我刚刚开始使用 R 并想使用该trend包对Mann-Kendall多口井的年度地下水位数据进行测试。这是我的数据帧中的一个示例pri.csv：
dput(head(GL, 10))
structure(list(WLCODE = structure(c(8L, 5L, 3L, 6L, 1L, 7L, 18L,
19L, 20L, 22L), .Label = c("W04039", "W04295", "W04299", "W04790",
"W04791", "W04797", "W04799", "W04824", "W25293", "W25298", "W25355",
"W25385", "W25404", "W25438", "W25445", "W25451", "W25476", "W25596",
"W25598", "W25599", "W25600", "W25612", "W25697", "W31151", "W31152"
), class = "factor"), YEAR_OBS = c(2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L), POMRB = c(4.14, 9.05,
8.87, 2.3, 6.05, 3.6, 2.05, 4.27, 7.9, 8.4)), .Names = c("WLCODE",
"YEAR_OBS", "POMRB"), row.names = c(NA, 10L), class = "data.frame")
这是我到目前为止所做的：
mydata<-read.csv("pri.csv")
sp <- split(mydata, mydata$WLCODE)
results_list <- lapply(sp, function(mydata){
tryCatch(mk.test(mydata[, 3]),
error = function(e) e)
})
bad <- sapply(results_list, inherits, "error")
results_list[!bad]
$W07792
Mann-Kendall trend test
data: DF[, 3]
z = 1.0355, n = 16, p-value = 0.3004
alternative hypothesis: true S is not equal to 0
sample estimates:
S varS tau
24.0000 493.3333 0.2000
$W07793
Mann-Kendall trend test
data: DF[, 3]
z = -0.68205, n = 20, p-value = 0.4952
alternative hypothesis: true S is not equal to 0
sample estimates:
S varS tau
-22.0000000 948.0000000 -0.1164037
我想使用此输出生成一个表，并将其导出到 excel 中WLCODE，每一行中都有Z, p-value, S, varS,tau列中的值。我怎样才能做到这一点？请帮忙。
一个选项是tidy从broom
library(broom)
library(purrr)
library(trend)
map_dfr(results_list[!bad], tidy)
或通过提取单个组件使用lapplyfrombase R
out <- do.call(rbind, lapply(results_list[!bad], function(x)
cbind(data.frame(p.value = x$p.value, statistic = x$statistic),
data.frame(as.list(x$estimates)))))
row.names(out) <- NULL
out
# p.value statistic S varS tau
#1 1.0000000 0.000000 1 3.666667 0.3333333
#2 1.0000000 0.000000 1 3.666667 0.3333333
#3 0.2962699 -1.044466 -3 3.666667 -1.0000000
#4 1.0000000 0.000000 -1 3.666667 -0.3333333
本文收集自互联网，转载请注明来源。
如有侵权，请联系 [email protected] 删除。
我来说两句