Python 年龄分布的Numpy梯形分布

我试图创建一个美国人口分布的粗略模型，以生成样本人口的随机年龄，并以下图作为来源

import numpy as np
import matplotlib.pyplot as p
%matplotlib inline

def agedistro(turn,end,size):
    pass
    totarea = turn + (end-turn)/2  # e.g. 50 + (90-50)/2
    areauptoturn = turn             # say 50
    areasloped = (end-turn)/2     # (90-50)/2
    size1= int(size*areauptoturn/totarea)
    size2= size- size1 
    s1 = np.random.uniform(low=0,high=turn,size= size1)  # (low=0.0, high=1.0, size=None)
    s2 = np.random.triangular(left=turn,mode=turn,right=end,size=size2) #(left, mode, right, size=None)
            # mode : scalar-  the value where the peak of the distribution occurs. 
            #The value should fulfill the condition left <= mode <= right.
    s3= np.concatenate((s1,s2)) # don't use add , it will add the numbers piecewise
    return s3

s3=agedistro(turn=50,end=90,size=1000000)    
p.hist(s3,bins=50)
p.show()

import numpy as np
import matplotlib.pyplot as p
%matplotlib inline

def agedistro(turn,end,size):
    pass
    totarea = turn + (end-turn)/2  # e.g. 50 + (90-50)/2
    areauptoturn = turn             # say 50
    areasloped = (end-turn)/2     # (90-50)/2
    size1= int(size*areauptoturn/totarea)
    size2= size- size1 
    s1 = np.random.uniform(low=0,high=turn,size= size1)  # (low=0.0, high=1.0, size=None)
    s2 = np.random.triangular(left=turn,mode=turn,right=end,size=size2) #(left, mode, right, size=None)
            # mode : scalar-  the value where the peak of the distribution occurs. 
            #The value should fulfill the condition left <= mode <= right.
    s3= np.concatenate((s1,s2)) # don't use add , it will add the numbers piecewise
    return s3

s3=agedistro(turn=50,end=90,size=1000000)    
p.hist(s3,bins=50)
p.show()

---

我觉得这可以最简单地用梯形分布来建模，梯形分布在50岁左右开始下降之前保持一致。然而，numpy似乎没有提供利用该分布函数的能力。因此，我想知道是否可以“组合”两个分布函数（在本例中，最大值为50的均匀分布函数和最小值为51、最大值为100的三角形分布函数）。这可能吗？有没有一种方法可以直接用python表示梯形分布函数

是的，您可以任意组合样本。只需使用

np.concatenate

import numpy as np
import matplotlib.pyplot as p
%matplotlib inline

def agedistro(turn,end,size):
    pass
    totarea = turn + (end-turn)/2  # e.g. 50 + (90-50)/2
    areauptoturn = turn             # say 50
    areasloped = (end-turn)/2     # (90-50)/2
    size1= int(size*areauptoturn/totarea)
    size2= size- size1 
    s1 = np.random.uniform(low=0,high=turn,size= size1)  # (low=0.0, high=1.0, size=None)
    s2 = np.random.triangular(left=turn,mode=turn,right=end,size=size2) #(left, mode, right, size=None)
            # mode : scalar-  the value where the peak of the distribution occurs. 
            #The value should fulfill the condition left <= mode <= right.
    s3= np.concatenate((s1,s2)) # don't use add , it will add the numbers piecewise
    return s3

s3=agedistro(turn=50,end=90,size=1000000)    
p.hist(s3,bins=50)
p.show()

将numpy导入为np
将matplotlib.pyplot导入为p
%matplotlib内联
def AGEDRISTRO（转弯、结束、大小）：
通过
总面积=转弯+（末端转弯）/2#例如50+（90-50）/2
areauptoturn=转身，比如说50
区域坡度=（末端转弯）/2#（90-50）/2
size1=int（大小*面积向上翻转/总面积）
size2=大小-大小1
s1=np.random.uniform（低=0，高=turn，大小=size1）#（低=0.0，高=1.0，大小=None）
s2=np.random.Triangal（左=转弯，模式=转弯，右=结束，大小=大小2）#（左，模式，右，大小=无）
#模式：标量-分布峰值出现的值。
#该值应满足以下条件，仅供参考：很快就会出现在您身边：@Warren在下一个版本之前，我能做些什么吗？哇，这真的很有用。谢谢