问题
I'm trying to run anova()
in R and running into some difficulty. This is what I've done up to now to help shed some light on my question.
Here is the str()
of my data to this point.
str(mhw)
'data.frame': 500 obs. of 5 variables:
$ r : int 1 2 3 4 5 6 7 8 9 10 ...
$ c : int 1 1 1 1 1 1 1 1 1 1 ...
$ grain: num 3.63 4.07 4.51 3.9 3.63 3.16 3.18 3.42 3.97 3.4 ...
$ straw: num 6.37 6.24 7.05 6.91 5.93 5.59 5.32 5.52 6.03 5.66 ...
$ Quad : Factor w/ 4 levels "NE","NW","SE",..: 2 2 2 2 2 2 2 2 2 2 ...
Column r is a numerical value indicating which row in the field an individual plot resides
Column c is a numerical value indicating which column an individual plot resides
Column Quad corresponds to the geographical location in the field to which each plot resides
Quad <- ifelse(mhw$c > 13 & mhw$r < 11, "NE",ifelse(mhw$c < 13 & mhw$r < 11,"NW", ifelse(mhw$c < 13 & mhw$r >= 11, "SW","SE")))
mhw <- cbind(mhw, Quad)
I have fit a lm()
as follows
nov.model <-lm(mhw$grain ~ mhw$straw)
anova(nov.model)
This is an anova()
for the entire field, which is testing grain yield against straw yield for each plot in the dataset.
My trouble is that I want to run an individual anova()
for the Quad column of my data to test grain yield and straw yield in each quadrant.
perhaps a with()
might fix that. I have never used it before and I am in the process of learning R currently. Any help would be greatly appreciated.
回答1:
I think you are looking for by
facility in R.
fit <- with(mhw, by(mhw, Quad, function (dat) lm(grain ~ straw, data = dat)))
Since you have 4 levels in Quad
, you end up with 4 linear models in fit
, i.e., fit
is a "by" class object (a type of "list") of length 4.
To get coefficient for each model, you can use
sapply(fit, coef)
To produce model summary, use
lapply(fit, summary)
To export ANOVA table, use
lapply(fit, anova)
As a reproducible example, I am taking the example from ?by
:
tmp <- with(warpbreaks,
by(warpbreaks, tension,
function(x) lm(breaks ~ wool, data = x)))
class(tmp)
# [1] "by"
mode(tmp)
# [1] "list"
sapply(tmp, coef)
# L M H
#(Intercept) 44.55556 24.000000 24.555556
#woolB -16.33333 4.777778 -5.777778
lapply(tmp, anova)
#$L
#Analysis of Variance Table
#
#Response: breaks
# Df Sum Sq Mean Sq F value Pr(>F)
#wool 1 1200.5 1200.50 5.6531 0.03023 *
#Residuals 16 3397.8 212.36
#---
#Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
#$M
#Analysis of Variance Table
#
#Response: breaks
# Df Sum Sq Mean Sq F value Pr(>F)
#wool 1 102.72 102.722 1.2531 0.2795
#Residuals 16 1311.56 81.972
#
#$H
#Analysis of Variance Table
#
#Response: breaks
# Df Sum Sq Mean Sq F value Pr(>F)
#wool 1 150.22 150.222 2.3205 0.1472
#Residuals 16 1035.78 64.736
I was aware of this option, but not familiar with it. Thanks to @Roland for providing code for the above reproducible example:
library(nlme)
lapply(lmList(breaks ~ wool | tension, data = warpbreaks), anova)
For your data I think it would be
fit <- lmList(grain ~ straw | Quad, data = mhw)
lapply(fit, anova)
You don't need to install nlme
; it comes with R as one of recommended packages.
来源:https://stackoverflow.com/questions/39769376/fitting-linear-model-anova-by-group