lookup table indicating whether
the central pixel should be deleted in that sub-iteration.
References
----------
.. [1] Z. Guo and R. W. Hall, "Parallel thinning with
two-subiteration algorithms," Comm. ACM, vol. 32, no. 3,
pp. 359-373, 1989.
.. [2] Lam, L., Seong-Whan Lee, and Ching Y. Suen, "Thinning
Methodologies-A Comprehensive Survey," IEEE Transactions on
Pattern Analysis and Machine Intelligence, Vol 14, No. 9,
September 1992, p. 879
Examples
--------
>>> square = np.zeros((7, 7), dtype=np.uint8)
>>> square[1:-1, 2:-2] = 1
>>> square[0,1] = 1
>>> square
array([[0, 1, 0, 0, 0, 0, 0],
[0, 0, 1, 1, 1, 0, 0],
[0, 0, 1, 1, 1, 0, 0],
[0, 0, 1, 1, 1, 0, 0],
[0, 0, 1, 1, 1, 0, 0],
[0, 0, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 0, 0]], dtype=uint8)
>>> skel = bwmorph_thin(square)
>>> skel.astype(np.uint8)
array([[0, 1, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 0],
[0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0]], dtype=uint8)
"""
# check parameters
if n_iter is None:
n = -1
elif n_iter <= 0:
raise ValueError('n_iter must be > 0')
else:
n = n_iter
# check that we have a 2d binary image, and convert it
# to uint8
skel = np.array(image).astype(np.uint8)
if skel.ndim != 2:
raise ValueError('2D array required')
if not np.all(np.in1d(image.flat,(0,1))):
raise ValueError('Image contains values other than 0 and 1')
# neighborhood mask
mask = np.array([[ 8, 4, 2],
[16, 0, 1],
[32, 64,128]],dtype=np.uint8)
# iterate either 1) indefinitely or 2) up to iteration limit
while n != 0:
before = np.sum(skel) # count points before thinning
# for each subiteration
for lut in [G123_LUT, G123P_LUT]:
# correlate image with neighborhood mask
N = ndi.correlate(skel, mask, mode='constant')
# take deletion decision from this subiteration's LUT
D = np.take(lut, N)
# perform deletion
skel[D] = 0
after = np.sum(skel) # coint points after thinning
if before == after:
# iteration had no effect: finish
break
# count down to iteration limit (or endlessly negative)
n -= 1
return skel.astype(np.bool)
"""
# here's how to make the LUTs
def nabe(n):
return np.array([n>>i&1 for i in range(0,9)]).astype(np.bool)
def hood(n):
return np.take(nabe(n), np.array([[3, 2, 1],
[4, 8, 0],
[5, 6, 7]]))
def G1(n):
s = 0
bits = nabe(n)
for i in (0,2,4,6):
if not(bits[i]) and (bits[i+1] or bits[(i+2) % 8]):
s += 1
return s==1
g1_lut = np.array([G1(n) for n in range(256)])
def G2(n):
n1, n2 = 0, 0
bits = nabe(n)
for k in (1,3,5,7):
if bits[k] or bits[k-1]:
n1 += 1
if bits[k] or bits[(k+1) % 8]:
n2 += 1
return min(n1,n2) in [2,3]
g2_lut = np.array([G2(n) for n in range(256)])
g12_lut = g1_lut & g2_lut
def G3(n):
bits = nabe(n)
return not((bits[1] or bits[2] or not(bits[7])) and bits[0])
def G3p(n):
bits = nabe(n)
return not((bits[5] or bits[6] or not(bits[3])) and bits[4])
g3_lut = np.array([G3(n) for n in range(256)])
g3p_lut = np.array([G3p(n) for n in range(256)])
g123_lut = g12_lut & g3_lut
g123p_lut = g12_lut & g3p_lut
"""`
You will have to implement those on your own since they aren't present in OpenCV or skimage as far as I know.
However, it should be straightforward to check MATLAB's code on how it works and write your own version in Python/NumPy.
Here is a guide describing in detail NumPy functions exclusively for MATLAB users, with hints on equivalent functions in MATLAB and NumPy:
Link
I'm writing a script that I'm going to use quite often, with datasets of different sizes, and I have to do some comparisons that I just can't get straight in Python.
There will be multiple lists (around 20 or more, but I've reduced them to three for example and testing purposes), all with the same number of integer items in a certain order. I want to compare items on the same position in every list to find differences.
For a defined number of lists, this is easy:
a = [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0]
b = [0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 4, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0]
c = [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 2, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
for x,y,z in zip(a,b,c):
if x != y != z:
print x, y, z
I've tried wrapping that loop in a function, so the number of arguments can vary, but there I got stuck.
def compare(*args):
for x in zip(args):
???
In the final script I will have not multiple single lists, but all together in one list of list. Would that help? If I loop through a list of lists, I won't get every list at once...
Forget the function, it's not really useful anyway as it will be part of a bigger script and it's too difficult defining the different arguments.
I'm now comparing two lists at a time, saving those that are identical. That way, I can later easily remove all those from my whole list and keep only the unique ones.
l_o_l = [[0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 4, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 2, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
for i in range(0, (len(l_o_l)-1)):
for j in range((i+1), len(l_o_l)):
if l_o_l[i] == l_o_l[j]:
duplicates.append(key_list[i])
duplicates.append(key_list[j])
dup = list(set(duplicates))
uniques = [x for x in key_list if x not in dup]
where the key_list contains, from a dictionary, identifiers for my lists.
Any suggestions for improvement?
Maybe something like this
def compare(*args):
for things in zip(*args):
yield all(x == things[0] for x in things)
You can then use it like this
a = range(10)
b = range(10)
c = range(10)
d = range(11, 20)
for match in compare(a,b,c):
print match
for match in compare(a,b,c,d):
print match
Here is a demo using your example (its a generator, so you have to iter over it or exhaust it using list)
a = [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0]
b = [0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 4, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0]
c = [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 2, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
print list(compare(a,b,c))
def compare(*args):
for x in zip(args):
values_list = list(x[0]) # x[0] because x is a tuple
different_values = set(values_list) # a set does not contain identical values
if len(different_values) != 1: # if you have more than 1 value you have different values in your list
print 'different values', values_list
gives you
a = [0, 0, 1]
b = [0, 1, 1]
c = [1, 1, 1]
compare(a, b, c)
>>> different values [0, 0, 1]
>>> different values [0, 1, 1]
Assuming the lists are similar to the ones in the example, I would use:
def compare(*args):
for x in zip(args):
if min(x) != max(x):
print x
def compare(elements):
return len(set(elements)) == bool(elements)
If you want to know whether all the lists are the same you can simply do:
all(compare(elements) for elements in zip(the_lists))
An alternative could be to transform the lists into tuples and use set there:
len(set(tuple(the_list) for the_list in the_lists) == bool(the_lists)
If you simply want to remove duplicates this should be faster:
the_lists = [list(elem) for elem in set(tuple(the_list) for the_list in the_lists)]
Example usage:
>>> a = range(100)
>>> b = range(100, 200)
>>> c = range(200, 300)
>>> d = a[:]
>>> e = b[:]
>>> the_lists = [a,b,c,d,e]
>>> the_lists2 = [list(elem) for elem in set(tuple(the_list) for the_list in the_lists)]
>>> [a,b,c] == sorted(the_lists2) #order is not maintained by set
True
It seems to be pretty fast:
>>> timeit.timeit('[list(elem) for elem in set(tuple(the_list) for the_list in the_lists)]', 'from __main__ import the_lists', number=1000000)
7.949447154998779
Less than 8 seconds for executing 1 million times. (Where the_lists is the same used before.)
Edit:
If you want to remove only the duplicated list then the simplest algorithm I can think of is sorting the list-of-lists and using itertools.groupby:
>>> a = range(100)
>>> b = range(100,200)
>>> c = range(200,300)
>>> d = a[:]
>>> e = b[:]
>>> the_lists = [a,b,c,d,e]
>>> the_lists.sort()
>>> import itertools as it
>>> for key, group in it.groupby(the_lists):
... if len(list(group)) == 1:
... print key
...
[200, 201, 202, ..., 297, 298, 299]
I think trying to get clever with *args and zip is just confusing the issue. I would write it something like this:
def compare(list_of_lists):
# assuming not an empty data set
inner_len = len(list_of_lists[0])
for index in range(inner_len):
expected = list_of_lists[0][index]
for inner_list in list_of_lists:
if inner_list[index] != expected:
# report difference at this index