I have the following range from a much larger matrix :
我有以下范围从一个更大的矩阵:
range(a)
范围(一个)
ans =
ans =
94 153 144 59 79 90 131 64
My professor is asking us to: Divide the range into N = 10 equal-length segments (hereafter called “bins”), and for each bin, find its bounds (aj, bj) as well as its center cj.
我的教授要求我们:将范围划分为N = 10个等长段(以后称为“bin”)，对于每个bin，找到它的界限(aj, bj)和它的中心cj。
(5) Place each measured bacterial count xi into that bin whose lower bound is less than or equal to xi and whose upper bound is greater than xi; thereafter, for each bin count the number of xi assigned to it (= nj).
(5)将每一个被测菌数xi放入下界小于或等于xi上界大于xi的bin中;此后，对于每个bin计算分配给它的xi的个数(= nj)。
(6) Plot a histogram of the measured bacterial counts using N = 10 bars. Try the MATLAB functions hist(x,N) and bar(c, n)
(6)用N = 10条，绘制测量细菌计数的直方图。试试MATLAB函数hist(x,N)和bar(c, N)
I know this is a lot, but I have absolutely no instruction from this guy and would really appreciate a helping hand :)
我知道这很重要，但我绝对没有这个人的指导，我真的很感激你的帮助。
2 个解决方案
#1
8
Consider the following example, it should solve all your points:
考虑下面的例子，它应该能解决你所有的问题:
%# random data vector of integers
M = randi([50 200], [100 1]);
%# compute bins
nbins = 10;
binEdges = linspace(min(M),max(M),nbins+1);
aj = binEdges(1:end-1); %# bins lower edge
bj = binEdges(2:end); %# bins upper edge
cj = ( aj + bj ) ./ 2; %# bins center
%# assign values to bins
[~,binIdx] = histc(M, [binEdges(1:end-1) Inf]);
%# count number of values in each bin
nj = accumarray(binIdx, 1, [nbins 1], @sum);
%# plot histogram
bar(cj,nj,'hist')
set(gca, 'XTick',binEdges, 'XLim',[binEdges(1) binEdges(end)])
xlabel('Bins'), ylabel('Counts'), title('histogram of measured bacterial')
Note that this correctly handles the last bin (read this related question for a discussion on those edge cases)
注意，这正确地处理了最后一个bin(请阅读这个相关问题，以便对这些边缘案例进行讨论)
#2
3
computing histogram:
计算直方图:
range = [94 153 144 59 79 90 131 64]
[n,xout] = hist(range, 10)
xout is bin centers, n bin counts.
xout是bin centers, n bin count。
plotting bar graph:
绘制柱状图:
bar(xout,n)
computing bin edges:
计算本边缘:
width = xout(2)-xout(1)
xoutmin = xout-width/2
xoutmax = xout+width/2
#1
8
Consider the following example, it should solve all your points:
考虑下面的例子，它应该能解决你所有的问题:
%# random data vector of integers
M = randi([50 200], [100 1]);
%# compute bins
nbins = 10;
binEdges = linspace(min(M),max(M),nbins+1);
aj = binEdges(1:end-1); %# bins lower edge
bj = binEdges(2:end); %# bins upper edge
cj = ( aj + bj ) ./ 2; %# bins center
%# assign values to bins
[~,binIdx] = histc(M, [binEdges(1:end-1) Inf]);
%# count number of values in each bin
nj = accumarray(binIdx, 1, [nbins 1], @sum);
%# plot histogram
bar(cj,nj,'hist')
set(gca, 'XTick',binEdges, 'XLim',[binEdges(1) binEdges(end)])
xlabel('Bins'), ylabel('Counts'), title('histogram of measured bacterial')
Note that this correctly handles the last bin (read this related question for a discussion on those edge cases)
注意，这正确地处理了最后一个bin(请阅读这个相关问题，以便对这些边缘案例进行讨论)
#2
3
computing histogram:
计算直方图:
range = [94 153 144 59 79 90 131 64]
[n,xout] = hist(range, 10)
xout is bin centers, n bin counts.
xout是bin centers, n bin count。
plotting bar graph:
绘制柱状图:
bar(xout,n)
computing bin edges:
计算本边缘:
width = xout(2)-xout(1)
xoutmin = xout-width/2
xoutmax = xout+width/2