Single Stacked Barplot with labelling - label

I managed to make a simple single stacked barplot (see code below and thanks for your help!), but now I'd need to make something more complex, please see attached figure (B) as an example. I'd like to plot the "Cell_number", and have a y-axis with numbers and small lines (right hand side of figure B) and next to it a barplot with cell fraction ("Percentage"). I do have 1 barplot per dataset. Many thanks in advance.
gplot(Data, aes(x = "", y = Percentage, fill = Cell_type)) +
geom_col(width = .25) +
scale_fill_manual(values = c("green3", "orange2", "lightskyblue2")) +
coord_flip() +
theme_minimal() +
labs(x= "metPAN TLS", y = "Cell types (%)") +
geom_text(aes(label = paste0(Percentage, "")),
position = position_stack(vjust = 0.5))enter code here

Related

Format Data Label as Percentage and Move it next to the other Data Label

I'm new here and also a beginner Python user. In this chart I was trying to make, I just need to format the blue labels into percentage with 2 decimal places and move them right next to the values at the end of the bars.
Here is the code:
index = np.arange(6)
fig, ax = plt.subplots(figsize=(10, 2))
list1x = list(dict(train['Outlet_Location_Type'].value_counts()).keys())
list1y = list(train['Outlet_Location_Type'].value_counts())
ax.barh(list1x, list1y, alpha=0.7,
# width = 0.5,
color=cm.Blues([i / 0.00525 for i in [ 0.00808, 0.0045, 0.00281]])
)
plt.rcParams.update({'font.size': 10})
rects = ax.patches
for i, label in enumerate(ii / 8325 * 100 for ii in list1y):
ax.text(label, i, str(label), size=10, ha='left', va='center', color = "blue")
for h, label in enumerate(list1y):
ax.text(label, h, label, size=10, ha='left', va='center')
ax.text(0, 1.02, 'Outlet Location Type Count', transform=ax.transAxes, size=12, weight=600, color='#777777')
ax.xaxis.set_ticks_position('bottom')
ax.tick_params(axis='x', colors='black', labelsize=9)
ax.set_axisbelow(True)
plt.xticks([])
ax.spines['right'].set_visible(False)
ax.spines['top'].set_visible(False)
plt.show()
I think there could be data type issue causing the blue values to stay at the left side. I can't find the formatting solution to change them into percentage.
Thank you so much in advance!

how to make smooth and remove the white edge? Besides, why the black lines arise in my answer? How to solve it?

Based on Shai and Biguri's codes and comments, I have finished a color picture like this:
A problem arises, how to remove the white edge and make it smooth? One solution may be to build 3x3 matrix or bigger and average. But the calculations should be large for every white-edge points. Or there may be some useful functions in Matlab to deal well with this problems?
If you have a license for the image processing toolbox, you can try using for example medfilt2 to apply a median filter on the image. A 11 by 11 median filter should do the trick. It is not very difficult to reimplement the filter yourself if you don't have the toolbox.
This is just one of the possibilities, you can use many different filters that will have different impacts on sharpness ang edge removal.
Edit:
Here is a quick median filter implementation (it may contain errors and could be optimized):
function ret = imageMedianFilter(im, np)
if(size(np,2) == 1)
npx = np;
npy = np;
else
npx = np(1);
npy = np(2);
end
ret = zeros(size(im,1),size(im,2));
for xpos = 1:size(im,1)
for ypos = 1:size(im,2)
curval = double(0);
if(xpos + npx - 1) > size(im,1)
npixels_x = size(im,1) - xpos + 1;
else
npixels_x = npx;
end
if(ypos + npy - 1) > size(im,2)
npixels_y = size(im,2) - ypos + 1;
else
npixels_y = npy;
end
a = im(xpos:xpos+npixels_x-1 , ypos:ypos+npixels_y-1);
a = reshape(a,1,size(a,1)*size(a,2));
curval = median(a);
ret(xpos , ypos) = curval;
end
end
ret = uint8(ret);
end
You can use it on R,G and B components as shown by Rotem below:
RGB = cat(3, imageMedianFilter(RGB(:,:,1), [11,11]), imageMedianFilter(RGB(:,:,2), [11,11]), imageMedianFilter(RGB(:,:,3), [11,11]));
(assuming your image is named RGB).
Here is my solution. I take n*n patch to average the near RGB. But there is a problem arising. Why the right down side of processed picture showing black lines?
clc;clf;close all;clear all;
img = imread('sample2color_t1.bmp'); %// read image
bw = img(:,:,1) > 128; %// convert to binary mask
[lb,lab] = bwlabel(bw,4); %// extract distinct regions
[a,b,c]=size(img);
R=ones(a,b);
G=ones(a,b);
B=ones(a,b);
%I have omitted other colors process codes. Below it is the white edges code.
r=[];c=[];
[r,c] = find(lb ==0);
for i=1:length(r)
R(r(i),c(i))=1;
G(r(i),c(i))=1;
B(r(i),c(i))=1;
end
scale=5;%步长1,8连通
for i=1:length(r)
sumR=0;sumG=0;sumB=0;
for j=0:2*scale
for k=0:2*scale
sumR=sumR+R(r(i)-scale+j,c(i)-scale+k);
sumG=sumG+G(r(i)-scale+j,c(i)-scale+k);
sumB=sumB+B(r(i)-scale+j,c(i)-scale+k);
end
end
R(r(i),c(i))=sumR/(2*scale+1)^2;
G(r(i),c(i))=sumG/(2*scale+1)^2;
B(r(i),c(i))=sumB/(2*scale+1)^2;
end
imPaint=cat(3,R,G,B);
figure;
imshow(imPaint);

Skipping some axis labels in a plot with imagesc

I have created a big heat map using matlab's imagesc command. It plots the error output for each combination of the values in x and y axes. As can be seen in the figure there are too many axes labels. This might become even denser as I plan to increase the number of points in both x and y axes - which means I will get more outputs on a finer grid.
I want to be flexible with the labels, and skip some of them. I want to do this for both X and Y. I also want to be flexible with the "ticks" and draw either all of them or maybe skip some of them. Keep in mind that both the X and Y values are not increasing in order, at first the increment is 0.01 for 9 points, then 0.1, then 1 or 3 or whatever. I will change these increments too.
I tried to show what I want the graph look like in the second image. I want roughly the labels shown in red boxes only. As I said these are not set values, and I will make the increments smaller which will lead to denser plot.
Thank you for your help.
OS: Windows 7, 8 (64 bit)
Matlab version: Matlab 2014 a
You can manipulate the ticks and labels like this:
ticksarray=[1 33 41 100 ...] % edit these to whatever you want
tickslabels={'1', '33', '41', '100'; ...} % match the size of both arrays
set(gca,'XTick',ticksarray)
set(gca,'XTickLabel',tickslabels)
The same thing applies to the y-axis.
Small working example:
x=1:100;
y=2*x.^2-3*x+2;
plot(x,y)
ticksarray=[1 33 41 100];
tickslabels={'1', '33', '41', '100'};
set(gca,'XTick',ticksarray)
set(gca,'XTickLabel',tickslabels)
Example:
figure(1)
load clown
subplot(211)
imagesc(X);
subplot(212)
imagesc(X);
h = gca;
Now you can either set a maximum number of labels per axis:
%// define maximum number of labels
maxLabel = 3;
h.XTick = linspace(h.xlim(1),h.xlim(2),maxLabel);
h.YTick = linspace(h.ylim(1),h.ylim(2),maxLabel);
or define how many labels should be skipped:
%// define number of labels to skip
skipLabel = 2;
h.XTick = h.XTick(1:skipLabel:end);
h.YTick = h.YTick(1:skipLabel:end)
You can also get a different number of ticks and labels, more complicated though:
maxLabel = 3;
maxTicks = 6;
h.XTick = linspace(h.xlim(1),h.xlim(2),maxTicks);
h.YTick = linspace(h.ylim(1),h.ylim(2),maxTicks);
h.XTickLabel( setdiff( 1:maxTicks, 1:maxTicks/maxLabel:maxTicks ) ) = repmat({''},1,maxTicks-maxLabel);
h.YTickLabel( setdiff( 1:maxTicks, 1:maxTicks/maxLabel:maxTicks ) ) = repmat({''},1,maxTicks-maxLabel);
If you use a prior version of Matlab 2014b, then you will need the set command to set all properties:
%// define maximum number of labels
maxLabel = 3;
Xlim = get(h,'Xlim');
Ylim = get(h,'Ylim');
set(h,'XTick', linspace(Xlim(1),Xlim(2),maxLabel));
set(h,'YTick', linspace(Ylim(1),Ylim(2),maxLabel));
%// or define number of labels to skip
skipLabel = 2;
XTick = get(h,'XTick');
YTick = get(h,'YTick');
set(h,'XTick', XTick(1:skipLabel:end));
set(h,'YTick', YTick(1:skipLabel:end));
%// or combined
maxLabel = 3;
maxTicks = 6;
Xlim = get(h,'Xlim');
Ylim = get(h,'Ylim');
set(h,'XTick', linspace(Xlim(1),Xlim(2),maxTicks));
set(h,'YTick', linspace(Ylim(1),Ylim(2),maxTicks));
XTickLabel = cellstr(get(h,'XTickLabel'));
YTickLabel = cellstr(get(h,'YTickLabel'));
XTickLabel( setdiff( 1:maxTicks, 1:maxTicks/maxLabel:maxTicks ),: ) = repmat({''},1,maxTicks-maxLabel);
YTickLabel( setdiff( 1:maxTicks, 1:maxTicks/maxLabel:maxTicks ),: ) = repmat({''},1,maxTicks-maxLabel);
set(h,'XTickLabel',XTickLabel);
set(h,'YTickLabel',YTickLabel);
After applying the second method proposed by #thewaywewalk I got the second figure below. Apparently the labels need to be structured as well, because they only take the first so many labels.
Then I tried to manipulate the labels as shown below, and got the third image.
skipLabel = 2;
XTick = get(h,'XTick');
YTick = get(h,'YTick');
set(h,'XTick', XTick(1:skipLabel:end));
set(h,'YTick', YTick(1:skipLabel:end));
XTickLabel = get(h,'XTickLabel');
labelsX = cell( length(1: skipLabel:length(XTick)) , 1);
j = 1;
for i = 1: skipLabel:length(XTick)
labelsX{j} = XTickLabel(i, :);
j = j + 1;
end
set(h,'XTickLabel', labelsX);
YTickLabel = get(h,'YTickLabel');
labelsY = cell( length(1: skipLabel:length(YTick)) , 1);
j = 1;
for i = 1: skipLabel:length(YTick)
labelsY{j} = YTickLabel(i, :);
j = j + 1;
end
set(h,'YTickLabel', labelsY);
The Y axis labels seem to be in place as before (right next to tick), however the X axis labels seem to be shifted to the left a little. How can I correct this?
Another note: How can I change the scientific values into normal numbers? Also, probably there is a better approach at manipulating the labels.

Using d3.js is there a way to zero align two Y Axes with positive and negative values

I am new to d3, learning a lot. I have an issue I cannot find an example for:
I have two y axes with positive and negative values with vastly different domains, one being large dollar amounts the other being percentages.
The resulting graph from cobbling together examples looks really awesome with one slight detail, the zero line for each y axis is in a slightly different position. Does anyone know of a way in d3 to get the zero line to be at the same x position?
I would like these two yScales/axes to share the same zero line
// define yScale
var yScale = d3.scale.linear()
.range([height, 0])
.domain(d3.extent(dataset, function(d) { return d.value_di1; }))
;
// define y2 scale
var yScale2 = d3.scale.linear()
.range([height, 0])
.domain(d3.extent(dataset, function(d) { return d.calc_di1_di2_percent; }))
;
Here is a link to a jsfiddle with sample data:
http://jsfiddle.net/jglover/XvBs3/1/
(the x-axis ticks look horrible in the jsfiddle example)
In general, there's unfortunately no way to do this neatly. D3 doesn't really have a concept of several things lining up and therefore no means of accomplishing it.
In your particular case however, you can fix it quite easily by tweaking the domain of the second y axis:
.domain([d3.min(dataset, function(d) { return d.calc_di1_di2_percent; }), 0.7])
Complete example here.
To make the 0 level the same position, a strategy is to equalize the length/proportion of the y axes.
Here are the concepts to the solution below:
The alignment of baseline depends on the length of the y axes.
To let all value shown in the bar, we need to extend the shorter side of the dimension, which compares to the other, to make the proportion of the two axes equal.
example:
// dummy data
const y1List = [-1000, 120, -130, 1400],
y2List = [-0.1, 0.2, 0.3, -0.4];
// get proportion of the two y axes
const totalY1Length = Math.abs(d3.min(y1List)) + Math.abs(d3.max(y1List)),
totalY2Length = Math.abs(d3.min(y2List)) + Math.abs(d3.max(y2List)),
maxY1ToY2 = totalY2Length * d3.max(y1List) / totalY1Length,
minY1ToY2 = totalY2Length * d3.min(y1List) / totalY1Length,
maxY2ToY1 = totalY1Length * d3.max(y2List) / totalY2Length,
minY2ToY1 = totalY1Length * d3.min(y2List) / totalY2Length;
// extend the shorter side of the upper dimension with corresponding value
let maxY1Domain = d3.max(y1List),
maxY2Domain = d3.max(y2List);
if (maxY1ToY2 > d3.max(y2List)) {
maxY2Domain = d3.max(y2List) + maxY1ToY2 - d3.max(y2List);
} else {
maxY1Domain = d3.max(y1List) + maxY2ToY1 - d3.max(y1List);
}
// extend the shorter side of the lower dimension with corresponding value
let minY1Domain = d3.min(y1List),
minY2Domain = d3.min(y2List);
if (minY1ToY2 < d3.min(y2List)) {
minY2Domain = d3.min(y2List) + minY1ToY2 - d3.min(y2List);
} else {
minY1Domain = d3.min(y1List) + minY2ToY1 - d3.min(y1List);
}
// finally, we get the domains for our two y axes
const y1Domain = [minY1Domain, maxY1Domain],
y2Domain = [minY2Domain, maxY2Domain];

Image with accordion effect

I have read in an image file to MATLAB and I am trying to stretch it in one direction, but a variable amount (sinusoidal). This would create an accordion affect on the image. I have toyed around with imresize, however that only resizes the image linearly. I would like the amount of "stretch" to vary for each image line. I tried to convey this with the following code:
periods = 10; % Number of "stretch" cycles
sz = size(original_image,2)/periods;
s = 0;
x = 0;
for index = 1:periods
B = original_image(:,round(s+1:s+sz));
if mod(index,2) == 0
amp = 1.5;
else
amp = 0.75;
end
xi = size(B,2)*amp;
new_image(:,x+1:x+xi) = imresize(B, [size(B,1) size(B,2)*amp]);
s = s + sz;
x = x+xi;
end
You can see that segments of the image are stretched, then compressed, then stretched, etc, like an accordion. However, each segment has a uniform amount of stretch, whereas I'd like it to be increasing then decreasing as you move along the image.
I have also looked at MATLAB's example of Applying a Sinusoidal Transformation to a Checkerboard which seems very applicable to my problem, however I have been trying and I cannot get this to produce the desired result for my image.
Any help is much appreciated.
UPDATE:
Thank you for Answer #1. I was unable to get it to work for me, but also realized it would resulted in loss of data, as the code only called for certian lines in the original image, and other lines would have been ignored.
After experimenting further, I developed the code below. I used a checkerboard as an example. While combersome, it does get the job done. However, upon trying the script with an actual high-resolution image, it was extremely slow and ended up failing due to running out of memory. I believe this is because of the excessive number of "imresize" commands that are used in loop.
I = checkerboard(10,50);
I = imrotate(I,90);
[X Y] = size(I);
k = 4; % Number of "cycles"
k = k*2;
x = 1;
y = 2;
z = 2;
F = [];
i = 1;
t = 0;
s = 0;
for j = 1:k/2
t = t + 1;
for inc = round(s+1):round(Y/k*t)
Yi = i + 1;
F(:,(x:y)) = imresize(I(:,(inc:inc)),[X Yi]);
x = y + 1;
y = x + z;
z = z + 1;
i = i + 1;
end
y = y - 2;
z = z - 4;
for inc = round(Y/k*t+1):round(Y/k*(t+1))
Yi = i - 1;
F(:,(x:y)) = imresize(I(:,(inc:inc)),[X Yi]);
x = y + 1;
y = x + z;
z = z - 1;
i = i - 1;
end
y = y + 2;
z = z + 4;
s = Y/k*(t+1);
t = t + 1;
end
Fn = imresize(F, [X Y]);
imshow(Fn);
Does anyone know of a simpler way to achieve this? If you run the code above, you can see the effect I am trying to achieve. Unfortunately, my method above does not allow me to adjust the amplitude of the "stretch" either, only the number of "cycles," or frequency. Help on this would also be appreciated. Much thanks!
Here is how I would approach it:
Determine how the coordinate of each point in your Final image F maps into your Initial image I of size (M,N)
Since you want to stretch horizontally only, given a point (xF,yF) in your final image, that point would be (xI,yI) in your initial image where xI and yI can be obtained as follows:
yI = yF;
xI = xF + Lsin(xFK);
Notes:
these equations do not guarantee that xI remains within the range [1:N] so cropping needs to be added
K controls the how many wrinkles you want to have in your accordion effect. For example, if you only want one wrinkle, K would be 2*pi/N
L controls how much stretching you want to apply
Then simply express your image F from image I with the transforms you have in 1.
Putting it all together, the code below creates a sample image I and generates the image F as follows:
% Generate a sample input image
N=500;
xF=1:N;
I=(1:4)'*xF/N*50;
% Set the parameters for your accordion transform
K=2*pi/N;
L=100;
% Apply the transform
F=I(:, round(min(N*ones(1,N), max(ones(1,N), (xF + L*sin(xF*K))))) );
% Display the input and output images side by side
image(I);
figure;
image(F);
If you run this exact code you get:
As you can see, the final image on the right stretches the center part of the image on the left, giving you an accordion effect with one wrinkle.
You can fiddle with K and L and adjust the formula to get the exact effect you want, but note how by expressing the transform in a matrix form MATLAB executes the code in a fraction of second. If there is one take away for you is that you should stay away from for loops and complex processing whenever you can.
Have fun!

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