标准化直方图Matplotlib

时间:2016-02-18 13:16:46

标签: python matplotlib histogram

嗨我正在绘制三种不同的直方图,这些直方图具有不同的总频率,但我想对它们进行归一化,使频率相同。

enter image description here

从图中可以看出,这三组具有不同的总频率,但我想对它们进行归一化,使它们具有相同的总频率,但我想保留x-每个值的频率比例。轴。

这是我用来绘制直方图的代码

setA = [22.972972972972972, 0.0, 0.0, 27.5, 25.0, 18.64406779661017, 8.88888888888889, 20.512820512820515, 11.11111111111111, 15.151515151515152, 17.741935483870968, 13.333333333333334, 16.923076923076923, 12.820512820512821, 27.77777777777778, 4.0, 0.0, 15.625, 14.814814814814815, 7.142857142857143, 15.384615384615385, 14.545454545454545, 38.095238095238095, 17.647058823529413, 21.951219512195124, 21.428571428571427, 32.432432432432435, 10.526315789473685, 36.8421052631579, 13.114754098360656, 17.91044776119403, 12.64367816091954, 16.0, 22.727272727272727, 18.181818181818183, 9.523809523809524, 17.105263157894736, 11.904761904761905, 20.58823529411765, 10.714285714285714, 15.686274509803921, 27.5, 16.129032258064516, 21.333333333333332, 40.90909090909091, 11.904761904761905, 13.157894736842104]
setB = [1.492537313432836, 3.5714285714285716, 17.94871794871795, 11.363636363636363, 13.513513513513514, 14.285714285714286, 15.686274509803921, 17.94871794871795, 9.090909090909092, 41.07142857142857, 10.714285714285714, 25.0, 20.0, 40.0, 13.333333333333334, 13.793103448275861, 3.5714285714285716, 17.073170731707318, 25.675675675675677, 15.625, 17.46031746031746, 8.333333333333334, 18.64406779661017, 14.285714285714286, 0.0, 6.0606060606060606, 6.976744186046512, 18.181818181818183, 26.785714285714285, 22.80701754385965, 6.666666666666667, 12.5]
setC = [13.846153846153847, 23.076923076923077, 25.0, 10.714285714285714, 16.666666666666668, 9.75609756097561, 10.0, 10.0, 17.857142857142858, 20.0, 9.75609756097561, 26.470588235294116, 12.5, 13.333333333333334, 4.3478260869565215, 5.882352941176471, 14.545454545454545, 13.333333333333334, 8.571428571428571, 11.764705882352942, 0.0]

plt.figure('sets')
n, bins, patches = plt.hist(setA, 20, alpha=0.40 , label = 'setA')  
n, bins, patches = plt.hist(setB, 20, alpha=0.40 , label = 'setB')
n, bins, patches = plt.hist(setC, 20, alpha=0.40 , label = 'setC')    
plt.xlabel('Set')
plt.ylabel('Frequency')
plt.title('Different Sets that need to be normalised')

plt.legend()
plt.grid(True)
plt.show()

作为一个加号,因为我的目标是能够比较三组的分布,是否有更好的直方图可视化,我可以用它来更好地图形化它们。

1 个答案:

答案 0 :(得分:4)

您可以使用normed=True选项对直方图进行标准化。这意味着所有直方图的面积将加起来为1.

您还可以通过对所有三个直方图使用相同的固定分档来使绘图看起来更整洁(例如,使用bins选项histbins = np.arange(0,48,2)。< / p>

试试这个:

import numpy as np

...

mybins = np.arange(0,48,2)

n, bins, patches = plt.hist(setA, bins=mybins, alpha=0.40 , label = 'setA', normed=True)  
n, bins, patches = plt.hist(setB, bins=mybins, alpha=0.40 , label = 'setB', normed=True)
n, bins, patches = plt.hist(setC, bins=mybins, alpha=0.40 , label = 'setC', normed=True)   

enter image description here

另一种选择是在一次调用中将所有三个直方图绘制到plt.hist,在这种情况下,您可以使用stacked=True选项,这可以进一步清理您的绘图。

注意:此方法将所有三个直方图标准化,因此总积分为1.它不会使所有三个直方图加起来相同。

n, bins, patches = plt.hist([setA,setB,setC], bins=mybins, 
                            label = ['setA','setB','setC'], 
                            normed=True, stacked=True)

enter image description here

或者,最后,如果堆积直方图不符合您的喜好,您可以通过在一次调用中再次绘制所有三个直方图来绘制彼此相邻的条形图,但从上面的行中删除stacked=True选项:

n, bins, patches = plt.hist([setA,setB,setC], bins=mybins, 
                            label = ['setA','setB','setC'], 
                            normed=True)

enter image description here

正如评论中所讨论的,当使用stacked=True时,normed选项只意味着所有三个直方图的总和将等于1,因此它们可能不会像其他方法那样被标准化

要解决此问题,我们可以使用np.histogram,并使用plt.bar绘制结果。

例如,使用上面相同的数据集:

mybins = np.arange(0,48,2)

nA,binsA = np.histogram(setA,bins=mybins,normed=True)
nB,binsB = np.histogram(setB,bins=mybins,normed=True)
nC,binsC = np.histogram(setC,bins=mybins,normed=True)

# Since the sum of each of these will be 1., lets divide by 3.,
# so the sum of the stacked histogram will be 1.
nA/=3.
nB/=3.
nC/=3.

# Use bottom= to set where the bars should begin
plt.bar(binsA[:-1],nA,width=2,color='b',label='setA')
plt.bar(binsB[:-1],nB,width=2,color='g',label='setB',bottom=nA)
plt.bar(binsC[:-1],nC,width=2,color='r',label='setC',bottom=nA+nB)

enter image description here