CODEWITHSUNDEEP
rggplot2
Gianluca
Gianluca

Reputation: 53

ggplot2, fitting data with log2 or log10 doesn't affect the plot

I wanted a display a geom_smooth with a natural log and this code works fine: 

    df <- iris
iris_logplot <- ggplot(df, aes(Sepal.Length, Sepal.Width, colour = Species))

iris_logplot + stat_summary(fun.y =median, geom = "point") + stat_summary(fun.data = mean_cl_boot, aes(group = Species), geom = "errorbar", width = 0.2) + 
  geom_smooth(method="lm", formula=y~log(x))

now I want to display a geom_smooth with a log whose base is 2 and I apply this code: 

df <- iris
iris_logplot <- ggplot(df, aes(Sepal.Length, Sepal.Width, colour = Species))

iris_logplot + stat_summary(fun.y =median, geom = "point") +
  stat_summary(fun.data = mean_cl_boot, aes(group = Species), geom = "errorbar", width = 0.2) + geom_smooth(method="lm", formula=y~log2(x))

Why the plots are the same?

Thanks

Upvotes: 1

Views: 271

Answers (1)

Gregor Thomas
Gregor Thomas

Reputation: 145825

The lines are the same because multiplying a feature in a linear model by a constant does not change the fit, the coefficients are just divided by the same constant. The "change of base" formula tells us that log_b(x) = log_a(x) / log_a(b).

We can verify this by examining the models:

m_log_e = lm(Sepal.Width ~ log(Sepal.Length) * Species, data = iris)
m_log_2 = lm(Sepal.Width ~ log2(Sepal.Length) * Species, data = iris)

summary(m_log_e)
# Call:
# lm(formula = Sepal.Width ~ log(Sepal.Length) * Species, data = iris)
# 
# Residuals:
#      Min       1Q   Median       3Q      Max 
# -0.71398 -0.15310 -0.00419  0.16595  0.60237 
# 
# Coefficients:
#                                     Estimate Std. Error t value Pr(>|t|)    
# (Intercept)                          -2.9663     0.8872  -3.343 0.001055 ** 
# log(Sepal.Length)                     3.9760     0.5512   7.214 2.86e-11 ***
# Speciesversicolor                     2.3355     1.1899   1.963 0.051595 .  
# Speciesvirginica                      3.0464     1.1639   2.617 0.009807 ** 
# log(Sepal.Length):Speciesversicolor  -2.0626     0.7087  -2.910 0.004186 ** 
# log(Sepal.Length):Speciesvirginica   -2.4373     0.6811  -3.579 0.000471 ***
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
# 
# Residual standard error: 0.272 on 144 degrees of freedom
# Multiple R-squared:  0.6237,  Adjusted R-squared:  0.6106 
# F-statistic: 47.73 on 5 and 144 DF,  p-value: < 2.2e-16

summary(m_log_2)
# Call:
# lm(formula = Sepal.Width ~ log2(Sepal.Length) * Species, data = iris)
# 
# Residuals:
#      Min       1Q   Median       3Q      Max 
# -0.71398 -0.15310 -0.00419  0.16595  0.60237 
# 
# Coefficients:
#                                      Estimate Std. Error t value Pr(>|t|)    
# (Intercept)                           -2.9663     0.8872  -3.343 0.001055 ** 
# log2(Sepal.Length)                     2.7560     0.3820   7.214 2.86e-11 ***
# Speciesversicolor                      2.3355     1.1899   1.963 0.051595 .  
# Speciesvirginica                       3.0464     1.1639   2.617 0.009807 ** 
# log2(Sepal.Length):Speciesversicolor  -1.4297     0.4913  -2.910 0.004186 ** 
# log2(Sepal.Length):Speciesvirginica   -1.6894     0.4721  -3.579 0.000471 ***
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
# 
# Residual standard error: 0.272 on 144 degrees of freedom
# Multiple R-squared:  0.6237,  Adjusted R-squared:  0.6106 
# F-statistic: 47.73 on 5 and 144 DF,  p-value: < 2.2e-16

Comparing the summaries, you can convince yourself that the fits are the same - the residuals are the same, the statistics are the same, the intercepts are the same, the only difference are the coefficients for terms including Sepal.Length. We can divide the coefficients:

coef(m_log_e) / coef(m_log_2)
#                         (Intercept)                   log(Sepal.Length)                   Speciesversicolor                    Speciesvirginica 
#                            1.000000                            1.442695                            1.000000                            1.000000 
# log(Sepal.Length):Speciesversicolor  log(Sepal.Length):Speciesvirginica 
#                            1.442695                            1.442695

And see that the terms involving Sepal.Length are off by a fixed ratio. And what is that ratio?

1 / log(2)
# [1] 1.442695

It is 1 /log(2), because of the change of base formula referenced at the start of this answer.

Upvotes: 1

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