布里渊多样性指数

Question

我试图用Python计算Brillouin的多样性指数，根据使用pandas和numpy的列“a”。但出了点事。你知道吗

import pandas as pd
import numpy as np
def Brillouin_Index(x):
    for i in range(len(x)):
        x["Brillouin_Index"] = (np.log10(np.math.factorial(np.sum(x))) - np.sum(np.log10(np.math.factorial(x[i])))) / np.sum(x)
        return x
a = list("ABCDEADECS")
b = [12,23,12,12,32,34,21,2,10,5]
c = {"a":a,"b":b}
data = pd.DataFrame(c)
data
data.groupby("a").apply(Brillouin_Index)

我执行了上面的代码，有两个错误。你知道吗

TypeError: cannot convert the series to <class 'int'>
AttributeError: 'int' object has no attribute 'log10'

具体公式见以下链接Brillouin’s Diversity Index

我用其他软件计算每组的数值

1. HƏA=0.2965
2. H_B=0
3. HƏC=0.264
4. 高度=0.259
5. HƏE=0.08085
6. H_S=0

非常感谢！你知道吗

Answer 1

我用R来计算布里渊的分集指数组。组代码如下：

Brillouin_Diversity_Index <- function(x)
{  N <- sum(x)

 (log10(factorial(N)) - sum(log10(factorial(x)))) / N

}

dt <- data.table(x = c("A","B","C","D","E","A","D","E","C","S"),
y = c(12,23,12,12,32,34,21,2,10,5))
dt[,Brillouin_Diversity_Index(y),by = .(x)]

1. 第1版
2. 甲0.23021887
3. 亿英镑
4. 0.2641221加元
5. 直径0.25909105
6. 传真：0.08085185
7. 0.00000000新加坡元