SME
Ecology
Community interactions
Use this for “A plot holds 50, 30 and 20 plants of three species. What are H′ and the evenness?”
Chart
50 + 30 + 20 + 0 = 100.
This is an example. Type your own values to start a new problem.
n₁
Species 1 count
example
n₂
Species 2 count
example
n₃
Species 3 count
example
n₄
Species 4 count
example
N
Individuals in all
worked out (not typed)
S
Species present
worked out (not typed)
H′
Shannon index
worked out (not typed)
J
Evenness
worked out (not typed)
Formulas
N
=
n₁
+
n₂
+
n₃
+
n₄
100 = 50 + 30 + 20 + 0
S
=
species
among
n₁
,
n₂
,
n₃
,
n₄
above
0
3 = species among 50, 30, 20, 0 above 0
H′
=
−((
n₁
N
)
×
ln(
n₁
N
)
+
(
n₂
N
)
×
ln(
n₂
N
)
+
(
n₃
N
)
×
ln(
n₃
N
)
+
(
n₄
N
)
×
ln(
n₄
N
))
1.03
=
−((
50
100
)
×
ln(
50
100
)
+
(
30
100
)
×
ln(
30
100
)
+
(
20
100
)
×
ln(
20
100
)
+
(
0
100
)
×
ln(
0
100
))
J
=
H′
ln(
S
)
0.9372
=
1.03
ln(3)
Assumptions
•
Each species’ share is pᵢ = nᵢ ÷ N, and H′ = −Σ pᵢ ln pᵢ adds one term a species.
•
Evenness J = H′ ÷ ln S is 1 when every species is equally common and near 0 when one dominates.
•
A species with a count of 0 is absent: it adds no term and is not counted in S.
•
The count is a fair sample, so its shares stand for the whole community’s.
Step-by-step
WE KNOW
n₁ = 50, n₂ = 30, n₃ = 20, n₄ = 0
FIND
individuals in all (N), species present (S), Shannon index (H′), evenness (J)
Step 1 · Find individuals in all (N)
Use N = n₁ + n₂ + n₃ + n₄
Add the counts of the four species.
N
=
50
+
30
+
20
+
0
N
=
80
+
20
N
=
100
Step 2 · Find species present (S)
Use S = species among n₁, n₂, n₃, n₄ above 0
Each species with at least one individual counts 1; a count of 0 counts nothing.
S
=
1
+
1
+
1
+
0
S
=
3
Step 3 · Find Shannon index (H′)
Use H′ = −((n₁ ÷ N) × ln(n₁ ÷ N) + (n₂ ÷ N) × ln(n₂ ÷ N) + (n₃ ÷ N) × ln(n₃ ÷ N) + (n₄ ÷ N) × ln(n₄ ÷ N))
Each species adds its share times the ln of its share; the minus makes the sum positive.
H′
=
−((
n₁
N
)
×
ln(
n₁
N
)
+
(
n₂
N
)
×
ln(
n₂
N
)
+
(
n₃
N
)
×
ln(
n₃
N
))
H′
=
−((
50
100
)
×
ln(
50
100
)
+
(
30
100
)
×
ln(
30
100
)
+
(
20
100
)
×
ln(
20
100
))
H′ = −(0.5 × ln 0.5 + 0.3 × ln 0.3 + 0.2 × ln 0.2)
H′ = −((−0.3466) + (−0.3612) + (−0.3219))
H′ = −(−1.03)
H′ = 1.02965
Step 4 · Find evenness (J)
Use J = H′ ÷ ln(S)
ln
S
is
the
largest
H′
that
S
species
can
give:
J
is
the
share
of
it
reached.
J
=
1.02965
ln(3)
J
=
1.0297
1.0986
J
=
0.9372
Check
Put the numbers back in. Both sides should match.
100 = 50 + 30 + 20 + 0 ✓
3 = 1 + 1 + 1 + 0 ✓
1.03 = −(0.5 × ln 0.5 + 0.3 × ln 0.3 + 0.2 × ln 0.2) ✓
0.9372
=
1.02965
ln(3)
✓
Topic check
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