Biodiversity Index Calculator
Calculate Shannon-Wiener diversity index (H'), Simpson's diversity index (1−D), species richness (S), and Pielou's evenness (J) from any species count data — with full step-by-step working, per-species contribution table, and community comparison.
Accepted formats:
Comma-separated counts: 10, 5, 8, 3, 12
Named species: Oak: 15, Pine: 8, Birch: 12
Line-separated (one per line) or Excel paste (tab-separated)
| # | Species Name (optional) | Count (nᵢ) | Del |
|---|
Species Richness
—S = number of distinct species
—Shannon-Wiener H′
—H′ = −Σ(pᵢ × ln pᵢ)
—Simpson's 1−D
—D = Σpᵢ² | 1/D = —
—Pielou's Evenness J
—J = H′ / ln(S)
—Key insight: Each species' contribution to H′ is pᵢ × ln(pᵢ). This is maximised at intermediate pᵢ values — very rare species (pᵢ → 0) and very dominant species (pᵢ → 1) both contribute less to Shannon diversity. Species evenness, not just richness, drives a high Shannon index.
| Species | Count (nᵢ) | Relative Abundance (pᵢ) | pᵢ × ln(pᵢ) (Shannon contrib.) | pᵢ² (Simpson contrib.) |
|---|
Enter species count data for two communities to compare all four diversity indices simultaneously. Use the pre-filled example to see how an old-growth forest compares with a plantation monoculture.
Community A
Name: count format, comma-separated or one per line
Community B
Name: count format, comma-separated or one per line
| Index | Community A | Community B | Higher Diversity |
|---|
| Index | Formula | Range | High diversity = | Common Use |
|---|---|---|---|---|
| Species Richness S | Count of species | 0 to ∞ | High S | Simplest measure |
| Shannon H′ (nats) | −Σpᵢ ln(pᵢ) | 0 to ln(S) | High H′ | Most widely used |
| Shannon H′ (bits) | −Σpᵢ log₂(pᵢ) | 0 to log₂(S) | High H′ | Information theory |
| Simpson D | Σpᵢ² | 0 to 1 | LOW D (dominance) | Dominance focus |
| Simpson 1−D | 1 − Σpᵢ² | 0 to 1 | High 1−D | Diversity focus |
| Simpson 1/D | 1 / Σpᵢ² | 1 to S | High 1/D | Intuitive scale |
| Pielou J | H′ / ln(S) | 0 to 1 | High J (max = 1) | Evenness only |
Shannon H′ Typical Values by Ecosystem
Use this chart to compare your calculated Shannon H′ values against expected ranges for different ecosystem types. Shannon diversity index values above 3 are rare and indicate exceptional biodiversity.
Shannon H′ Interpretation Guide
| H′ Value | Interpretation | Typical Context |
|---|---|---|
| H′ = 0 | Only 1 species — no diversity | Complete monoculture |
| H′ < 1 | Low diversity | Heavily disturbed habitat |
| H′ 1–2 | Moderate diversity | Agricultural land, degraded habitat |
| H′ 2–3 | Good diversity | Temperate forests, healthy grasslands |
| H′ > 3 | High diversity (rare) | Tropical rainforest, coral reef |
Simpson 1−D Interpretation Guide
| 1−D Value | D Value | Interpretation |
|---|---|---|
| 0 – 0.3 | 0.7 – 1.0 | Low diversity — one or few species dominate |
| 0.3 – 0.6 | 0.4 – 0.7 | Moderate diversity |
| 0.6 – 0.8 | 0.2 – 0.4 | Good diversity |
| 0.8 – 1.0 | 0 – 0.2 | High diversity — species well distributed |
What Is the Shannon-Wiener Diversity Index — Definition and Formula
This biodiversity index calculator computes the Shannon-Wiener diversity index (H′), Simpson's diversity index (1−D), species richness (S), and Pielou's evenness (J) from any species count data entered as a table or pasted text. The Shannon diversity index calculator above shows every calculation step and per-species contribution, making it the ideal tool for ecology students, researchers, and field biologists.
The Shannon index is widely used because it captures two components of biodiversity simultaneously: species richness (how many species) and species evenness (how equally they are distributed). A community with ten species all equally abundant has a higher Shannon index than one where ten species exist but a single species accounts for 90% of individuals.
Shannon-Wiener Diversity Index — Formula and Calculation
Variable definitions for the Shannon-Wiener index formula:
- H′ = Shannon-Wiener diversity index (nats when using natural log)
- pᵢ = proportion of species i = nᵢ / N
- nᵢ = number of individuals of species i
- N = total number of individuals = Σnᵢ
- Σ = sum over all S species
- ln = natural logarithm (can also use log₂ for bits or log₁₀)
The Shannon diversity index combines species richness AND evenness into a single number. The maximum possible Shannon index is H′_max = ln(S) — achieved only when all S species have perfectly equal abundance. The closer H′ is to ln(S), the more even the community.
Log Base Choices for the Shannon Index
The Shannon-Wiener index formula can use any logarithm base. The choice of log base changes the numerical value but preserves all rankings between communities:
- Natural log (nats) — most common in ecology, used by default in this Shannon diversity index calculator
- Log₂ (bits) — information theory context; H′_max = log₂(S)
- Log₁₀ — occasionally used; H′_max = log₁₀(S)
Critical: Never compare Shannon H′ values calculated with different log bases — nats and bits are not numerically comparable. Always state the log base when reporting Shannon diversity index values.
Shannon Diversity Index Interpretation
| H′ (nats) | Interpretation |
|---|---|
| H′ = 0 | Only one species present — no diversity (not "no species") |
| H′ = 1 – 2 | Moderate diversity |
| H′ = 2 – 3 | Good diversity |
| H′ > 3 | High diversity (rare in most ecosystems) |
| H′ = ln(S) | Maximum possible — all species equally abundant (J = 1) |
Simpson's Diversity Index — D, 1−D, and 1/D Forms
Critical note: Confusion between the three forms of Simpson's diversity index is one of the most common errors in biodiversity reporting. Always specify which form you are using.
Form A — Simpson's D (dominance index): D = Σpᵢ² ranges from 0 (infinite diversity) to 1 (complete dominance by one species). A HIGH D means LOW diversity. D measures the probability that two randomly selected individuals belong to the same species. For finite samples: D = Σ(nᵢ(nᵢ−1)) / (N(N−1)).
Form B — Simpson's Diversity Index 1−D: 1−D = 1 − Σpᵢ² ranges from 0 (no diversity) to 1 (maximum diversity). A HIGH 1−D means HIGH diversity. This is the most commonly reported form in modern ecology because it is intuitive.
Form C — Simpson's Reciprocal Index 1/D: 1/D = 1/Σpᵢ² ranges from 1 (no diversity, one species) to S (all species equally abundant). The reciprocal gives an intuitive value: 1/D = 5 means diversity equivalent to 5 equally abundant species.
Worked Example — All Three Simpson Forms from Bird Survey Data
Species: Robin=15, Sparrow=22, Blackbird=8, Blue tit=5, Wren=10 | N=60
- Calculate proportions: p₁=15/60=0.250, p₂=22/60=0.367, p₃=8/60=0.133, p₄=5/60=0.083, p₅=10/60=0.167
- D = Σpᵢ² = 0.250²+0.367²+0.133²+0.083²+0.167² = 0.0625+0.1346+0.0178+0.0069+0.0278 = D = 0.2496
- 1−D = 1 − 0.2496 = 1−D = 0.7504 (good diversity)
- 1/D = 1 / 0.2496 = 1/D = 4.006 (equivalent to ~4 equally abundant species)
Species Richness vs Species Evenness — The Difference
Species richness is the simplest biodiversity measure — it is simply the count of distinct species present in a sample (S). Species richness tells you how many species exist but nothing about how equally they are distributed.
Species evenness measures how equally individuals are distributed among species. Two communities can have identical species richness but radically different evenness — and thus very different Shannon diversity index values.
Same Species Richness, Very Different Shannon Index
| Community | Species counts | S (richness) | H′ (Shannon) | J (evenness) |
|---|---|---|---|---|
| Community A | 90, 5, 5 | 3 | 0.33 nats | 0.30 |
| Community B | 33, 33, 34 | 3 | 1.10 nats ≈ ln(3) | ≈ 1.00 |
Both communities have S=3 (identical species richness). But Community B has nearly three times higher Shannon H′ because species are evenly distributed. Community A is dominated by one species (90 individuals) — low species evenness collapses the Shannon index.
How to Calculate Species Evenness — Pielou's J
Pielou's J measures species evenness independently of species richness. J = 1 means all species are equally abundant (maximum possible Shannon diversity for S species). J = 0 means one species completely dominates. To find species evenness: divide the observed H′ by the maximum possible H′ = ln(S).
How to find species richness: Count the number of distinct species in your sample. Species richness S does not involve any formula — it is a simple count. Every other index (Shannon H′, Simpson D, Pielou J) depends on species richness S.
How to Measure Biodiversity — Which Index to Use?
There is no single universally "best" biodiversity index — each answers a slightly different ecological question. Ecologists typically report multiple indices together for a complete picture of how to measure biodiversity in an ecosystem.
- Use S (species richness) when you want the pure count of species — useful for rapid biodiversity surveys and conservation planning. The simplest way to measure biodiversity.
- Use Shannon H′ when you want to balance species richness AND species evenness — this is the most common choice in ecology. The Shannon diversity index gives more weight to rare species than Simpson does.
- Use Simpson 1−D when you want to emphasise dominant species and are less concerned with rare species. Simpson is more robust to small sample sizes than Shannon.
- Use Pielou's J when you specifically want to measure species evenness independent of species richness — J allows evenness comparison between communities of different size.
Ways of measuring biodiversity span from simple species richness counts to information-theoretic Shannon indices to probability-based Simpson indices. No single measure captures all aspects of biodiversity. The biodiversity index calculator above computes all four simultaneously.
Shannon-Wiener Index — Worked Examples Step by Step
Example 1 — 3 Species, Equal Abundances (Maximum Evenness)
- Species A=10, B=10, C=10. N=30, S=3
- p₁=p₂=p₃ = 10/30 = 0.333
- Each pᵢ×ln(pᵢ) = 0.333×ln(0.333) = 0.333×(−1.099) = −0.366
- Σ(pᵢ×ln(pᵢ)) = 3 × (−0.366) = −1.099
- H′ = −(−1.099) = 1.099 = ln(3) ✓ — equal abundances achieve maximum Shannon index
- J = H′/ln(S) = 1.099/1.099 = 1.000 — perfect evenness
Example 2 — 3 Species, One Dominant (Low Evenness)
- Oak=90, Birch=3, Pine=2. N=95, S=3
- p₁=0.947, p₂=0.032, p₃=0.021
- Shannon contributions: 0.947×ln(0.947)=−0.052; 0.032×ln(0.032)=−0.110; 0.021×ln(0.021)=−0.082
- H′ = −(−0.052−0.110−0.082) = −(−0.244) = 0.244 nats — very low Shannon diversity
- J = 0.244/ln(3) = 0.244/1.099 = 0.222 — very uneven community
Example 3 — Real Bird Survey (5 species)
- Robin=15, Sparrow=22, Blackbird=8, Blue tit=5, Wren=10. N=60, S=5
- Proportions: 0.250, 0.367, 0.133, 0.083, 0.167
- Shannon contributions: −0.347, −0.368, −0.268, −0.207, −0.299
- Σ(pᵢ×ln(pᵢ)) = −1.489 → H′ = 1.489 nats
- H′_max = ln(5) = 1.609; J = 1.489/1.609 = 0.925
- D = Σpᵢ² = 0.2496; 1−D = 0.750; 1/D = 4.006
Example 4 — Comparing Communities: S Rankings vs H′ Rankings Can Differ
Community X: 5 species (A=50, B=5, C=5, D=5, E=5) vs Community Y: 3 species (P=20, Q=20, R=20)
- Community X: S=5 (higher richness), H′ = 0.981 nats, J = 0.610
- Community Y: S=3 (lower richness), H′ = 1.099 nats, J = 1.000
- Community Y has higher Shannon H′ despite lower species richness S — because it has perfect species evenness while X is dominated by species A.
- This demonstrates why reporting Shannon H′ alongside S is essential — richness alone can be misleading.
Common Mistakes in Biodiversity Index Calculations
Mistake 1 — Using D instead of 1−D and concluding "high diversity": Simpson's D=0.8 means LOW diversity (one species dominates). Simpson's 1−D=0.8 means HIGH diversity. This is the most common error in student reports — always check which Simpson form you are using.
Mistake 2 — Using counts nᵢ directly in the Shannon formula instead of proportions pᵢ: H′ = −Σ(pᵢ×ln(pᵢ)) requires proportions (0 to 1), not raw counts. Always divide each nᵢ by total N first.
Mistake 3 — Mixing log bases between comparisons: Shannon H′(nats) = 1.609 and Shannon H′(bits) = 2.322 describe the same community (5 equally abundant species) but are numerically different. Never compare Shannon values from different log bases.
Mistake 4 — Treating H′=0 as "no species present": H′=0 means only ONE species is present (p₁=1, ln(1)=0, so H′=0). It does not mean the sample is empty. A monoculture of 10,000 individuals of one species gives H′=0.
Mistake 5 — Forgetting to divide by ln(S) when computing Pielou's J evenness: J = H′/ln(S) — NOT H′/log₂(S) or H′/S. The denominator must be H′_max = ln(S), using the same log base as H′. If you calculated H′ using log₂, then H′_max = log₂(S).
Worked Examples — 6 Complete Step-by-Step Problems
Problem 1 — Simple 3-species community with Shannon and Simpson
Species A=20, B=30, C=10. N=60, S=3
- Proportions: pA=0.333, pB=0.500, pC=0.167
- Shannon: −(0.333×ln(0.333) + 0.500×ln(0.500) + 0.167×ln(0.167)) = −(−0.366−0.347−0.298) = H′ = 1.011
- J = 1.011/ln(3) = 1.011/1.099 = 0.920
- D = 0.333²+0.500²+0.167² = 0.111+0.250+0.028 = 0.389; 1−D = 0.611; 1/D = 2.570
Problem 2 — Maximum species evenness verification
5 species each with 10 individuals. N=50, S=5, each pᵢ=0.2
- Each pᵢ×ln(pᵢ) = 0.2×ln(0.2) = 0.2×(−1.6094) = −0.3219
- Σ = 5×(−0.3219) = −1.6094
- H′ = 1.6094 = ln(5) ✓ — confirms maximum Shannon index equals ln(S)
- J = 1.6094/1.6094 = 1.000 — perfect evenness
- D = 5×(0.2²) = 5×0.04 = 0.200; 1−D = 0.800; 1/D = 5.000 = S ✓
Problem 3 — Pielou's evenness calculation with 4 species
Frog=45, Toad=30, Salamander=15, Newt=10. N=100
- p = 0.45, 0.30, 0.15, 0.10
- Shannon contributions: 0.45×ln(0.45)=−0.361; 0.30×ln(0.30)=−0.361; 0.15×ln(0.15)=−0.285; 0.10×ln(0.10)=−0.230
- H′ = −(−1.237) = 1.237 nats
- H′_max = ln(4) = 1.386; J = 1.237/1.386 = 0.893
Problem 4 — Low diversity, dominant species
Dandelion=180, Clover=12, Grass=8. N=200
- p = 0.900, 0.060, 0.040
- H′ = −(0.900×ln(0.900) + 0.060×ln(0.060) + 0.040×ln(0.040)) = −(−0.095−0.170−0.147) = 0.412 nats
- D = 0.81+0.0036+0.0016 = 0.815; 1−D = 0.185 (Low diversity — one species dominates)
- J = 0.412/ln(3) = 0.412/1.099 = 0.375 — uneven community
Problem 5 — Community comparison showing H′ vs S disagreement
Forest A: 6 species (100,5,5,5,5,5). Forest B: 4 species (30,30,30,30).
- Forest A: S=6, H′=0.647, J=0.361 — high richness, very low evenness
- Forest B: S=4, H′=1.386=ln(4), J=1.000 — lower richness, perfect evenness
- Shannon ranks B higher despite lower species richness — evenness matters more here
Problem 6 — Full calculation including Simpson finite sample formula
Species X=8, Y=12, Z=5. N=25, S=3
- p = 0.320, 0.480, 0.200
- D (proportion formula) = 0.102+0.230+0.040 = 0.372
- D (finite sample) = Σnᵢ(nᵢ−1)/N(N−1) = (8×7+12×11+5×4)/(25×24) = (56+132+20)/600 = 208/600 = 0.347
- Using proportion formula: 1−D = 0.628; 1/D = 2.688
- H′ = −(0.320×ln(0.320)+0.480×ln(0.480)+0.200×ln(0.200)) = 1.075 nats; J = 0.978
Frequently Asked Questions
Related Calculators
| H′ = 0 | 1 species only |
| H′ < 1 | Low diversity |
| H′ 1–2 | Moderate |
| H′ 2–3 | Good diversity |
| H′ > 3 | High (rare) |
| = ln(S) | Perfect evenness |
Simpson 1−D Interpretation
| 0 – 0.3 | Low diversity |
| 0.3 – 0.6 | Moderate |
| 0.6 – 0.8 | Good diversity |
| 0.8 – 1.0 | High diversity |