Age-Adjusted Rates in R

Here is just some random code for future use.

Standard population

#https://seer.cancer.gov/stdpopulations/stdpopdic.html

standard_pop <- read.fwf("https://seer.cancer.gov/stdpopulations/stdpop.18ages.txt", widths = c(3,3,8),  
                          header = 0, col.names = c('Standard_ID','Age','Standard_Population')) %>% 
                filter(Standard_ID==009 ) %>% ## 009 = World (WHO 2000-2025) Std Million (18 age groups) 
                mutate(AGE = (Age-1)*5,  ## as the numbers are 1 to 18, the formula X-1 * 5 will give us the value.
                       Proportion = Standard_Population/sum(Standard_Population)) %>% 
                select(AGE, Standard_Population, Proportion) %>% 
      mutate(AGE=ifelse(AGE>=80,80, AGE)) %>%
      group_by(AGE) %>% 
      summarise(Standard_Population=sum(Standard_Population), Proportion=sum(Proportion)) %>% 
      ungroup()

standard_pop %>% 
  #filter(AGE>5) %>% 
  mutate(prop_test = Standard_Population / sum(Standard_Population)) %>% 
  mutate(sum(Proportion), sum(prop_test))

Data Sample

location	AGE	counts	population
Guatemala	0	35	10250
Guatemala	5	25	12859
Panama	5	50	80253
Panama	5	38	21224
Costa Rica	20	25	15351

You must have the same number of age groups as your standard_pop to be abble to merge, and remember to replace all the missings with 0 in the counts column.

data  %>%
 mutate(rate=counts/population) %>% 
  left_join(standard_pop, by = "AGE") %>% 
  group_by(location) %>% 
  summarise(adjusted1= weighted.mean(rate, Standard_Population), 
			adjusted2= sum(rate*Proportion))

In the standard_pop you can have proportions or total numbers to adjust your rates. Both work fine, if you choice to work with total numbers you must use weighted.mean(), if you choice to use proportions you can use sum(rate*Proportion)

Written on August 11, 2020