我正试图在RStudio中使用名为"化石"的包的spp.est函数。我创建了一个名为"akimiskibb"的丰度数据矩阵，以物种为列，以地点为行。当我尝试使用函数spp.est时，我键入以下内容：
spp.est(akimiskibb，rand=10，abund=TRUE，counter=FALSE，max.est="all"(
问题来了，因为我的丰度数据有很多零，所以我得到了这个错误消息：
if(max(x(==1(警告中的错误("无法将关联数据用于基于丰度的分析。如果数据是基于关联的，请再次运行此函数，并选择"abund=FALSE"(：缺少值，其中TRUE/FALSE需要
该函数过去曾用于具有大量零的矩阵(也是丰度数据，而不是存在/不存在(。我不知道我做错了什么。
有人经历过类似的事情并找到了解决办法吗？
谢谢你，
Kayla
数据：矩阵格式：
*sp1 sp2 sp3 sp4 sp5 sp6 sp7 sp8 sp9 sp10 sp11 sp12 sp13 sp14 sp15 sp16 sp17
sample1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1
0 0
sample 2 0 0 0 1 0 0 1 25 7 0 18 12 0 0 0
1 1
sample3 0 0 0 0 0 0 0 3 0 0 3 1 0 0 0
5 4
sp18 sp19 sp20 sp21 sp22 sp23 sp24 sp25 sp26 sp27 sp28 sp29 sp30 sp31
sp32
sample1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0
sample 2 1 0 1 0 0 0 0 0 0 0 0 3 2
0 3
sample3 0 0 1 0 0 11 0 0 0 0 0 0 0
0 1
sp33 sp34 sp35 sp36 sp37 sp38 sp39 sp40 sp41 sp42 sp43 X
sample1 0 0 0 0 0 0 0 0 0 0 0 NA
sample 2 0 0 3 2 1 0 0 1 8 0 0 NA
sample3 0 0 0 0 0 0 0 0 0 0 0 NA*
dput:
*structure(list(sp1 = c(0L, 0L, 0L), sp2 = c(0L, 0L, 0L), sp3 = c(0L,
0L, 0L), sp4 = c(0L, 1L, 0L), sp5 = c(0L, 0L, 0L), sp6 = c(0L,
0L, 0L), sp7 = c(0L, 1L, 0L), sp8 = c(1L, 25L, 3L), sp9 = c(0L,
7L, 0L), sp10 = c(0L, 0L, 0L), sp11 = c(0L, 18L, 3L), sp12 = c(0L,
12L, 1L), sp13 = c(0L, 0L, 0L), sp14 = c(0L, 0L, 0L), sp15 = c(1L,
0L, 0L), sp16 = c(0L, 1L, 5L), sp17 = c(0L, 1L, 4L), sp18 = c(0L,
1L, 0L), sp19 = c(0L, 0L, 0L), sp20 = c(0L, 1L, 1L), sp21 = c(0L,
0L, 0L), sp22 = c(0L, 0L, 0L), sp23 = c(0L, 0L, 11L), sp24 = c(0L,
0L, 0L), sp25 = c(0L, 0L, 0L), sp26 = c(0L, 0L, 0L), sp27 = c(0L,
0L, 0L), sp28 = c(0L, 0L, 0L), sp29 = c(0L, 3L, 0L), sp30 = c(0L,
2L, 0L), sp31 = c(0L, 0L, 0L), sp32 = c(0L, 3L, 1L), sp33 = c(0L,
0L, 0L), sp34 = c(0L, 0L, 0L), sp35 = c(0L, 3L, 0L), sp36 = c(0L,
2L, 0L), sp37 = c(0L, 1L, 0L), sp38 = c(0L, 0L, 0L), sp39 = c(0L,
0L, 0L), sp40 = c(0L, 1L, 0L), sp41 = c(0L, 8L, 0L), sp42 = c(0L,
0L, 0L), sp43 = c(0L, 0L, 0L), X = c(NA, NA, NA)), .Names = c("sp1",
"sp2", "sp3", "sp4", "sp5", "sp6", "sp7", "sp8", "sp9", "sp10",
"sp11", "sp12", "sp13", "sp14", "sp15", "sp16", "sp17", "sp18",
"sp19", "sp20", "sp21", "sp22", "sp23", "sp24", "sp25", "sp26",
"sp27", "sp28", "sp29", "sp30", "sp31", "sp32", "sp33", "sp34",
"sp35", "sp36", "sp37", "sp38", "sp39", "sp40", "sp41", "sp42",
"sp43", "X"), class = "data.frame", row.names = c("sample1",
"sample 2", "sample3"))*
使用的包：
化石(R版本3.4.4制造(
R:R x64 3.4.1版本
好吧，我不熟悉这个包，所以谢谢你把它介绍给我。看到R是如何在这么多学科中使用的，真是太好了。
我发现你的申请有两个问题。首先，根据spp.est的fossil文档，您的数据需要将样本作为列，将物种作为行。第二个问题是具有NA值的"物种"X。你需要去掉这些，因为函数无法处理它们。
这是代码：
library(fossil)
library(tidyverse)
akimiskibb <- as.data.frame(t(akimiskibb)) %>%
filter(!is.na(sample1) == T)
spp.est(akimiskibb, rand = 10, abund = TRUE, counter = FALSE, max.est = 'all')
# which results in
N.obs S.obs S.obs(+95%) S.obs(-95%) Chao1 Chao1(upper) Chao1(lower)
[1,] 1 11.8 25.30995 -1.709948 20.90 28.38331 13.41669
[2,] 2 16.0 25.46770 6.532301 27.15 33.37680 20.92320
[3,] 3 20.0 20.00000 20.000000 26.00 29.24037 22.75963
ACE ACE(upper) ACE(lower) Jack1 Jack1(upper) Jack1(lower)
[1,] 14.56286 31.16289 -2.037175 16.82501 36.43129 -2.781268
[2,] 19.19286 30.64000 7.745713 22.21008 35.28448 9.135677
[3,] 22.28571 22.28571 22.285714 25.95082 25.95082 25.950820
attr(,"data.type")
[1] "abundance"
Warning messages:
1: In ests[[k]](b) :
This data appears to be presence/absence based, but this estimator is for abundance data only
2: In ests[[k]](b) :
This data appears to be presence/absence based, but this estimator is for abundance data only