Loading data

install.packages("NHANES")
  1. Do not include an evaluated install.package() in your Rmd program. What can happen if you do?

Either the Rmd will not knit and throw an error, or it will knit. However if it does, the package would reinstall with every execution of the Rmd program. Not only does this waste CPU and RAM, but also could potentially introduce bugs when the package has been updated, or not installed successfully.

  1. What is in the “Description” section of the help page for the “NHANES” object?

“This is survey data collected by the US National Center for Health Statistics (NCHS) which has conducted a series of health and nutrition surveys since the early 1960’s. Since 1999 approximately 5,000 individuals of all ages are interviewed in their homes every year and complete the health examination component of the survey. The health examination is conducted in a mobile examination centre (MEC).” 

library(NHANES)
library(dplyr)

Viewing the data

What can happen if you have an evaluated View() command in your .Rmd file? > It will throw an error and not knit since view command is attempting to open the entire dataset, which is not possible in a non-interactive html or pdf.

Summarizing the data

summary(NHANES$Weight)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.    NA's 
##    2.80   56.10   72.70   70.98   88.90  230.70      78
summary(NHANES$Height) 
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.    NA's 
##    83.6   156.8   166.0   161.9   174.5   200.4     353

How many missing values are there for weight and height?

There are 78 missing values for Weight and 353 missing values for Height.

Writing a Function

#Creating mini dataframe
WtHt <- NHANES %>%
  select(Weight,Height)

#creating custom function
clean.mean.dif <- function(df,i) {
  
 subset <-  slice(df,i)%>%
          na.omit()
 
 # need to use "[[" operator to extract as a vector 
 
 mean_col_1 <- mean(subset[[1]]) 
 
 mean_col_2 <- mean(subset[[2]])
 
 mean_diff <- mean_col_2 - mean_col_1
 
 mean_diff
 
}

clean.mean.dif(WtHt,1:10)
## [1] 87.37

Debugging

After debugging, I figured out I needed to use “[[” operator to extract specified dataframe column into a vector, which R can then calculate mean.

Write another function

#creating new mini dataframe
WtGender <- NHANES %>%
  select(Weight,Gender)


#defining function that will take the difference of means from the values in Col 1 associated with col 2's factor level 2 and those associated with level 1

mean.diff.factor <- function(df,i) {
  
 subset <-  slice(df,i)%>%
          na.omit()

 #filtering df subset on factor level 1 in col 2 
 subset_L1 <- subset %>%
   filter(subset[[2]] == levels(subset[[2]])[1])
 
  #filtering df subset on factor level 2 in col 2

 subset_L2<- subset %>%
   filter(subset[[2]] == levels(subset[[2]])[2])

 
 
 mean_subset_L1 <- mean(subset_L1[[1]]) 
 
 mean_subset_L2 <- mean(subset_L2[[1]])
 
 
 # in this case we want to find males - females.  
 # males is level 2, females is level 1
 
 mean_diff <-  mean_subset_L2 - mean_subset_L1
 
 mean_diff
 
}


mean.diff.factor(WtGender,1:10000)
## [1] 9.583432

*Side note, realizing it would be a lot more useful if you can pass which levels and columns to use 

mean.diff.ab <- function(df,i,mean_col_index,fac_col_index,level_a,level_b) {
  
 subset <-  slice(df,i)%>%
          na.omit()

 #filtering df subset on factor level 1 in col 2 
 subset_level_a <- subset %>%
   filter(subset[[fac_col_index]] == levels(subset[[2]])[level_a])
 
  #filtering df subset on factor level 2 in col 2

 subset_level_b<- subset %>%
   filter(subset[[fac_col_index]] == levels(subset[[2]])[level_b])

 
 
 mean_subset_La <- mean(subset_level_a[[mean_col_index]]) 
 
 mean_subset_Lb <- mean(subset_level_b[[mean_col_index]])
 
 
 # in this case we want to find males - females.  
 # males is level 2, females is level 1
 
 mean_diff <-  mean_subset_La - mean_subset_Lb
 
 mean_diff
 
}


#this argument is asking to find means in col[1] based on factors in col[2].
#It is then specifying to find the difference between level 2 associated means and level 1 associated means

mean.diff.ab(WtGender,1:10000,1,2,2,1)
## [1] 9.583432