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I wish to draw lines on a square graph.
The scales of x-axis and y-axis should be the same.
e.g. x ranges from 0 to 10 and it is 10cm on the screen. y has to also range from 0 to 10 and has to be also 10 cm.
The square shape has to be maintained, even if I mess around with the window size.
Currently, my graph scales together with the window size.
How may I achieve this?
UPDATE:
I tried the following, but it did not work.
plt.xlim(-3, 3)
plt.ylim(-3, 3)
plt.axis("equal")
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How to equalize the scales of x-axis and y-axis in matplotlib? around: Questions
Removing white space around a saved image in matplotlib
2 answers
I need to take an image and save it after some process. The figure looks fine when I display it, but after saving the figure, I got some white space around the saved image. I have tried the "tight" option for savefig method, did not work either. The code:
import matplotlib.image as mpimg
import matplotlib.pyplot as plt
fig = plt.figure(1)
img = mpimg.imread(path)
plt.imshow(img)
ax=fig.add_subplot(1,1,1)
extent = ax.get_window_extent().transformed(fig.dpi_scale_trans.inverted())
plt.savefig("1.png", bbox_inches=extent)
plt.axis("off")
plt.show()
I am trying to draw a basic graph by using NetworkX on a figure and save it. I realized that without a graph it works, but when added a graph I get white space around the saved image;
import matplotlib.image as mpimg
import matplotlib.pyplot as plt
import networkx as nx
G = nx.Graph()
G.add_node(1)
G.add_node(2)
G.add_node(3)
G.add_edge(1,3)
G.add_edge(1,2)
pos = {1:[100,120], 2:[200,300], 3:[50,75]}
fig = plt.figure(1)
img = mpimg.imread("image.jpg")
plt.imshow(img)
ax=fig.add_subplot(1,1,1)
nx.draw(G, pos=pos)
extent = ax.get_window_extent().transformed(fig.dpi_scale_trans.inverted())
plt.savefig("1.png", bbox_inches = extent)
plt.axis("off")
plt.show()
Answer #1
You can remove the white space padding by setting bbox_inches="tight" in savefig:
plt.savefig("test.png",bbox_inches="tight")
You"ll have to put the argument to bbox_inches as a string, perhaps this is why it didn"t work earlier for you.
Possible duplicates:
Matplotlib plots: removing axis, legends and white spaces
Answer #2
I cannot claim I know exactly why or how my “solution” works, but this is what I had to do when I wanted to plot the outline of a couple of aerofoil sections — without white margins — to a PDF file. (Note that I used matplotlib inside an IPython notebook, with the -pylab flag.)
plt.gca().set_axis_off()
plt.subplots_adjust(top = 1, bottom = 0, right = 1, left = 0,
hspace = 0, wspace = 0)
plt.margins(0,0)
plt.gca().xaxis.set_major_locator(plt.NullLocator())
plt.gca().yaxis.set_major_locator(plt.NullLocator())
plt.savefig("filename.pdf", bbox_inches = "tight",
pad_inches = 0)
I have tried to deactivate different parts of this, but this always lead to a white margin somewhere. You may even have modify this to keep fat lines near the limits of the figure from being shaved by the lack of margins.
How to remove convexity defects in a Sudoku square?
5 answers
I was doing a fun project: Solving a Sudoku from an input image using OpenCV (as in Google goggles etc). And I have completed the task, but at the end I found a little problem for which I came here.
I did the programming using Python API of OpenCV 2.3.1.
Below is what I did :
- Read the image
- Find the contours
- Select the one with maximum area, ( and also somewhat equivalent to square).
Find the corner points.
e.g. given below:
(Notice here that the green line correctly coincides with the true boundary of the Sudoku, so the Sudoku can be correctly warped. Check next image)
warp the image to a perfect square
eg image:
Perform OCR ( for which I used the method I have given in Simple Digit Recognition OCR in OpenCV-Python )
And the method worked well.
Problem:
Check out this image.
Performing the step 4 on this image gives the result below:
The red line drawn is the original contour which is the true outline of sudoku boundary.
The green line drawn is approximated contour which will be the outline of warped image.
Which of course, there is difference between green line and red line at the top edge of sudoku. So while warping, I am not getting the original boundary of the Sudoku.
My Question :
How can I warp the image on the correct boundary of the Sudoku, i.e. the red line OR how can I remove the difference between red line and green line? Is there any method for this in OpenCV?
Answer #1
I have a solution that works, but you"ll have to translate it to OpenCV yourself. It"s written in Mathematica.
The first step is to adjust the brightness in the image, by dividing each pixel with the result of a closing operation:
src = ColorConvert[Import["http://davemark.com/images/sudoku.jpg"], "Grayscale"];
white = Closing[src, DiskMatrix[5]];
srcAdjusted = Image[ImageData[src]/ImageData[white]]
The next step is to find the sudoku area, so I can ignore (mask out) the background. For that, I use connected component analysis, and select the component that"s got the largest convex area:
components =
ComponentMeasurements[
[email protected][srcAdjusted], {"ConvexArea", "Mask"}][[All,
2]];
largestComponent = Image[SortBy[components, First][[-1, 2]]]
By filling this image, I get a mask for the sudoku grid:
mask = FillingTransform[largestComponent]
Now, I can use a 2nd order derivative filter to find the vertical and horizontal lines in two separate images:
lY = ImageMultiply[MorphologicalBinarize[GaussianFilter[srcAdjusted, 3, {2, 0}], {0.02, 0.05}], mask];
lX = ImageMultiply[MorphologicalBinarize[GaussianFilter[srcAdjusted, 3, {0, 2}], {0.02, 0.05}], mask];
I use connected component analysis again to extract the grid lines from these images. The grid lines are much longer than the digits, so I can use caliper length to select only the grid lines-connected components. Sorting them by position, I get 2x10 mask images for each of the vertical/horizontal grid lines in the image:
verticalGridLineMasks =
SortBy[ComponentMeasurements[
lX, {"CaliperLength", "Centroid", "Mask"}, # > 100 &][[All,
2]], #[[2, 1]] &][[All, 3]];
horizontalGridLineMasks =
SortBy[ComponentMeasurements[
lY, {"CaliperLength", "Centroid", "Mask"}, # > 100 &][[All,
2]], #[[2, 2]] &][[All, 3]];
Next I take each pair of vertical/horizontal grid lines, dilate them, calculate the pixel-by-pixel intersection, and calculate the center of the result. These points are the grid line intersections:
centerOfGravity[l_] :=
ComponentMeasurements[Image[l], "Centroid"][[1, 2]]
gridCenters =
Table[centerOfGravity[
ImageData[Dilation[Image[h], DiskMatrix[2]]]*
ImageData[Dilation[Image[v], DiskMatrix[2]]]], {h,
horizontalGridLineMasks}, {v, verticalGridLineMasks}];
The last step is to define two interpolation functions for X/Y mapping through these points, and transform the image using these functions:
fnX = ListInterpolation[gridCenters[[All, All, 1]]];
fnY = ListInterpolation[gridCenters[[All, All, 2]]];
transformed =
ImageTransformation[
srcAdjusted, {fnX @@ Reverse[#], fnY @@ Reverse[#]} &, {9*50, 9*50},
PlotRange -> {{1, 10}, {1, 10}}, DataRange -> Full]
All of the operations are basic image processing function, so this should be possible in OpenCV, too. The spline-based image transformation might be harder, but I don"t think you really need it. Probably using the perspective transformation you use now on each individual cell will give good enough results.
Answer #2
Nikie"s answer solved my problem, but his answer was in Mathematica. So I thought I should give its OpenCV adaptation here. But after implementing I could see that OpenCV code is much bigger than nikie"s mathematica code. And also, I couldn"t find interpolation method done by nikie in OpenCV ( although it can be done using scipy, i will tell it when time comes.)
1. Image PreProcessing ( closing operation )
import cv2
import numpy as np
img = cv2.imread("dave.jpg")
img = cv2.GaussianBlur(img,(5,5),0)
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
mask = np.zeros((gray.shape),np.uint8)
kernel1 = cv2.getStructuringElement(cv2.MORPH_ELLIPSE,(11,11))
close = cv2.morphologyEx(gray,cv2.MORPH_CLOSE,kernel1)
div = np.float32(gray)/(close)
res = np.uint8(cv2.normalize(div,div,0,255,cv2.NORM_MINMAX))
res2 = cv2.cvtColor(res,cv2.COLOR_GRAY2BGR)
Result :
2. Finding Sudoku Square and Creating Mask Image
thresh = cv2.adaptiveThreshold(res,255,0,1,19,2)
contour,hier = cv2.findContours(thresh,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)
max_area = 0
best_cnt = None
for cnt in contour:
area = cv2.contourArea(cnt)
if area > 1000:
if area > max_area:
max_area = area
best_cnt = cnt
cv2.drawContours(mask,[best_cnt],0,255,-1)
cv2.drawContours(mask,[best_cnt],0,0,2)
res = cv2.bitwise_and(res,mask)
Result :
3. Finding Vertical lines
kernelx = cv2.getStructuringElement(cv2.MORPH_RECT,(2,10))
dx = cv2.Sobel(res,cv2.CV_16S,1,0)
dx = cv2.convertScaleAbs(dx)
cv2.normalize(dx,dx,0,255,cv2.NORM_MINMAX)
ret,close = cv2.threshold(dx,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,kernelx,iterations = 1)
contour, hier = cv2.findContours(close,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
for cnt in contour:
x,y,w,h = cv2.boundingRect(cnt)
if h/w > 5:
cv2.drawContours(close,[cnt],0,255,-1)
else:
cv2.drawContours(close,[cnt],0,0,-1)
close = cv2.morphologyEx(close,cv2.MORPH_CLOSE,None,iterations = 2)
closex = close.copy()
Result :
4. Finding Horizontal Lines
kernely = cv2.getStructuringElement(cv2.MORPH_RECT,(10,2))
dy = cv2.Sobel(res,cv2.CV_16S,0,2)
dy = cv2.convertScaleAbs(dy)
cv2.normalize(dy,dy,0,255,cv2.NORM_MINMAX)
ret,close = cv2.threshold(dy,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,kernely)
contour, hier = cv2.findContours(close,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
for cnt in contour:
x,y,w,h = cv2.boundingRect(cnt)
if w/h > 5:
cv2.drawContours(close,[cnt],0,255,-1)
else:
cv2.drawContours(close,[cnt],0,0,-1)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,None,iterations = 2)
closey = close.copy()
Result :
Of course, this one is not so good.
5. Finding Grid Points
res = cv2.bitwise_and(closex,closey)
Result :
6. Correcting the defects
Here, nikie does some kind of interpolation, about which I don"t have much knowledge. And i couldn"t find any corresponding function for this OpenCV. (may be it is there, i don"t know).
Check out this SOF which explains how to do this using SciPy, which I don"t want to use : Image transformation in OpenCV
So, here I took 4 corners of each sub-square and applied warp Perspective to each.
For that, first we find the centroids.
contour, hier = cv2.findContours(res,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)
centroids = []
for cnt in contour:
mom = cv2.moments(cnt)
(x,y) = int(mom["m10"]/mom["m00"]), int(mom["m01"]/mom["m00"])
cv2.circle(img,(x,y),4,(0,255,0),-1)
centroids.append((x,y))
But resulting centroids won"t be sorted. Check out below image to see their order:
So we sort them from left to right, top to bottom.
centroids = np.array(centroids,dtype = np.float32)
c = centroids.reshape((100,2))
c2 = c[np.argsort(c[:,1])]
b = np.vstack([c2[i*10:(i+1)*10][np.argsort(c2[i*10:(i+1)*10,0])] for i in xrange(10)])
bm = b.reshape((10,10,2))
Now see below their order :
Finally we apply the transformation and create a new image of size 450x450.
output = np.zeros((450,450,3),np.uint8)
for i,j in enumerate(b):
ri = i/10
ci = i%10
if ci != 9 and ri!=9:
src = bm[ri:ri+2, ci:ci+2 , :].reshape((4,2))
dst = np.array( [ [ci*50,ri*50],[(ci+1)*50-1,ri*50],[ci*50,(ri+1)*50-1],[(ci+1)*50-1,(ri+1)*50-1] ], np.float32)
retval = cv2.getPerspectiveTransform(src,dst)
warp = cv2.warpPerspective(res2,retval,(450,450))
output[ri*50:(ri+1)*50-1 , ci*50:(ci+1)*50-1] = warp[ri*50:(ri+1)*50-1 , ci*50:(ci+1)*50-1].copy()
Result :
The result is almost same as nikie"s, but code length is large. May be, better methods are available out there, but until then, this works OK.
Regards ARK.
Is there a library function for Root mean square error (RMSE) in python?
5 answers
I know I could implement a root mean squared error function like this:
def rmse(predictions, targets):
return np.sqrt(((predictions - targets) ** 2).mean())
What I"m looking for if this rmse function is implemented in a library somewhere, perhaps in scipy or scikit-learn?
Answer #1
sklearn >= 0.22.0
sklearn.metrics has a mean_squared_error function with a squared kwarg (defaults to True). Setting squared to False will return the RMSE.
from sklearn.metrics import mean_squared_error
rms = mean_squared_error(y_actual, y_predicted, squared=False)
sklearn < 0.22.0
sklearn.metrics has a mean_squared_error function. The RMSE is just the square root of whatever it returns.
from sklearn.metrics import mean_squared_error
from math import sqrt
rms = sqrt(mean_squared_error(y_actual, y_predicted))
