What is Simpson’s diversity index?
Type how many individuals of each species you counted. The Simpson’s diversity index calculator gives Simpson’s index D, the index of diversity 1 − D, and the reciprocal index 1 ÷ D, worked from the exact counts.
- Simpson’s index of diversity (1 − D)
- 0.722222
Simpson’s index of diversity is 0.722222, with D = 0.277778, for 4 species.
- Simpson’s index (D)
- 0.277778
- Simpson’s reciprocal index (1 ÷ D)
- 3.6
- Species (S)
- 4
- Total (N)
- 100
Simpson’s index of diversity (1 − D): 0.722222. Simpson’s index of diversity is 0.722222, with D = 0.277778, for 4 species.
How to calculate
Computes Simpson’s diversity index from the number of individuals of each species: Simpson’s index D, the index of diversity 1 − D and the reciprocal index 1 ÷ D, with the species count.
Example with the default inputs (Individuals of each species [25, 15, 40, 20], Formula Σ n(n − 1) ÷ N(N − 1)): Simpson’s index of diversity is 0.722222, with D = 0.277778, for 4 species.
Method: D = Σ n(n − 1) ÷ (N(N − 1)), or D = Σ (n ÷ N)² for proportions; Simpson’s index of diversity = 1 − D; reciprocal index = 1 ÷ D.
- Each count is the number of individuals of one species; N is their sum and S the number of species with a count above 0.
- Σ n(n − 1) ÷ N(N − 1) is Simpson’s estimate from a sample (two individuals drawn without replacement); Σ p² treats the counts as proportions of the whole community and also accepts decimals.
- Higher 1 − D and 1 ÷ D mean more diversity; D itself is lower when diversity is higher.
Worked examples
Each example is checked against the calculator on every build.
- Individuals of each species 0.25, 0.15, 0.4, 0.2, Formula Σ p² gives Simpson’s index (D) 0.285, Simpson’s index of diversity (1 − D) 0.715, Simpson’s reciprocal index (1 ÷ D) 3.508772, Species (S) 4.Source: NIST, Dataplot Reference Manual, SIMPSON DIVERSITY INDEX (D = Σ (f_i ÷ n)², or Σ p_i² for proportions; lower values mean more diversity; Simpson 1949; proportions 0.25, 0.15, 0.40, 0.20 give 0.2850000), https://www.itl.nist.gov/div898/software/dataplot/refman2/auxillar/simpson.htm (retrieved 2026-10-18)
- Individuals of each species 25, 15, 40, 20, Formula Σ n(n − 1) ÷ N(N − 1) gives Simpson’s index (D) 0.277778, Simpson’s index of diversity (1 − D) 0.722222, Simpson’s reciprocal index (1 ÷ D) 3.6, Species (S) 4, Total (N) 100.Source: Simpson, E. H. (1949), Measurement of diversity, Nature 163, 688 (the index Σ π_i² and its estimate Σ n_i(n_i − 1) ÷ (N(N − 1))), https://doi.org/10.1038/163688a0; Tiffeau-Mayer, A. (2024), Unbiased estimation of sampling variance for Simpson’s diversity index, arXiv:2310.03439, Eqs. 1 and 2 (the probability that two individuals drawn at random are the same species, Σ p_i², and Simpson’s estimate Σ n_i(n_i − 1) ÷ (N(N − 1)) for pairs drawn without replacement), https://arxiv.org/abs/2310.03439 (retrieved 2026-10-18)
- Individuals of each species 10, 0, 10, Formula Σ n(n − 1) ÷ N(N − 1) gives Simpson’s index (D) 0.473684, Simpson’s index of diversity (1 − D) 0.526316, Simpson’s reciprocal index (1 ÷ D) 2.111111, Species (S) 2.Source: Simpson, E. H. (1949), Measurement of diversity, Nature 163, 688 (the index Σ π_i² and its estimate Σ n_i(n_i − 1) ÷ (N(N − 1))), https://doi.org/10.1038/163688a0
- Individuals of each species 1, 1, 1, Formula Σ n(n − 1) ÷ N(N − 1) gives Simpson’s index (D) 0, Simpson’s index of diversity (1 − D) 1, Species (S) 3.Source: Simpson, E. H. (1949), Measurement of diversity, Nature 163, 688 (the index Σ π_i² and its estimate Σ n_i(n_i − 1) ÷ (N(N − 1))), https://doi.org/10.1038/163688a0
How it works
Let n be the number of individuals of each species, N = Σ n the total, and S the number of species with n above 0.
Simpson’s index, with the default formula (a sample, drawn without replacement):
D = Σ n(n − 1) ÷ (N(N − 1))
With the Σ p² formula (proportions of the whole community):
D = Σ (n ÷ N)²
From D:
- Simpson’s index of diversity = 1 − D
- Simpson’s reciprocal index = 1 ÷ D (not shown when D is 0)
D is the chance that two individuals picked at random belong to the same species, so a smaller D, a larger 1 − D, and a larger 1 ÷ D all mean more diversity.
Rules. Type 1 to 1,000 counts; none may be negative. With Σ n(n − 1), every count must be a whole number and the total N must be at least 2. With Σ p², counts may be decimals (such as proportions or percent cover) and the total must be above 0. Counts of 0 add nothing and do not count toward S. A total above about 1.8 × 10³⁰⁸ (too large for a 64-bit float) has no answer. So does a 1 − D above 0 but below about 10⁻³²³ (too small for a 64-bit float), which only Σ p² with one count far below another gives; the page does not show it as 0. Typed values are read as exact decimals, so D, 1 − D and 1 ÷ D are exact fractions until they are rounded for display.
Output format. D, 1 − D and 1 ÷ D show 6 significant figures; S is a whole number.
Worked examples by hand
NIST’s proportions, Σ p². p = 0.25, 0.15, 0.40, 0.20. D = 0.0625 + 0.0225 + 0.16 + 0.04 = 0.285 (NIST: 0.2850000). 1 − D = 0.715; 1 ÷ D = 3.50877.
The same community as 100 counted individuals (the default). Counts 25, 15, 40, 20, so N = 100. Σ n(n − 1) = 25 × 24 + 15 × 14 + 40 × 39 + 20 × 19 = 600 + 210 + 1,560 + 380 = 2,750. D = 2,750 ÷ (100 × 99) = 0.277778; 1 − D = 0.722222; 1 ÷ D = 9,900 ÷ 2,750 = 3.6. S = 4.
Two species and an empty one. Counts 10, 0, 10, so N = 20 and S = 2. D = (90 + 0 + 90) ÷ (20 × 19) = 180 ÷ 380 = 0.473684; 1 − D = 0.526316; 1 ÷ D = 2.11111.
Three singletons. Counts 1, 1, 1: Σ n(n − 1) = 0, so D = 0 and 1 − D = 1. 1 ÷ D has no value.
Other questions people ask
How do I calculate Simpson’s diversity index?
For each species, multiply its count n by n − 1 and add the results. Divide by N(N − 1), where N is the total count: that is D. For counts of 25, 15, 40 and 20: (600 + 210 + 1,560 + 380) ÷ (100 × 99) = 2,750 ÷ 9,900 = 0.278. The index of diversity is 1 − D = 0.722.
What is the difference between D, 1 − D and 1 ÷ D?
D is the chance that two individuals picked at random are the same species, so a lower D means more diversity. 1 − D, Simpson’s index of diversity, is the chance that they differ, so higher means more diverse. 1 ÷ D, the reciprocal index, runs from 1 up to the number of species.
Should I use n(n − 1) or p²?
Use Σ n(n − 1) ÷ N(N − 1) for counts from a sample: Simpson showed it is an unbiased estimate of the chance that two individuals drawn without replacement match. Σ p² uses the proportions n ÷ N directly, as NIST’s Dataplot does, and suits proportions or percent cover. The two are close for large samples.
What is a good Simpson’s diversity index?
There is no fixed scale; compare sites counted the same way. 1 − D is 0 when there is only one species and gets closer to 1 as there are more species with more even numbers. A community of 4 species in the proportions 25%, 15%, 40% and 20% has 1 − D = 0.715 (Σ p²).
Why is there no reciprocal index for some counts?
When every species has a count of 1, no two individuals can be the same species, so Σ n(n − 1) is 0 and D is 0. Then 1 ÷ D has no value, and the page leaves it out.
How is Simpson’s index different from the Shannon index?
Both grow with the number of species and with how evenly the individuals are spread among them. Simpson’s D adds up p², so the most common species make up most of it. The Shannon index adds up −p × ln p, which gives rare species more weight.