📚 Simpson’s Index of Diversity | 辛普森多样性指数
Biodiversity is a cornerstone of modern biology, and the ability to measure it quantitatively is essential for any ecologist. Simpson’s Index of Diversity (D) offers a robust tool that accounts for both the number of species and their relative abundance, helping students and researchers compare habitats with confidence. This article unpacks the index step by step, fully aligned with the Cambridge A-Level Biology syllabus.
生物多样性是现代生物学的基石，而定量测量生物多样性的能力对任何生态学家都至关重要。辛普森多样性指数（D）提供了一个强有力的工具，能够同时考虑物种数量和它们的相对丰度，帮助学生和研究人员自信地比较不同栖息地。本文将逐步解析这一指数，完全符合剑桥A-Level生物学课程大纲。
1. Understanding Biodiversity | 理解生物多样性
Biodiversity can be studied at genetic, species, and ecosystem levels. At A-Level, the focus is on species diversity, which has two key components: species richness (the total number of different species) and species evenness (how similar the abundances of the different species are). A community with high species diversity tends to be more resilient and productive.
生物多样性可以从遗传、物种和生态系统三个层次研究。A-Level的重点在于物种多样性，它包含两个关键组成部分：物种丰富度（不同物种的总数）和物种均匀度（各个物种的丰度有多相似）。具有高物种多样性的群落往往更具恢复力和生产力。
2. Species Richness vs Evenness | 物种丰富度与均匀度
Two habitats may have the same species richness yet support very different diversity patterns. For instance, imagine Community A with five species but one species dominates 90% of individuals, while Community B has the same five species each accounting for 20% of individuals. Richness alone would treat them as equal, but clearly Community B is more balanced. Simpson’s Index captures this difference.
两个栖息地可能具有相同的物种丰富度，但却呈现出截然不同的多样性模式。例如，假设群落A有五个物种，但其中一个物种占了个体总数的90%，而群落B拥有相同的五个物种，每个物种各占20%。仅凭丰富度会将它们视为相同，但显然群落B更加均衡。辛普森指数能够捕捉这种差异。
| Species | Community A (n) | Community B (n) |
|---|---|---|
| Sp. 1 | 90 | 20 |
| Sp. 2 | 4 | 20 |
| Sp. 3 | 3 | 20 |
| Sp. 4 | 2 | 20 |
| Sp. 5 | 1 | 20 |
| Total (N) | 100 | 100 |
Both communities have identical species richness (5 species), but their evenness is dramatically different. Diversity indices must therefore account for relative abundance, not just the count of species.
两个群落的物种丰富度完全相同（都是5种），但均匀度却大相径庭。因此，多样性指数必须考虑相对丰度，而不仅仅是物种的数目。
3. Introducing Simpson’s Index | 辛普森指数简介
Simpson’s Index was proposed by Edward H. Simpson in 1949. It calculates the probability that any two individuals taken at random from the community belong to the same species. To turn this into a diversity measure, we subtract this probability from 1. The resulting value, D, ranges from 0 (virtually no diversity – a single species dominates completely) to a value approaching 1 (very high diversity, with many species and great evenness).
辛普森指数由爱德华·H·辛普森于1949年提出。它计算的是从群落中随机抽取的两个个体属于同一物种的概率。为了将其转化为多样性度量，我们用1减去这个概率。所得值D的范围从0（几乎没有多样性，单一物种完全占优势）到接近1（多样性极高，物种众多且均匀度极好）。
4. The Formula Explained | 公式解释
The most commonly used formula in Cambridge A-Level Biology is the simplified Simpson’s Index of Diversity:
剑桥A-Level生物学中最常用的公式是简化版辛普森多样性指数：
D = 1 − Σ (n / N)²
where n = number of individuals of a particular species, N = total number of individuals of all species, and Σ means ‘sum of’. For greater accuracy, especially with small samples, the adjusted formula is:
其中n = 某个特定物种的个体数，N = 所有物种的个体总数，Σ表示“求和”。为了更高的准确性，特别是样本较小时，可使用修正公式：
D = 1 − [ Σ n(n − 1) / N(N − 1) ]
Both versions give very similar results when N is large. In exams, you will usually be given the formula, but you must be able to use it correctly.
当N很大时，这两个公式的结果非常接近。考试中通常会给出公式，但你必须能够正确使用它。
5. Step-by-Step Calculation | 逐步计算
Step 1: Identify each species and record its number of individuals, n.
步骤1：鉴定每个物种并记录其个体数量，n。
Step 2: Calculate the total number of individuals, N = Σ n.
步骤2：计算个体总数，N = Σ n。
Step 3: For each species, calculate n / N, then square the result to get (n / N)².
步骤3：对每个物种，计算n / N，然后将结果平方得到 (n / N)²。
Step 4: Add up all the (n / N)² values to obtain Σ (n / N)².
步骤4：将所有 (n / N)² 值相加，求得 Σ (n / N)²。
Step 5: Subtract this sum from 1: D = 1 − Σ (n / N)².
步骤5：从1中减去该总和：D = 1 − Σ (n / N)²。
6. Interpreting D Values | 解读D值
A high value of D (e.g. 0.8 or above) indicates high diversity – many species and a relatively even distribution. A low value (e.g. 0.2 or below) points to low diversity, often because one or two species are overwhelmingly dominant. If a community contains only one species, then n / N = 1, Σ (n / N)² = 1, and D = 1 − 1 = 0.
D值较高（如0.8或以上）表示多样性高——物种多且分布相对均匀。D值较低（如0.2或以下）则意味着多样性低，通常是由于一两个物种占据了绝对优势。如果一个群落只有一个物种，那么n / N = 1，Σ (n / N)² = 1，因此D = 1 − 1 = 0。
Values of D are dimensionless and can be compared directly between different habitats, provided the sampling method and effort are consistent.
D值是无量纲的，只要采样方法和努力程度一致，就可以在不同栖息地之间直接比较。
7. Worked Example: A Woodland Plot | 实例：林地样方
A student surveys a small woodland plot and records the following tree counts: Oak 15, Beech 10, Birch 5. Use Simpson’s Index to calculate the diversity of this plot.
一名学生调查了一小块林地样方，记录了以下树木数量：橡树15，山毛榉10，桦树5。请使用辛普森指数计算该样方的多样性。
| Species | Number (n) | n / N | (n / N)² |
|---|---|---|---|
| Oak | 15 | 15/30 = 0.500 | 0.250 |
| Beech | 10 | 10/30 ≈ 0.333 | 0.111 |
| Birch | 5 | 5/30 ≈ 0.167 | 0.028 |
| Totals | N = 30 | Σ = 0.389 |
N = 15 + 10 + 5 = 30. Sum of (n/N)² = 0.250 + 0.111 + 0.028 = 0.389. Therefore, D = 1 − 0.389 = 0.611. The woodland has a moderate level of diversity.
N = 15 + 10 + 5 = 30。 (n/N)² 总和 = 0.250 + 0.111 + 0.028 = 0.389。因此，D = 1 − 0.389 = 0.611。这片林地具有中等程度的多样性。
8. Comparing Communities Using D | 用D值比较群落
Returning to the earlier Communities A and B (from Section 2), let us compute Simpson’s Index for each. For Community A: n/N values are 0.9, 0.04, 0.03, 0.02, 0.01; their squares sum to 0.81 + 0.0016 + 0.0009 + 0.0004 + 0.0001 = 0.813. Hence Dₐ = 1 − 0.813 = 0.187. For Community B, each species has n/N = 0.2, so (0.2)² = 0.04; sum = 5 × 0.04 = 0.20. Thus D_B = 1 − 0.20 = 0.80.
回到前文群落A和B（第2节），我们来计算各自的辛普森指数。群落A：n/N分别为0.9、0.04、0.03、0.02、0.01；它们的平方和为0.81 + 0.0016 + 0.0009 + 0.0004 + 0.0001 = 0.813。因此Dₐ = 1 − 0.813 = 0.187。群落B：每个物种的n/N = 0.2，故(0.2)² = 0.04；总和 = 5 × 0.04 = 0.20。于是D_B = 1 − 0.20 = 0.80。
Despite having the same species richness, Community B’s diversity is far higher (0.80 versus 0.19) because its evenness is much greater. This demonstrates the power of Simpson’s Index to discriminate between habitats that mere species counts would judge as equal.
尽管物种丰富度相同，群落B的多样性却远高于群落A（0.80对0.19），因为它的均匀度大得多。这表明辛普森指数能够区分那些仅凭物种计数会被视为等同的栖息地，效果十分强大。
9. Limitations and Assumptions | 局限性与假设
Like any ecological tool, Simpson’s Index has limitations:
与任何生态学工具一样，辛普森指数也存在局限性：
• Sampling must be random and adequately sized – biased or small samples distort D.
• 采样必须随机且样本量足够——有偏差或过小的样本会扭曲D值。
• The index
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