UniFrac Calculations

Introduction

The different UniFrac algorithms are listed below, along with examples for calculating them.


Input Data

library(ecodive)

counts <- matrix(
  data     = c(0, 0, 9, 3, 3, 1, 4, 2, 8, 0), 
  ncol     = 2, 
  dimnames = list(paste0('Species_', 1:5), c('Sample_A', 'Sample_B')) )

tree <- read_tree(
  underscores = TRUE,
  newick      = "
    (((Species_1:0.8,Species_2:0.5):0.4,Species_3:0.9):0.2,(Species_4:0.7,Species_5:0.3):0.6);" )

L <- tree$edge.length
A <- c(9,0,0,0,9,6,3,3)
B <- c(7,5,1,4,2,8,8,0)

# local({ # man/figures/unifrac-tree.png
#     
#   par(xpd = NA)
#   ape::plot.phylo(
#     x          = tree, 
#     direction  = 'downwards', 
#     srt        = 90, 
#     adj        = 0.5, 
#     no.margin  = TRUE,
#     underscore = TRUE,
#     x.lim      = c(0.5, 5.5) )
#   
#   ape::edgelabels(tree$edge.length, bg = 'white', frame = 'none', adj = -0.4)
# })
# 
# local({ # man/figures/unifrac-weights.png
#   
#   tree$edge.length <- c(1, 1, 1, 1, 2, 1, 2, 2)
#   
#   par(xpd = NA)
#   ape::plot.phylo(
#     x               = tree, 
#     direction       = 'downwards', 
#     srt             = 90, 
#     adj             = 0.5, 
#     no.margin       = TRUE,
#     underscore      = TRUE,
#     x.lim           = c(.8, 6) )
#   
#   ape::edgelabels(1:8, frame = 'circle')
#   
#   ape::edgelabels(paste('A =', A), bg = 'white', frame = 'none', adj = c(-0.4, -1.2))
#   ape::edgelabels(paste('B =', B), bg = 'white', frame = 'none', adj = c(-0.4,  0.0))
#   ape::edgelabels(paste('L =', L), bg = 'white', frame = 'none', adj = c(-0.3,  1.2))
# })
Sample_A Sample_B
Species_1 `r counts[1,1]` `r counts[1,2]`
Species_2 `r counts[2,1]` `r counts[2,2]`
Species_3 `r counts[3,1]` `r counts[3,2]`
Species_4 `r counts[4,1]` `r counts[4,2]`
Species_5 `r counts[5,1]` `r counts[5,2]`

  knitr::include_graphics('../man/figures/unifrac-tree.png')


Definitions

The branch indices (green circles) are used for ordering the $L$, $A$, and $B$ arrays. Values for $L$ are drawn from the input phylogenetic tree. Values for $A$ and $B$ are the total number of species observations descending from that branch; $A$ for Sample_A, and $B$ for Sample_B.

knitr::include_graphics('../man/figures/unifrac-weights.png')


$n = 8$ Number of branches
$A = \{`r A`\}$ Branch **weights** for Sample_A.
$B = \{`r B`\}$ Branch **weights** for Sample_B.
$A_T = 15$ Total observations for Sample_A.
$B_T = 15$ Total observations for Sample_B.
$L = \{`r L`\}$ The branch **lengths**.


Unweighted

First, transform A and B into presence (1) and absence (0) indicators.

\begin{align*} A &= \{`r A`\} \\ A' &= \{`r as.numeric(A > 0)`\} \end{align*} \begin{align*} B &= \{`r B`\} \\ B' &= \{`r as.numeric(B > 0)`\} \end{align*}

Then apply the formula:

\begin{align} U &= \displaystyle \frac{\sum_{i = 1}^{n} L_i(|A'i - B'_i|)}{\sum{i = 1}^{n} L_i(max(A'_i,B'_i))} \ \ U &= \displaystyle \frac{L_1(|A'_1-B'_1|) + L_2(|A'_2-B'_2|) + \cdots + L_n(|A'_n-B'_n|)}{L_1(max(A'_1,B'_1)) + L_2(max(A'_2,B'_2)) + \cdots + L_n(max(A'_n,B'_n))} \ \ U &= \displaystyle \frac{0.2(|1-1|) + 0.4(|0-1|) + \cdots + 0.3(|1-0|)}{0.2(max(1,1)) + 0.4(max(0,1)) + \cdots + 0.3(max(1,0))} \ \ U &= \displaystyle \frac{0.2(0) + 0.4(1) + 0.8(1) + 0.5(1) + 0.9(0) + 0.6(0) + 0.7(0) + 0.3(1)}{0.2(1) + 0.4(1) + 0.8(1) + 0.5(1) + 0.9(1) + 0.6(1) + 0.7(1) + 0.3(1)} \ \ U &= \displaystyle \frac{0.4 + 0.8 + 0.5 + 0.3}{0.2 + 0.4 + 0.8 + 0.5 + 0.9 + 0.6 + 0.7 + 0.3} \ \ U &= \displaystyle \frac{2}{4.4} \ \ U &= 0.4545455 \end{align}

Weighted

\begin{align} W &= \sum_{i = 1}^{n} L_i|\frac{A_i}{A_T} - \frac{B_i}{B_T}| \ \ W &= L_1|\frac{A_1}{A_T} - \frac{B_1}{B_T}| + L_2|\frac{A_2}{A_T} - \frac{B_2}{B_T}| + \cdots + L_n|\frac{A_n}{A_T} - \frac{B_n}{B_T}| \ \ W &= 0.2|\frac{9}{15} - \frac{7}{15}| + 0.4|\frac{0}{15} - \frac{5}{15}| + \cdots + 0.3|\frac{3}{15} - \frac{0}{15}| \ \ W &= 0.02\overline{6} + 0.1\overline{3} + 0.05\overline{3} + 0.1\overline{3} + 0.42 + 0.08 + 0.2\overline{3} + 0.06 \ \ W &= 1.14 \end{align}

Normalized

\begin{align*}

N &= \displaystyle \frac {\sum_{i = 1}^{n} L_i|\frac{A_i}{A_T} - \frac{B_i}{B_T}|} {\sum_{i = 1}^{n} L_i(\frac{A_i}{A_T} + \frac{B_i}{B_T})} \ \

N &= \displaystyle \frac {L_1|\frac{A_1}{A_T} - \frac{B_1}{B_T}| + L_2|\frac{A_2}{A_T} - \frac{B_2}{B_T}| + \cdots + L_n|\frac{A_n}{A_T} - \frac{B_n}{B_T}|} {L_1(\frac{A_1}{A_T} + \frac{B_1}{B_T}) + L_2(\frac{A_2}{A_T} + \frac{B_2}{B_T}) + \cdots + L_n(\frac{A_n}{A_T} + \frac{B_n}{B_T})} \ \

N &= \displaystyle \frac {0.2|\frac{9}{15} - \frac{7}{15}| + 0.4|\frac{0}{15} - \frac{5}{15}| + \cdots + 0.3|\frac{3}{15} - \frac{0}{15}|} {0.2(\frac{9}{15} + \frac{7}{15}) + 0.4(\frac{0}{15} + \frac{5}{15}) + \cdots + 0.3(\frac{3}{15} + \frac{0}{15})} \ \

N &= \displaystyle \frac {0.02\overline{6} + 0.1\overline{3} + 0.05\overline{3} + 0.1\overline{3} + 0.42 + 0.08 + 0.2\overline{3} + 0.06} {0.21\overline{3} + 0.1\overline{3} + 0.05\overline{3} + 0.1\overline{3} + 0.66 + 0.56 + 0.51\overline{3} + 0.06} \ \

N &= \displaystyle \frac{1.14}{2.326667} \ \ N &= 0.4899713 \end{align*}

Generalized

\begin{align*}

G &= \displaystyle \frac {\sum_{i = 1}^{n} L_i(\frac{A_i}{A_T} + \frac{B_i}{B_T})^{\alpha} |\displaystyle \frac {\frac{A_i}{A_T} - \frac{B_i}{B_T}} {\frac{A_i}{A_T} + \frac{B_i}{B_T}} |} {\sum_{i = 1}^{n} L_i(\frac{A_i}{A_T} + \frac{B_i}{B_T})^{\alpha}} \ \

G &= \displaystyle \frac { L_1(\frac{A_1}{A_T} + \frac{B_1}{B_T})^{0.5} |\displaystyle \frac {\frac{A_1}{A_T} - \frac{B_1}{B_T}} {\frac{A_1}{A_T} + \frac{B_1}{B_T}}| + \cdots + L_n(\frac{A_n}{A_T} + \frac{B_n}{B_T})^{0.5} |\displaystyle \frac {\frac{A_n}{A_T} - \frac{B_n}{B_T}} {\frac{A_n}{A_T} + \frac{B_n}{B_T}}| }{ L_1(\frac{A_1}{A_T} + \frac{B_1}{B_T})^{0.5} + \cdots + L_n(\frac{A_n}{A_T} + \frac{B_n}{B_T})^{0.5} }
\ \

G &= \displaystyle \frac { 0.2(\frac{9}{15} + \frac{7}{15})^{0.5} |\displaystyle \frac {\frac{9}{15} - \frac{7}{15}} {\frac{9}{15} + \frac{7}{15}}| + \cdots + 0.3(\frac{3}{15} + \frac{0}{15})^{0.5} |\displaystyle \frac {\frac{3}{15} - \frac{0}{15}} {\frac{3}{15} + \frac{0}{15}}| }{ 0.2(\frac{9}{15} + \frac{7}{15})^{0.5} + \cdots + 0.3(\frac{3}{15} + \frac{0}{15})^{0.5} }
\ \

G &\approx \displaystyle \frac {0.03 + 0.23 + 0.21 + 0.26 + 0.49 + 0.08 + 0.27 + 0.13} {0.21 + 0.23 + 0.21 + 0.26 + 0.77 + 0.58 + 0.60 + 0.13} \ \

G &= \displaystyle \frac{1.701419}{2.986235} \ \ G &= 0.569754

\end{align*}

Variance Adjusted

\begin{align*}

V &= \displaystyle \frac {\sum_{i = 1}^{n} L_i\displaystyle \frac {|\frac{A_i}{A_T} - \frac{B_i}{B_T}|} {\sqrt{(A_i + B_i)(A_T + B_T - A_i - B_i)}} } {\sum_{i = 1}^{n} L_i\displaystyle \frac {\frac{A_i}{A_T} + \frac{B_i}{B_T}} {\sqrt{(A_i + B_i)(A_T + B_T - A_i - B_i)}} } \ \

V &= \displaystyle \frac { L_1\displaystyle \frac {|\frac{A_1}{A_T} - \frac{B_1}{B_T}|} {\sqrt{(A_1 + B_1)(A_T + B_T - A_1 - B_1)}} + \cdots + L_n\displaystyle \frac {|\frac{A_n}{A_T} - \frac{B_n}{B_T}|} {\sqrt{(A_n + B_n)(A_T + B_T - A_n - B_n)}} }{ L_1\displaystyle \frac {\frac{A_1}{A_T} + \frac{B_1}{B_T}} {\sqrt{(A_1 + B_1)(A_T + B_T - A_1 - B_1)}} + \cdots + L_n\displaystyle \frac {\frac{A_n}{A_T} + \frac{B_n}{B_T}} {\sqrt{(A_n + B_n)(A_T + B_T - A_n - B_n)}} }
\ \

V &= \displaystyle \frac { 0.2\displaystyle \frac {|\frac{9}{15} - \frac{7}{15}|} {\sqrt{(9 + 7)(15 + 15 - 9 - 7)}} + \cdots + 0.3\displaystyle \frac {|\frac{3}{15} - \frac{0}{15}|} {\sqrt{(3 + 0)(15 + 15 - 3 - 0)}} }{ 0.2\displaystyle \frac {\frac{9}{15} + \frac{7}{15}} {\sqrt{(9 + 7)(15 + 15 - 9 - 7)}} + \cdots + 0.3\displaystyle \frac {\frac{3}{15} + \frac{0}{15}} {\sqrt{(3 + 0)(15 + 15 - 3 - 0)}} }
\ \

V &\approx \displaystyle \frac {0.002 + 0.012 + 0.010 + 0.013 + 0.029 + 0.005 + 0.016 + 0.007} {0.014 + 0.012 + 0.010 + 0.013 + 0.046 + 0.037 + 0.036 + 0.007} \ \

V &= \displaystyle \frac{4.09389}{4.174402} \ \ V &= 0.9807128

\end{align*}



Try the ecodive package in your browser

Any scripts or data that you put into this service are public.

ecodive documentation built on Aug. 23, 2025, 1:13 a.m.