0

我有如下两列。 

ABC =

4.1103   25.5932
5.0852   31.2679
6.0021   15.9020
5.8495   21.4804
4.3245   19.9674
5.9378   38.3452
6.9460    8.8233
7.4568   44.7429
5.7358   32.7608
5.3510   35.2645
5.1657   54.6566
5.1381   44.1870
4.1566  101.8947
5.7310   -3.0565
5.5496   28.3637
4.5672   -1.7736
4.5805   11.8384
4.7948   33.7640
3.9901    6.0607
4.4203   17.7308
4.2712   -1.5834
4.8808   -2.3123
5.9004   -0.4623
5.3929    1.1477
5.6594    6.9741
5.5114   11.3982
5.4715    5.9189
5.0021    6.2561
4.1576   10.3207
6.1025    3.4654
3.9960    6.6892
5.6938    3.8429
5.2416    7.7513
7.0922    2.6871
5.3277   14.0617
6.1350    4.0316
6.0211  -20.3587
6.7399   14.0224
5.0818  102.6360
5.6444   24.3167
6.2542   19.8522
6.2862   24.3430
5.6452   -6.4020
5.4561   14.7813
4.7934    9.4639
3.8523   32.0766
3.9878    8.5313
4.5232   42.0309
4.2489  -12.0325
6.0413   -5.5464
4.9334   -3.2520
4.1349   20.9038
4.2329   20.6303
4.2009   31.8840
4.0624   48.5402
4.7674   28.6595
4.0767    4.7767
4.0971   34.8460
3.8442   24.0209
5.2471   38.8815
6.0241   59.3785
6.9743    6.5027
7.8732    4.5422
4.3094   68.4340
4.5601   -4.2946
4.6140  109.4510
4.5862   71.8387
5.2210   66.1310
4.3835   32.7592
6.1432   36.3832
5.4624   13.7891
5.2129   40.1301
3.8987   67.2705
6.6328   15.0286
8.0786   -7.3078
4.8968   -6.7754
4.1200    4.5333
4.1098   -3.3204
4.0373   26.4890
3.8467   48.8121
7.7795   -2.3606
6.9553   21.3609
6.2635   24.4985
6.1518   -1.4200
4.9115   11.5784
5.5908   13.1351
7.0117   -2.8297
5.2193   38.6937
6.0786   16.9453
6.8229   14.0907
8.0385   13.6228
8.6596   -1.4478
6.3257    8.0361
6.9223  -14.2179
3.8337   15.5773
4.0039  -24.1494
4.6332   17.9308
6.3684   11.3398
5.8592    4.0367
6.9040   12.1495
7.8524   -0.0432
8.3545   10.8865
9.3946   20.4614
4.3015   25.9674
4.4782   21.9045
4.1994   39.2286
4.3499   22.1004
4.3652   33.6220
4.2026   -5.8153
5.1330    6.4996
5.3118   33.7835
4.2002   -3.1917
3.8285   32.1016
3.9485   21.6358
3.8688   21.7830
4.0494   24.7914
4.0869   10.6577
4.6699    8.4756
5.1199   11.1885
5.1831    8.6163
4.5560    8.2806
4.4886    4.8017
4.5618    5.9434
4.1135   12.8942
4.1377   22.1423

我等于没有。来自“x”的 bin 和相应的平均 bin 值“yy”。如下所示

x=ABC(:,1);
y=ABC(:,2);
counter=1
    for i=min(x):0.3:max(x)     
         bin= x>i &  x<= i+0.3;       
         xbin(counter,1)  = mean(x(bin)); 
         yy(counter,1)    = mean(y(bin));
         counter          = counter+1
    end

plot(x,y,'ro'); hold on
plot(xbin,yy,'bo-');

其中为“x”的某个范围定义了“bin”（请参见循环）。现在输出包含来自“x”的“xbin”和来自“y”对应的“xbin”的数据“yy”的平均值。我担心应该从大约获得的平均值“yy”。等号 的数据点。如果 'bin' 中没有足够的 'y' 数据点，则平均值 'yy' 应该是 NaN。请有人在这方面提供帮助。谢谢

4

3 回答 3

1

检查-loop每次迭代的1s数。如果该数字低于某个阈值，则分配给： binforNaNyy

x=ABC(:,1);
y=ABC(:,2);
counter=1;

nbinmin = 5; % this is the threshold

for i=min(x):0.3:max(x)
    bin= x>i &  x<= i+0.3;
    xbin(counter,1)  = mean(x(bin));

    % check if the number of 1s in bin is less than the threshold
    if length(bin(bin==1)) < nbinmin
        yy(counter,1)    = NaN;
    else
        yy(counter,1)    = mean(y(bin));
    end
    counter = counter+1;
end
于 2013-08-27T17:31:39.960 回答
1

这个问题并不完全清楚，但是您是否尝试过使用直方图函数hist？它似乎可以为你做很多工作

% choose the bin locations
xcenters = min(x):0.3:max(x);

% compute counts in each bin
[counts, ctrs] = hist(y, xcenters);

% set any with too few samples to NaN
count_min = 3;
counts(counts < count_min) = NaN;

% plot -- either as a histogram, 
figure(1)
bar(ctrs, counts)
%or as a line plot (note that the line won't join up if too many NaN segments)
figure(2)
plot(ctrs, counts)

您可以在此处指定输入 bin 中心，但要改为定义 bin 的边缘，请查看histc.

于 2013-08-27T17:39:06.963 回答
1

您基本上是在寻找具有非均匀箱的直方图或具有相等计数的直方图。

对于非均匀直方图，最简单的情况是对其中的N值进行排序x并将排序后的向量分隔到kbin 中，即每个 bin 都有N/k样本（您也可以通过指定 来设置比率N = ck）。

您不是对范围域 x 进行线性间距，而是对有序向量进行线性分割（因此对原始范围进行非线性、非均匀分离）。

在您的情况下，它看起来像这样：

[sortedX, indeX] = sort(x);
nVals = length(x); % N
nBins = nVals/10;  % k = N/c

% linear split of the sorted vector
stepX = (1:nVals/nBins:nVals);
if stepX(end)~=nVals, stepX = [stepX nVals+1]; end

% counting and bining on the indexed vector
for i = 1 : length(stepX)-1    
    bin = indeX(stepX(i):stepX(i+1)-1);
    xbin(i,1) = mean(x(bin));
    yy(i,1) = mean(y(bin));  

end

要计算实际范围（即直方图的边缘），您可以使用bin 中的最大值和 bin中的最小值之间的中点ii+1。您可以在循环中添加类似以下内容： 

% calculate the range
maxX(i) = max(x(bin));
minX(i) = min(x(bin));

所需的（非线性）范围是：

rangeX = [min(x) maxX(1:end-1) + (minX(2:end) - maxX(1:end-1))/2 max(x)];

而您的原始（线性）范围是：

rangeX_OP = min(x):0.3:max(x);

You can use histc to verify the equal counts (for rangeX) and non-equal counts (for rangeX_OP). This is how the counts would look (for random x in similar range to yours and c = 10 counts per bin). Top is the linear spacing if range, bottom is the non-linear.

enter image description here

于 2013-08-27T18:09:27.000 回答