I am trying to get my profile picture to display as a circular view using swift 3. This is my code:
self.view.layoutIfNeeded()
self.profileImageView.image = image
self.profileImageView.layer.cornerRadius = self.profileImageView.frame.width/2.0
self.profileImageView.clipsToBounds = true
self.profileImageView.layer.masksToBounds = true
It works well on square images. But once the image is not square this doesn't display the image as circular. What do I need to do in order to get it to be display the imageview as a circle? Or is this feature only limited to square images?
Your code is making the corner radius half the width. This works fine when height == width (so radius also == height/2), but otherwise it won't work.
To fix this, add constraints to make your profileImageView square, then set the profileImageView.contentMode = .aspectFill.
Add self.view.layoutIfNeeded() line before you are set corner radius.
self.view.layoutIfNeeded()
self.profileImageView.layer.cornerRadius = self.profileImageView.frame.width/2.0
self.profileImageView.clipsToBounds = true
Related
I have the following sample of handwriting taken with three different writing instruments:
Looking at the writing, I can tell that there is a distinct difference between the first two and the last one. My goal is to determine an approximation of the stroke thickness for each letter, allowing me to group them based on being thin or thick.
So far, I have tried looking into stroke width transform, but I have struggled to translate it to my example.
I am able to preprocess the image such that I am just left with just the contours of the test in question. For example, here is thick from the last line:
I suggest detecting contours with cv::findContours as you are doing and then compare bounding rectangle area and contour area. The thicker writing the greater coefficent (contourArea/boundingRectArea) will be.
This approach will help you. This will calcuate the stroke width.
from skimage.feature import peak_local_max
from skimage import img_as_float
def adaptive_thresholding(image):
output_image = cv2.adaptiveThreshold(image,255,cv2.ADAPTIVE_THRESH_GAUSSIAN_C,cv2.THRESH_BINARY,21,2)
return output_image
def stroke_width(image):
dist = cv2.distanceTransform(cv2.subtract(255,image), cv2.DIST_L2, 5)
im = img_as_float(dist)
coordinates = peak_local_max(im, min_distance=15)
pixel_strength = []
for element in coordinates:
x = element[0]
y = element[1]
pixel_strength.append(np.asarray(dist)[x,y])
mean_pixel_strength = np.asarray(pixel_strength).mean()
return mean_pixel_strength
image = cv2.imread('Small3.JPG', 0)
process_image = adaptive_thresholding(image)
stroke_width(process_image)
A python implementation for this might go something like this, using Stroke Width Transform implementation of SWTloc.
Full Disclosure: I am the author of this library.
EDIT : Post v2.0.0
Transforming The Image
import swtloc as swt
imgpath = 'images/path_to_image.jpeg'
swtl = swt.SWTLocalizer(image_paths=imgpath)
swtImgObj = swtl.swtimages[0]
# Perform SWT Transformation with numba engine
swt_mat = swtImgObj.transformImage(auto_canny_sigma=1.0, gaussian_blurr=False,
minimum_stroke_width=3, maximum_stroke_width=50,
maximum_angle_deviation=np.pi/3)
Localize Letters
localized_letters = swtImgObj.localizeLetters()
Plot Histogram of Each Letters Strokes Widths
import seaborn as sns
import matplotlib.pyplot as plt
all_sws = []
for letter_label, letter in localized_letters.items():
all_sws.append(letter.stroke_widths_mean)
sns.displot(all_sws, bins=31)
From the distribution plot, it can be inferred that there might be three fontsize of the text available in the image - [3, 15, 27]
How to insert an image from file into a PlotWidget (plt1 = pg.PlotWidget(w);)?
The image is for a layout on which some calculated and plotted points should appear.
I tried to insert the image in a Qlabel behind the PlotWidget and make the PlotWidget transparent, but didn't work as transparency takes the color of the window not real transparency.
Thanks
Use QGraphicsPixmapItem:
plt1 = pg.PlotWidget(w)
img = pg.QtGui.QGraphicsPixmapItem(pg.QtGui.QPixmap(fileName))
plt1.addItem(img)
# depending on your preference, you probably want to invert the image:
img.scale(1, -1)
# OR invert the entire view:
plt.invertY(True)
I am trying to fit part of in image into a raphael object.
Scaling the image works perfectly, but when I try to translate it, it ends up returning the wrong part of the image.
I am scalling the image using "S1.5,1.5,0,0", that is, I am not scalling it around the middle point, so scalling it works beautifully.
But, as I try to offset the image, the resulting image fragment is offset.
Maybe there's another way to do it in Raphael.
What I am trying to accomplish is use a fragment of an image as an image object in Raphael and I need to copy a rectangle from an external image into it.
Something like:
copy original image fragment (x0 = 100, y0 = 120, width = 300, height = 250) to the image object, which has dimensions (width = 150 and 125).
I have been looking for an answer for some time, but nothing that really helps.
Edit:
The fiddle is
/w9XSf/12/
In the example above, I am grabbing a 100 x 60px area from the original image (which is 612 x 325px), and trying to display it on the output image, which is 500 x 300px.
The scale works, but the area it is grabbing is not the one I need.
It does work, if I grab from 0, 0.
But, as I move from the top left corner of the originsl image, the actual area it gives me is farther away from what I actually need :(.
Any ideas? (I have already tried swapping the order of the T and the S in the transform string).
Thanks.
Using Raphael, the following code creates a container, to be used to display an image, duly translated and scaled. A live version of the solution is also available at http://jsfiddle.net/s6DHf/. This is a forked version of the actual problem.
var outputW = 525,
outputH = 300;
sourceX = 100,
sourceY = 100,
scaleX = 1.5,
scaleY = 1.5,
paper = new Raphael("image", outputW, outputH),
bgImg = paper.image("http://cdn3.whatculture.com/wp-content/uploads/2013/04/MAN-OF-STEEL-e1365755036183.jpg", 0, 0, 350, 200)
.transform("t" + sourceX + "," + sourceY + "s" + scaleX +","+ scaleY + ",0,0");
Check the use of "s" and "t" (in lowercase), which denotes relative scaling and relative translation, respectively. The problem was due to the use of "S" and "T" (in uppercase), which is all about absolute scaling and translation, respectively.
Raphael reference: http://raphaeljs.com/reference.html#Element.transform
Hope this helps.
I have stitched two images together using OpenCV functions and C++. Now I am facing a problem that the final image contains a large black part.
The final image should be a rectangle containing the effective part.
My image is the following:
How can I remove the black section?
mevatron's answer is one way where amount of black region is minimised while retaining full image.
Another option is removing complete black region where you also loose some part of image, but result will be a neat looking rectangular image. Below is the Python code.
Here, you find three main corners of the image as below:
I have marked those values. (1,x2), (x1,1), (x3,y3). It is based on the assumption that your image starts from (1,1).
Code :
First steps are same as mevatron's. Blur the image to remove noise, threshold the image, then find contours.
import cv2
import numpy as np
img = cv2.imread('office.jpg')
img = cv2.resize(img,(800,400))
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
gray = cv2.medianBlur(gray,3)
ret,thresh = cv2.threshold(gray,1,255,0)
contours,hierarchy = cv2.findContours(thresh,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)
Now find the biggest contour which is your image. It is to avoid noise in case if any (Most probably there won't be any). Or you can use mevatron's method.
max_area = -1
best_cnt = None
for cnt in contours:
area = cv2.contourArea(cnt)
if area > max_area:
max_area = area
best_cnt = cnt
Now approximate the contour to remove unnecessary points in contour values found, but it preserve all corner values.
approx = cv2.approxPolyDP(best_cnt,0.01*cv2.arcLength(best_cnt,True),True)
Now we find the corners.
First, we find (x3,y3). It is farthest point. So x3*y3 will be very large. So we find products of all pair of points and select the pair with maximum product.
far = approx[np.product(approx,2).argmax()][0]
Next (1,x2). It is the point where first element is one,then second element is maximum.
ymax = approx[approx[:,:,0]==1].max()
Next (x1,1). It is the point where second element is 1, then first element is maximum.
xmax = approx[approx[:,:,1]==1].max()
Now we find the minimum values in (far.x,xmax) and (far.y, ymax)
x = min(far[0],xmax)
y = min(far[1],ymax)
If you draw a rectangle with (1,1) and (x,y), you get result as below:
So you crop the image to correct rectangular area.
img2 = img[:y,:x].copy()
Below is the result:
See, the problem is that you lose some parts of the stitched image.
You can do this with threshold, findContours, and boundingRect.
So, here is a quick script doing this with the python interface.
stitched = cv2.imread('stitched.jpg', 0)
(_, mask) = cv2.threshold(stitched, 1.0, 255.0, cv2.THRESH_BINARY);
# findContours destroys input
temp = mask.copy()
(contours, _) = cv2.findContours(temp, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
# sort contours by largest first (if there are more than one)
contours = sorted(contours, key=lambda contour:len(contour), reverse=True)
roi = cv2.boundingRect(contours[0])
# use the roi to select into the original 'stitched' image
stitched[roi[1]:roi[3], roi[0]:roi[2]]
Ends up looking like this:
NOTE : Sorting may not be necessary with raw imagery, but using the compressed image caused some compression artifacts to show up when using a low threshold, so that is why I post-processed with sorting.
Hope that helps!
You can use active contours (balloons/snakes) for selecting the black region accurately. A demonstration can be found here. Active contours are available in OpenCV, check cvSnakeImage.
I am using the setViewBox() function in Raphael 2. The width and height is multiplied by a value like (1.2, 1.3 ...). This changes the magnification/ zooming properly but the x and y which I have given as 0,0 makes the paper display its contents after some offset. If i modify the x and y to some positive value after the rendering( using firebug!!) then the top left of the paper moves back and above to its right position. I want to know how will the value be calculated. I have no idea about how the x,y affect the viewbox. If anybody can give me any pointers for this it will be a real help.
I have tried giving the difference between the width/ height divided by 2. Also I must mention that I am not rendering an image but various raphael shapes e.g. rects, paths text etc. in my paper.
Looking forward to some help!
Kavita
this is an example showing how to calculate the setViewBox values, I included jquery (to get my SVG cocntainer X and Y : $("#"+map_name).offset().left and $("#"+map_name).offset().top) and after that I calculated how much zoom I need :
var original_width = 777;
var original_height = 667;
var zoom_width = map_width*100/original_width/100;
var zoom_height = map_height*100/original_height/100;
if(zoom_width<=zoom_height)
zoom = zoom_width;
else
zoom = zoom_height;
rsr.setViewBox($("#"+map_name).offset().left, $("#"+map_name).offset().top, (map_width/zoom), (map_height/zoom));
did you put the center of your scaling to 0,0 like:
element.scale(1.2,1.2,0,0);
this can scale your element without moving the coordinates of the top left corner.