📚 CIE A-Level Biology: Simpson’s Diversity Index and Its Applications | CIE A-Level 生物：辛普森多样性指数及其应用
Biodiversity is a word that appears throughout ecology, but how do we actually measure it? Simply counting the number of species, known as species richness, is not enough to describe how a community is structured. The Simpson’s Diversity Index (SDI) combines species richness and species evenness into a single value, making it one of the most useful quantitative tools in the CIE A-Level Biology syllabus.
“生物多样性”一词在生态学中处处可见，但究竟如何测量它？仅仅数一数物种数量（即物种丰富度）还不足以描述群落的组成结构。辛普森多样性指数（SDI）将物种丰富度与物种均匀度合并为一个数值，因而成为 CIE A-Level 生物学课程中最常用的定量工具之一。
1. Species Richness and Evenness | 物种丰富度与均匀度
Species richness is simply the number of different species found in a sample. For example, a pond containing 6 different species of invertebrates has a species richness of 6. However, richness ignores how many individuals each species has.
物种丰富度就是指样本中发现的物种数量。例如，一个池塘中有 6 种无脊椎动物，其物种丰富度就是 6。然而，丰富度忽略了一个问题：每个物种各有多少个体？
Species evenness describes how evenly individuals are distributed among the species. A community with 10 individuals each of two species is more even than a community with 19 individuals of species A and only 1 individual of species B, even though both communities have the same richness.
物种均匀度描述的是个体在物种间的分布均匀程度。一个含两个物种、各 10 个个体的群落，比一个含 19 个物种 A 个体、只有 1 个物种 B 个体的群落更均匀，尽管两者的丰富度相同。
The power of Simpson’s Diversity Index is that it captures both richness and evenness at the same time, which makes it far more sensitive to ecological change than richness alone.
辛普森多样性指数的强大之处在于它同时体现了丰富度和均匀度，因此其对生态变化的敏感度远高于单独使用丰富度。
2. Definition of Simpson’s Diversity Index | 辛普森多样性指数的定义
Simpson’s Diversity Index was proposed by the statistician Edward H. Simpson in 1949. It measures the probability that two individuals chosen independently and at random from a community belong to different species. If the probability is high, the community is regarded as highly diverse.
辛普森多样性指数是由统计学家爱德华·H·辛普森于 1949 年提出的。它衡量的是：从群落中随机且独立地抽取两个个体，它们属于不同物种的概率。如果该概率高，就认为这个群落具有较高的多样性。
In CIE A-Level Biology, the index is commonly written in the following form:
在 CIE A-Level 生物学中，该指数通常写成以下形式：
D = 1 − Σ (n/N)²
Here, n is the number of individuals of each species, N is the total number of individuals of all species, and Σ means “sum over all species”. The closer the value of D is to 1, the greater the diversity.
其中，n 为每个物种的个体数，N 为所有物种的个体总数，Σ 表示“对所有物种求和”。D 的值越接近 1，多样性越高。
3. The Formula and Its Symbols | 公式与符号解析
Let us break the formula down. For each species, we calculate n/N, which is the proportion of all individuals that belong to that species. We then square this proportion to obtain (n/N)². Squaring gives more weight to common species and reduces the effect of rare species.
我们来拆解一下这个公式。对每个物种，先计算 n/N，也就是该物种个体占全部个体的比例；然后将该比例平方，得到 (n/N)²。平方使常见物种获得更大权重，同时降低稀有物种的影响。
Summing the squared proportions gives the probability that two randomly chosen individuals belong to the same species. Subtracting this sum from 1 gives the probability that they belong to different species.
将所有平方比例相加，得到的是“随机抽取两个个体属于同一物种”的概率。用 1 减去这个总和，就得到“它们属于不同物种”的概率。
D = 1 − [ (n₁/N)² + (n₂/N)² + (n₃/N)² + … ]
This is why the index lies between 0 and 1. A value near 0 means very low diversity, while a value near 1 means very high diversity.
因此该指数取值在 0 到 1 之间。数值接近 0 表示多样性很低，接近 1 表示多样性很高。
4. Worked Example: Step-by-Step Calculation | 计算实例：分步骤计算
Consider two habitats, A and B, each containing 30 individuals distributed among three species. The data are shown below.
设有两个生境 A 和 B，各含 30 个个体，均分属于 3 个物种。数据如下表所示。
| Habitat | 生境 | Species 1 | 物种 1 | Species 2 | 物种 2 | Species 3 | 物种 3 | Total N | 总数 N |
| A | 10 | 10 | 10 | 30 |
| B | 25 | 3 | 2 | 30 |
For habitat A, each species contributes (10/30)². The sum is 3 × (10/30)² = 3 × 0.111 = 0.333. Therefore:
对生境 A，每个物种贡献 (10/30)²。总和为 3 × (10/30)² = 3 × 0.111 = 0.333。因此：
D A = 1 − 0.333 = 0.667
For habitat B, the calculation is (25/30)² + (3/30)² + (2/30)² = 0.694 + 0.010 + 0.004 = 0.708. Therefore:
对生境 B，计算为 (25/30)² + (3/30)² + (2/30)² = 0.694 + 0.010 + 0.004 = 0.708。因此：
D B = 1 − 0.708 = 0.292
Both habitats have the same species richness, but habitat A has a much higher Simpson index because its individuals are distributed more evenly among the species.
两个生境的物种丰富度相同，但生境 A 的辛普森指数远高于生境 B，因为 A 中个体在各物种间的分布更加均匀。
5. Interpreting the Index Values | 解读指数数值
The Simpson’s Diversity Index produces a value between 0 and 1. A value of 0 indicates that only one species is present, whereas a value of 1 indicates infinite diversity, in which every individual belongs to a different species.
辛普森多样性指数的取值在 0 到 1 之间。0 表示群落中只有一种物种；1 表示多样性无穷大，即每一个个体都属于不同的物种。
In practice, ecological communities fall somewhere between these extremes. A high value, for example 0.8 or above, indicates a complex community with many species and no single dominant species. A low value, for example below 0.3, indicates a community dominated by one or very few species.
在实际生态系统中，群落通常介于这两个极端之间。高值（例如 0.8 以上）表示群落结构复杂、物种丰富且没有单一优势种；低值（例如 0.3 以下）表示群落由某一个或极少数物种主导。
When comparing sites, students should not just quote the numbers; they should explain what a higher or lower D value means in ecological terms, such as greater habitat complexity, more stable food webs, or the effects of pollution and disturbance.
在比较不同地点时，学生不应只罗列数字，还应用生态学语言解释 D 值高或低的意义，例如生境复杂程度更高、食物网更稳定，或反映了污染和干扰的影响。
6. Simpson’s Diversity vs Simpson’s Dominance | 辛普森多样性指数与优势度指数的区别
A very common source of confusion in A-Level exams is the difference between the diversity index and the dominance index. Some textbooks define the Simpson index simply as D = Σ(n/N)² without subtracting from 1. This quantity is actually the Simpson’s Dominance Index.
A-Level 考试中一个常见的混淆点在于多样性指数与优势度指数的区别。有些教材把辛普森指数直接定义为 D = Σ(n/N)²，不进行 1 减运算。但这个量实际上是辛普森优势度指数。
Under the dominance form, a higher value means lower diversity, because a single species dominates the community. Under the CIE form D = 1 − Σ(n/N)², a higher value means higher diversity.
在优势度形式下，数值越高表示多样性越低，因为某个单一物种主导了群落。在 CIE 形式 D = 1 − Σ(n/N)² 中，数值越高表示多样性越高。
To avoid losing marks, always write down the formula you are using before substituting numbers. If the question does not specify a formula, use the CIE syllabus form D = 1 − Σ(n/N)² and clearly show your working.
为避免失分，代入数据前一定要先写出所使用的公式。若题目未指定公式，应使用 CIE 考纲
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