This method formats analysis of variance (ANOVA) statistics from the class
aov. The default output is APA formatted, but this function allows control
over numbers of digits, leading zeros, italics, degrees of freedom,
and output format of Markdown or LaTeX.
Usage
# S3 method for class 'aov'
format_stats(
x,
term,
digits = 1,
pdigits = 3,
pzero = FALSE,
italics = TRUE,
dfs = "par",
type = "md",
...
)Arguments
- x
An
aovobject fromstats::aov().- term
Character string for row name of term to extract statistics for. This must be the exact string returned in the
summary()output from theaovobject.- digits
Number of digits after the decimal for means, confidence intervals, and test statistics.
- pdigits
Number of digits after the decimal for p-values, ranging between 1-5 (also controls cutoff for small p-values).
- pzero
Logical value (default = FALSE) for whether to include leading zero for p-values.
- italics
Logical value (default = TRUE) for whether p label should be italicized.
- dfs
Formatting for degrees of freedom ("par" = parenthetical, "sub" = subscript, "none" = do not print degrees of freedom).
- type
Type of formatting ("md" = markdown, "latex" = LaTeX).
- ...
Additional arguments passed to methods.
See also
Other functions for printing statistical objects:
format_bf(),
format_corr(),
format_stats(),
format_stats.BFBayesFactor(),
format_stats.easycorrelation(),
format_stats.htest(),
format_stats.lm(),
format_stats.lmerModLmerTest(),
format_stats.merMod(),
format_ttest()
Examples
test_aov <- aov(mpg ~ cyl * hp, data = mtcars)
# Format ANOVA
format_stats(test_aov, term = "cyl")
#> [1] "_F_(1, 28) = 92.5, _p_ < .001"
# Remove italics and make degrees of freedom subscripts
format_stats(test_aov, term = "cyl", italics = FALSE, dfs = "sub")
#> [1] "F~1,28~ = 92.5, p < .001"
# Change digits and add leading zero to p-value
format_stats(test_aov, term = "hp", digits = 3, pdigits = 4, pzero = TRUE)
#> [1] "_F_(1, 28) = 1.850, _p_ = 0.1846"
# Format for LaTeX
format_stats(test_aov, term = "hp", type = "latex")
#> [1] "$F$(1, 28) = 1.9, $p$ = .185"
