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License: Creative Commons Zero v1.0 Universal
Bias correction for richness in abundance data
License: Creative Commons Zero v1.0 Universal
Hi guys!
Running from IDLE 3.11.13 for Python 3.11.3 TK 8.6.12 on Mac M1, 2020, I cannot get the survival_rate function to complete. It runs and runs and never completes.
I tied unsuccessfully running it in many shapes, such as:
survival_ratio(abundance, method='chao1')
survival_ratio(abundance, method=“chao1”, n_iter=100, n_jobs=4)
survival_ratio(abundance, method=“chao1”, n_iter=10)
Anyone faces the same problem?
Create generic plot function, that determines what and how to plot on the basis of the keys in the dict provided.
All diversity functions use the same assertion statement:
Line 29 in 47c8804
If we hide these functions from the user, and direct them to only use the main interface function diversity()
, we can remove all assertions and add one to the diversity function.
and test whether they are set correctly
Decide on signatures and **kwargs for the richness functions. (Jackknife as exception?)
When using the idiom from X import *
we have to ensure that we specify what is meant by *
.
Implemnt extension of chao1 that also looks at f3 and f4 (specifically for Middle English).
Title says it all.
I think the estimates from these methods are overly pessimistic for my data because some of the assumptions in the equations don't apply, and want to mitigate by calculating "species" estimate based on a separate estimate of missing activity ("sightings").
Is there a way to get this from the species_accumulation
method or accumulation curve? Please advise.
Seems deprecation error?
Thanks for this great library, it was very timely for me.
I would like to be able to customize the species/sightings labels in the generated charts to reflect my actual data; didn't see anything in the docs about how this might be done.
What it says, including minsample bootstrap (optional).
Plots + evenness profile
runtime checks: assert that you get back an ax
Currently we use two ways to count the number of true elements in a NumPy array:
Line 32 in 47c8804
and
Line 79 in 47c8804
The first one using indexing is about twice as fast as the second one. However, an even faster solution is to use something like the following:
x = np.random.randint(0, 100, 1000)
np.count_nonzero(x > 0)
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