# Preferring a preference index

First I like to use a quantitative framework (you can use ranks-based indices as in Williams et al 2011, which has nice properties too). The simplest quantitative index is the forage ratio:

``````#obs: observed frequency of visits to item 1
#exp: expected frequency of visits to item 1 (i.e. resource abundance/availability)
fr <- function(obs, exp){
p_obs <- obs/sum(obs)
p_exp <- exp/sum(exp)
out <- p_obs/p_exp
out
}``````

But it ranges from 0 to infinity, which makes it hard to interpret.

Second, the most used index is “Electivity” (Ivlev 1961), which is nice because is bounded between -1 and 1. “Jacobs” index is similar, but correcting for item depletion, which do not apply to my question, but see below for completness.

``````electicity <- function(obs, exp){
p_obs <- obs/sum(obs)
p_exp <- exp/sum(exp)
out <- (p_obs-p_exp)/(p_obs+p_exp)
}

jacobs <- function(obs, exp){
p_obs <- obs/sum(obs)
p_exp <- exp/sum(exp)
out <- (p_obs-p_exp)/(p_obs+p_exp-2*p_obs*p_exp)
out
}``````

However, those indexes do not tell you if pollinators have general preferences over several items, and which of this items are preferred or not. To solve that we can use a simple chi test. Chi statistics can be used to assess if there is an overall preference. The Chi test Pearson residuals (obs-exp/√exp) estimate the magnitude of preference or avoidance for a given item based on deviation from expected values. Significance can be assessed by building Bonferroni confidence intervals (see Neu et al. 1974 and Beyers and Steinhorst 1984). That way you know which items are preferred or avoided.

``````chi_pref <- function(obs, exp, alpha = 0.05){
chi <- chisq.test(obs, p = exp, rescale.p = TRUE)
print(chi) #tells you if there is an overall preference. (sig = pref)
res <- chi\$residuals
#res <- (obs-exp)/sqrt(exp) #hand calculation, same result.
#calculate bonferoni Z-statistic for each plant.
alpha <- alpha
k <- length(obs)
n <- sum(obs)
p_obs <- obs/n
ak <- alpha/(2*k)
Zak <- abs(qnorm(ak))
low_interval <- p_obs - (Zak*(sqrt(p_obs*(1-p_obs)/n)))
upper_interval <- p_obs + (Zak*(sqrt(p_obs*(1-p_obs)/n)))
p_exp <- exp/sum(exp)
sig <- ifelse(p_exp >= low_interval & p_exp <= upper_interval, "ns", "sig")
plot(c(0,k+1), c(min(low_interval),max(upper_interval)), type = "n",
ylab = "Preference", xlab = "items", las = 1)
arrows(x0 = c(1:k), y0 = low_interval, x1 = c(1:k), y1 = upper_interval, code = 3
,angle = 90)
points(p_exp, col = "red")
out <- data.frame(chi_test_p = rep(chi\$p.value, length(res)),
chi_residuals = res, sig = sig)
out
}``````

And we wrap up all indexes…

``````#wrap up for all indexes
preference <- function(obs, exp, alpha = 0.05){
f <- fr(obs, exp)
e <- electicity(obs, exp)
#j <- jacobs(obs, exp)
c <- chi_pref(obs, exp, alpha = alpha)
out <- data.frame(exp = exp, obs = obs, fr = f, electicity = e, c)
out
}``````

It works well for preferences among items with similar availability, and when all are used to some extent. The plot shows expected values in red, and the observed confidence interval in black. If is not overlapping the expected, indicates a significant preference.

``````x <- preference(obs = c(25, 22,30,40), exp = c(40,12,12,53)) ``````
``````##  Chi-squared test for given probabilities
##
## data:  obs
## X-squared = 44.15, df = 3, p-value = 1.404e-09``````
``````##   exp obs     fr electicity chi_test_p chi_residuals sig
## 1  40  25 0.6250    -0.2308  1.404e-09        -2.372 sig
## 2  12  22 1.8333     0.2941  1.404e-09         2.887  ns
## 3  12  30 2.5000     0.4286  1.404e-09         5.196 sig
## 4  53  40 0.7547    -0.1398  1.404e-09        -1.786 sig``````

We can see that indices are correlated by simulating lots of random values.

``````x <- preference(obs = sample(c(0:100),100), exp = sample(c(0:100),100))
``````
``````##  Chi-squared test for given probabilities
##
## data:  obs
## X-squared = Inf, df = 99, p-value < 2.2e-16``````
``pairs(x[,c(-5,-7)]) #more or less same patern, across indexes.`` But the indexes do not behave well in two situations. First, when an item is not used, it is significanly un-preferred regardless of its abundance. I don’t like that, because a very rare plant is expected to get 0 visits from a biological point of view. This is an issue when building bonferroni intervals around 0, which are by definition 0, and any value different from 0 appears as significant.

``````x <- preference(obs = c(0,25,200), exp = c(0.00003,50,100)) ``````
``````##  Chi-squared test for given probabilities
##
## data:  obs
## X-squared = 50, df = 2, p-value = 1.389e-11``````
``````##     exp obs     fr electicity chi_test_p chi_residuals sig
## 1 3e-05   0 0.0000    -1.0000  1.389e-11     -0.006708 sig
## 2 5e+01  25 0.3333    -0.5000  1.389e-11     -5.773502 sig
## 3 1e+02 200 1.3333     0.1429  1.389e-11      4.082486 sig``````

The second issue is that if you have noise due to sampling (as always), rare items are more likely to be wrongly assessed than common items.

``````#assume no preferences
x <- preference(obs = c(5,50,100), exp = c(5,50,100)) ``````
``````##  Chi-squared test for given probabilities
##
## data:  obs
## X-squared = 0, df = 2, p-value = 1``````
``````##   exp obs fr electicity chi_test_p chi_residuals sig
## 1   5   5  1          0          1             0  ns
## 2  50  50  1          0          1             0  ns
## 3 100 100  1          0          1             0  ns``````
``````#now assume some sampling error of +-4 observations
x <- preference(obs = c(1,54,96), exp = c(5,50,100)) ``````
``````##  Chi-squared test for given probabilities
##
## data:  obs
## X-squared = 3.671, df = 2, p-value = 0.1595``````
``````##   exp obs     fr electicity chi_test_p chi_residuals sig
## 1   5   1 0.2053  -0.659341     0.1595       -1.7539 sig
## 2  50  54 1.1086   0.051508     0.1595        0.7580  ns
## 3 100  96 0.9854  -0.007338     0.1595       -0.1438  ns``````

As a conclusion, If you have all items represented (not highly uneven distributions) or all used to some extent (not 0 visits) this is a great tool.

Happy to know better options in the comments, if available.

Gist with the code of this post here.

## 12 thoughts on “Preferring a preference index”

1. Pingback: Condensed News | Data Analytics & R

2. Linn on said:

What is alpha in the code?

• ibartomeus on said:

alpha is the statistical significance threshold, usually fixed at 0.05.

3. alexis.paumier@irstea.fr on said:

Hi,
thank for the post. What if the availability is very different among items?

4. Stefan on said:

Hello!
Thank you very much for these helpfull functions!
What if i want to have categorial variables in my x-axis? (Like “grassland”, “agriculture” etc.? Instead of “item 2,4,6,8″…? I know i have to change the “c(0, k+1)” behind the plot command, but my R-Skills are very basic.

• ibartomeus on said:

Items are in fact categories, so you can use any categorical variable as input.

5. Tyler on said:

Hello Dr. Bartomeus,
I was wondering if you have a paper where you demonstrate this method? I am using this method and was hoping to cite your work.

6. Ive on said:

Hi,
Thank you for the post.
I was looking at a way to calculate the preference for a time period (diel activity)
Thus I have the number of hours per time period (and I calculated the proportion for 24h), and calculated the proportion of detection of each species in each period.
I was going to use your preference function, but you said that it worked best for similar availability, and I have: day ~ 11h (0.469), night ~9h (0.364) and twilight 4h (0.167) so I am wondering wether there is too much of a difference.

Also, in your comments, you said that you are recommending the null model, However I am not sure how to take into account the different availabilities…

• ibartomeus on said:

I am not an expert on this, just a user. Probably for a simple approach, the indexes are fine with your categories, but you can play with simulated data to see if you trust them.

For the null model, you can build your own. For example, by assuming an equal probability to use any hour of the day, and then pooling occurrences at your three time periods. You can then compare the distribution of usage of your null’s versus your observed values.

Best,