Column chart with two variables that assume a positive response and represented by opposite sides, such as dry mass of the area and dry mass of the root

barplot_positive(
a,
b,
ylab = "Response",
var_name = c("Var1", "Var2"),
legend.title = "Variable",
fill_color = c("darkgreen", "brown")
)

## Arguments

a

Object of DIC, DBC or DQL functions

b

Object of DIC, DBC or DQL functions

ylab

Y axis names

var_name

Name of the variable

legend.title

Legend title

fill_color

Bar fill color

## Value

The function returns a column chart with two positive sides

## Note

When there is only an effect of the isolated factor in the case of factorial or subdivided plots, it is possible to use the barplot_positive function.

## Author

Gabriel Danilo Shimizu, shimizu@uel.br

## Examples

data("passiflora")
attach(passiflora)
a=with(passiflora, DBC(trat, bloco, MSPA))
#>
#> -----------------------------------------------------------------
#> Normality of errors
#> -----------------------------------------------------------------
#>                          Method Statistic   p.value
#>  Shapiro-Wilk normality test(W) 0.9429864 0.2728826
#>
#> As the calculated p-value is greater than the 5% significance level, hypothesis H0 is not rejected. Therefore, errors can be considered normal
#>
#> -----------------------------------------------------------------
#> Homogeneity of Variances
#> -----------------------------------------------------------------
#>                               Method Statistic  p.value
#>  Bartlett test(Bartlett's K-squared)  4.070401 0.396562
#>
#> As the calculated p-value is greater than the 5% significance level, hypothesis H0 is not rejected. Therefore, the variances can be considered homogeneous
#>
#> -----------------------------------------------------------------
#> Independence from errors
#> -----------------------------------------------------------------
#>                  Method Statistic    p.value
#>  Durbin-Watson test(DW)  1.572708 0.03849032
#>
#> As the calculated p-value is less than the 5% significance level, H0 is rejected. Therefore, errors are not independent
#>
#> -----------------------------------------------------------------
#> -----------------------------------------------------------------
#>
#> CV (%) =  13.73
#> MStrat/MST =  0.84
#> Mean =  0.025
#> Median =  0.0241
#> Possible outliers =  No discrepant point
#>
#> -----------------------------------------------------------------
#> Analysis of Variance
#> -----------------------------------------------------------------
#>           Df       Sum Sq      Mean.Sq   F value       Pr(F)
#> trat       4 3.413717e-04 8.534292e-05 7.2705205 0.003259252
#> bloco      3 1.506425e-05 5.021417e-06 0.4277837 0.736759950
#> Residuals 12 1.408586e-04 1.173821e-05
#>
#> As the calculated p-value, it is less than the 5% significance level. The hypothesis H0 of equality of means is rejected. Therefore, at least two treatments differ
#>
#> -----------------------------------------------------------------
#> Multiple Comparison Test: Tukey HSD
#> -----------------------------------------------------------------
#>                   resp groups
#> carolina    0.03250000      a
#> Vermiculita 0.02615625     ab
#> Mc normal   0.02283750      b
#> areia       0.02215000      b
#> Mc organico 0.02113125      b
#>
#>
#> Your analysis is not valid, suggests using a non-parametric
#> test and try to transform the data

b=with(passiflora, DBC(trat, bloco, MSR))
#>
#> -----------------------------------------------------------------
#> Normality of errors
#> -----------------------------------------------------------------
#>                          Method Statistic   p.value
#>  Shapiro-Wilk normality test(W) 0.9836814 0.9723826
#>
#> As the calculated p-value is greater than the 5% significance level, hypothesis H0 is not rejected. Therefore, errors can be considered normal
#>
#> -----------------------------------------------------------------
#> Homogeneity of Variances
#> -----------------------------------------------------------------
#>                               Method Statistic  p.value
#>  Bartlett test(Bartlett's K-squared)  8.452343 0.076345
#>
#> As the calculated p-value is greater than the 5% significance level, hypothesis H0 is not rejected. Therefore, the variances can be considered homogeneous
#>
#> -----------------------------------------------------------------
#> Independence from errors
#> -----------------------------------------------------------------
#>                  Method Statistic   p.value
#>  Durbin-Watson test(DW)  1.826611 0.1152517
#>
#> As the calculated p-value is greater than the 5% significance level, hypothesis H0 is not rejected. Therefore, errors can be considered independent
#>
#> -----------------------------------------------------------------
#> -----------------------------------------------------------------
#>
#> CV (%) =  17.97
#> MStrat/MST =  0.49
#> Mean =  0.0116
#> Median =  0.0112
#> Possible outliers =  No discrepant point
#>
#> -----------------------------------------------------------------
#> Analysis of Variance
#> -----------------------------------------------------------------
#>           Df       Sum Sq      Mean.Sq   F value     Pr(F)
#> trat       4 2.263485e-05 5.658712e-06 1.3123500 0.3203367
#> bloco      3 4.997083e-06 1.665694e-06 0.3863024 0.7648958
#> Residuals 12 5.174271e-05 4.311892e-06
#>
#> As the calculated p-value is greater than the 5% significance level, H0 is not rejected

#>
#> -----------------------------------------------------------------
#> Multiple Comparison Test: Tukey HSD
#> -----------------------------------------------------------------
#> [1] "H0 is not rejected"
#>

barplot_positive(a, b, var_name = c("DMAP","DRM"), ylab = "Dry root (g)")