I tried different seaborn functions and pandas style functions. I want to apply a heatmap with center=0 (positive value gets red heatmap, negative values gets blue heatmap) while keeping all the categorical columns uncolored.
When I was using sns.diverging_palette together with pandas style.background_gradient(), I can specify subset of columns to be colored. But cannot specify the color center as 0.
When using sns.heatmap(), How to set center color in heatmap, it allows you to set center as 0 but the input has to be all numerical....
Is there a method to do both, so that I can apply heatmap to both positive/negative values while keeping the categorical columns uncolored?
Related
In D3.js choropleth map, we assign a color to each state or city. is there any solution or example that we can assign heatmap to each coordinate (my data are values from satellites so i have a value for each coordinate).In matlab or matplotlib I plot a figure and then apply borders but it seems it's not possible in D3. i want something like this
thanks.
folium python package have geojson (choropleth) mapping function that we can define rectangles (polygons, for example zip code regions). for more information interactive choropleth map
and popuo on map. my result
I have a bunch of images of clothes of many colors and I want to detect the colors of each image. Say that I have a blue skirt image in daylight conditions and I can get the correct color through RGB distributions. However, at night it's difficult to tell the color and the "blue" is recognized as "black". It's very hard to make a unified standard to specify colors through RGB distributions.
As such, I am wondering is there a way or algorithm to detect colors under different illuminations?
BTW: I also tried HSV color space and the results were not good.
That's a very hard problem and it's still trying to be solved today. The gist of it is to find a colour quantization using a representative set of basic colours of an image that is robust against different external stimuli... lighting, shade, poor illumination etc.
Unfortunately I can't suggest any one algorithm that would do the work for you for all cases. However, one algorithm that has worked for me in the past was when I was doing work in image retrieval. Specifically, the work by Jiebo Luo and David Crandall from Kodak Research Labs: http://vision.soic.indiana.edu/papers/compoundcolor2004cvpr.pdf
The basic algorithm is to take a look at the ISCC-NBS colour palette set. Also, this link is much more fruitful: https://www.w3schools.com/colors/colors_nbs.asp. It is a set of 267 colours that are representative of the colours that we see in modern society today. Usually when we describe colours, we have a set of one or more adjectives, followed by the dominant hue. For example, that shirt is a darkish pale blue, or a light bright yellow, etc. The beauty of this algorithm is that when the colour in question is subject to different external stimuli, we have all of these adjectives that give meaning to the colour, but at the end of the day, the last part of the colour - the dominant hue - is what we're after.
Each of these colours has an associated RGB value. These colours are transformed into the CIE Lab colour space which form a 267 CIE Lab lookup table.
To classify a particular input colour, you would transform this input's RGB values into the CIE Lab colour space, then determine the closest colour to this lookup table. It has been shown that the Euclidean distance between two colours in the CIE Lab colour space best represents the difference in human perception of colours. Once we determine which location in the lookup table the colour is closest to, we strip out all of the adjectives and see what the dominant hue is and we thus classify that colour accordingly.
For example, if we had a RGB pixel and we converted it to Lab, then found that the closest colour was bright yellow, we would remove the "bright" and the final colour that is representative of that RGB pixel would be yellow.
Therefore, the final algorithm is this:
Find the ISCC-NBS colour set's RGB values and convert to CIE Lab and create a lookup table, which I call LUT1. In Python for example, you could simply make this a 2D list or 2D NumPy array.
Create another lookup that stores the dominant hue for each of the colours in the ISCC-NBS colour set - so strip out all of the adjectives and leave the dominant hue, which I call LUT2. In Python for example, you could create a dictionary where the key is the corresponding row of LUT1 and the value would be the actual basic colour itself. Whether it's a string representation or a RGB triplet representing the basic colour is up to you.
For a pixel in question, find the closest ISCC-NBS colour that matches with LUT1 by the Euclidean distance between this pixel's CIE Lab components and the ones in LUT1.
Once we find this location in LUT1, use the same index to index into LUT2 and get the final colour to classify that input pixel's colour.
Hope this helps!
I recently asked a question about using matlab to reduce the number of colors in an image. However, when I attempted this, I was only able to get color approximations which then matched the pixel to the nearest color within the color map.
For example, using a color map with only three colors [red, green, blue], it would scan each color and then map either red green or blue. However, this process did not vary the RGB densities to create realistic looking color.
I'm curious if there is any sort of built in function that would use these three colors and vary the density of them to achieve the average color of a certain "pixel field".
I realize this would lose resolution, but I'm essentially trying to make realistic looking images, using only three colors by varying the amounts of RGB within a certain region.
You are looking for the function rgb2ind and its 'dither' option.
I have two datasets that I am plotting in a D3.js parallel coordinate chart. I want to highlight the differences between the two.
In my example the two datasets share all values except for the economy attribute.
Blue is the first dataset and red is the second.
How can I generate an area fill that represents the difference between the two svg line paths, in the example provided the area between the red and the blue paths?
I am also open to other 'design' considerations.
I have more then 1 week reading about selective color change of an image. It meand selcting a color from a color picker and then select a part of image in which I want to change the color and apply the changing of color form original color to color of color picker.
E.g. if I select a blue color in color picker and I also select a red part in the image I should be able to change red color to blue color in all the image.
Another example. If I have an image with red apples and oranges and if I select an apple on the image and a blue color in the color picket, then all apples should be changing the color from red to blue.
I have some ideas but of course I need something more concrete on how to do this
Thank you for reading
As a starting point, consider clustering the colors of your image. If you don't know how many clusters you want, then you will need methods to determine whether to merge or not two given clusters. For the moment, let us suppose that we know that number. For example, given the following image at left, I mapped its colors to 3 clusters, which have the mean colors as shown in the middle, and representing each cluster by its mean color gives the figure at right.
With the output at right, now what you need is a method to replace colors. Suppose the user clicks (a single point) somewhere in your image, then you know the positions in the original image that you will need to modify. For the next image, the user (me) clicked on a point that is contained by the "orange" cluster. Then he clicked on some blue hue. From that, you make a mask representing the points in the "orange" cluster and play with that. I considered a simple gaussian filter followed by a flat dilation 3x5. Then you replace the hues in the original image according to the produced mask (after the low pass filtering, the values on it are also considered as a alpha value for compositing the images).
Not perfect at all, but you could have a better clustering than me and also a much-less-primitive color replacement method. I intentionally skipped the details about clustering method, color space, and others, because I used only basic k-means on RGB without any pre-processing of the input. So you can consider the results above as a baseline for anything else you can do.
Given the image, a selected color, and a target new color - you can't do much that isn't ugly. You also need a range, some amount of variation in color, so you can say one pixel's color is "close enough" while another is clearly "different".
First step of processing: You create a mask image, which is grayscale and varying from 0.0 to 1.0 (or from zero to some maximum value we'll treat as 1.0), and the same size as the input image. For each input pixel, test if its color is sufficiently near the selected color. If it's "the same" or "close enough" put 1.0 in the mask. If it's different, put 0.0. If is sorta borderline, put an in-between value. Exactly how to do this depends on the details of the image.
This might work best in LAB space, and testing for sameness according to the angle of the A,B coordinates relative to their origin.
Once you have the mask, put it aside. Now color-transform the whole image. This might be best done in HSV space. Don't touch the V channel. Add a constant to S, modulo 360deg (or mod 256, if S is stored as bytes) and multiply S by a constant chosen so that the coordinates in HSV corresponding to the selected color is moved to the HSV coordinates for the target color. Convert the transformed S and H, with the unchanged L, back to RGB.
Finally, use the mask to blend the original image with the color-transformed one. Apply this to each channel - red, green, blue:
output = (1-mask)*original + mask*transformed
If you're doing it all in byte arrays, 0 is 0.0 and 255 is 1.0, and be careful of overflow and signed/unsigned problems.