data( ox )
ox <- Meth( ox )
# The simplest possible Bland-Altman plot
BA.plot( ox )
## With bells and whistles, comparing the naive and model
par( mfrow=c(2,2) )
BA.plot( ox, model=NULL, repl.conn=TRUE, col.lines="blue",
axlim=c(0,100), diflim=c(-50,50), xaxs="i", yaxs="i",
las=1, eqn=TRUE, dif.type="lin", pl.type="BA", sd.type="lin",
grid=1:9*10, digits=3,font.eqn=1)
par(new=TRUE)
BA.plot( ox, model="linked", repl.conn=TRUE, col.lines="red",
axlim=c(0,100), diflim=c(-50,50), xaxs="i", yaxs="i",
las=1, eqn=FALSE, dif.type="lin", pl.type="BA", sd.type="lin",
grid=1:0*10, digits=3)
BA.plot( ox, model=NULL, repl.conn=TRUE, col.lines="blue",
axlim=c(0,100), diflim=c(-50,50), xaxs="i", yaxs="i",
las=1, eqn=TRUE, dif.type="lin", pl.type="conv", sd.type="lin",
grid=1:9*10, digits=3,font.eqn=1)
par(new=TRUE)
BA.plot( ox, model="linked", repl.conn=TRUE, col.lines="red",
axlim=c(0,100), diflim=c(-50,50), xaxs="i", yaxs="i",
las=1, eqn=FALSE, dif.type="lin", pl.type="conv", sd.type="lin",
grid=1:9*10, digits=3)
# The same again, but now logit-transformed
BA.plot( ox, model=NULL, repl.conn=TRUE, col.lines="blue",
axlim=c(0,100), diflim=c(-50,50), xaxs="i", yaxs="i",
las=1, eqn=TRUE, dif.type="lin", pl.type="BA", sd.type="lin",
grid=1:9*10, digits=3,font.eqn=1,Transform="pctlogit")
par(new=TRUE)
BA.plot( ox, model="linked", repl.conn=TRUE, col.lines="red",
axlim=c(0,100), diflim=c(-50,50), xaxs="i", yaxs="i",
las=1, eqn=FALSE, dif.type="lin", pl.type="BA", sd.type="lin",
grid=1:0*10, digits=3,Transform="pctlogit")
BA.plot( ox, model=NULL, repl.conn=TRUE, col.lines="blue",
axlim=c(0,100), diflim=c(-50,50), xaxs="i", yaxs="i",
las=1, eqn=TRUE, dif.type="lin", pl.type="conv", sd.type="lin",
grid=1:9*10, digits=3,font.eqn=1,Transform="pctlogit")
par(new=TRUE)
BA.plot( ox, model="linked", repl.conn=TRUE, col.lines="red",
axlim=c(0,100), diflim=c(-50,50), xaxs="i", yaxs="i",
las=1, eqn=FALSE, dif.type="lin", pl.type="conv", sd.type="lin",
grid=1:9*10, digits=3,Transform="pctlogit")
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