When I tried dput function, the result was this:
> dput(x)
c(20, 200, 2000, 20000)
> dput(y)
c(0.45, 0.05, 0.5, 0.4, 0, 0.5, 0.4, 0.05, 0.4, 0.25, 0.35, 0.5,
0.05, 0.4, 0.5, 0.5, 0.5, 0.25, 0.85, 0.5, 0.5, 0.5, 0.25, 0.4,
0.25, 0.25, 0.4, 0.25, 0.5, 0.15, 0.25, 0.1, 0.25, 0.25, 0.015,
0.4, 0.5, 0.2, 0.25, 5e-05, 0.5, 0.005, 0.5, 0.25, 0.25, 0.4,
0.5, 0.4, 0.5, 0.5, 0.5, 0.5, 0.7142857143, 0.5, 0.005, 0.35,
0.5, 0.35, 0, 0.5, 0.25, 0.25, 1, 0.25, 0.1, 0.25, 0.5, 0.25,
0.55, NA, 0.25, 0.4, 0.35, 0.35, 0.25, 0, 0.8888888889, 0.5,
0.25, 0.5, 0.5, 0.5, 0.25, 0.2, 0.4, 0, 0.35, 0.025, 0.4, 0.5,
0.35, 0.25, 0.3, 0.25, 0.005, 0.5, 0.4, 0.05, 0.5, 0.4, 0.005,
0.45, 0.4, 0.35, 0.5, 0.005, 0.3, 0.05, 0.25, 0.35, 0.35, 0.75,
0.5, 0.375, 0.45, 0.1, 0.4, 0.25, 0.25, 0.25, 0.25, 0.5, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 0.2,
5e-04, 0.5, 0.5, 0.025, 0.25, 0.25, 0.01, 0.35, 0.15, 0.3, 0.5,
5e-04, 0.3, 0.4, 0.25, 0.4, 0.25, 0.85, 0.25, 0.375, 0.25, 0.1,
0.35, 0.05, 0.25, 0.2, 5000, 0.5, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, 0.05, 5e-05, 0.5, 0.6, 0.005,
0.25, 0.25, 0.0025, 0.4, 0.1, 0.25, 0.5, 0.001, 0.25, 0.4, 0.25,
0.45, 0.05, 0.6, 0.25, 0.4, 5e-05, 0.05, 0.35, 0.05, 0.15, 0.05,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA)
***
To have the same number of elements, I used the mean of each column to pair
with 20 ... 20 000; but this would affect the p-value, because R does not
know whar there were much more data than just four.
The result is this:
/> summary (lm (d~log(x)))
Call:
lm(formula = d ~ log(x))
Residuals:
1 3 4
-0.001108 0.010249 -0.009141
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.39008 0.02591 15.055 0.0422 *
log(x) 0.06184 0.01115 5.547 0.1135
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.01378 on 1 degrees of freedom
(1 observation deleted due to missingness)
Multiple R-squared: 0.9685, Adjusted R-squared: 0.937
F-statistic: 30.77 on 1 and 1 DF, p-value: 0.1135
Warning message:
In log(x) : NaNs produced/
***
I tried to handle this by not using just a single number (the mean of the
column), but compose the mean itself in the data:
> d3 <- c(mean(rdiktator20), mean(rDiktator200), mean(rDikt2000),
> mean(rDikt20000))
However, I did not ge any results from it:
> lm (d3~log(x))
Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
0 (non-NA) cases
So there are still NAs blocking the linear model, although I had used the
na.omit function...
--
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