The hardware and bandwidth for this mirror is donated by METANET, the Webhosting and Full Service-Cloud Provider.
If you wish to report a bug, or if you are interested in having us mirror your free-software or open-source project, please feel free to contact us at mirror[@]metanet.ch.
This vignette documents the method for calculating DSRs using the PHEindicatormethods::phe_dsr function which calculates DSRs and their confidence limits using the Dobson method.
The function can be used to calculate DSRs for grouping single or multiple geographic areas/ genders/ timeperiods/ indicators in a single execution and takes the following arguments as inputs:
Argument | Type | Definition | Default value |
---|---|---|---|
data | data.frame | data.frame containing the data to be standardised | none |
x | unquoted string | field name from data containing the observed number of events for each standardisation category (eg ageband) within each grouping set (eg area or indicator) | none |
n | unquoted string | field name from data containing the populations for each standardisation category (eg ageband) within each grouping set (eg area or indicator) | none |
stdpop | unquoted string | standard populations for each standardisation category (eg age band) specified as a field name from data or a vector. | esp2013 |
stdpoptype | quoted string | whether the stdpop argument has been specified as a vector or a field name | “vector” |
type | quoted string | defines the data and metadata columns to include in output. Can by ‘value’, ‘lower’, ‘upper’, ‘standard’ or ‘full’ | “full” |
confidence | numeric value | the required level of confidence expressed as a number between 0.9 and 1 or 90 and 100 | 0.95 |
multiplier | numeric value | the multiplier used to express the final values (eg 100,000 = rate per 100,000 | 100,000 |
Note that the European Standard Population 2013 divided into 19 five-year agebands (0-4, 5-9, 10-14, …..90+) is provided in vector format within the package and will be used as the default for the stdpop argument
If multiple DSRs are required from a single data frame then the data frame must be grouped prior to inputting to the function - this is demonstrated below
library(PHEindicatormethods)
library(dplyr)
#> Warning: package 'dplyr' was built under R version 4.2.3
In a real situation we’d most likely be sourcing our numerators and denominators from different places so let’s create them separately for now.
<- data.frame(indicator = rep(c("Ind1","Ind2","Ind3","Ind4"), each = 19 * 2 * 5),
pops period = rep(2012:2016, each = 19 * 2),
region = rep(rep(c("Area1","Area2"),each=19), times = 5),
ageband = rep(c(0,5,10,15,20,25,30,35,40,45,50,
55,60,65,70,75,80,85,90),times = 10),
pop = sample(10000:20000, 19 * 2 * 5 * 4, replace = TRUE))
head(pops)
#> indicator period region ageband pop
#> 1 Ind1 2012 Area1 0 10360
#> 2 Ind1 2012 Area1 5 10038
#> 3 Ind1 2012 Area1 10 13217
#> 4 Ind1 2012 Area1 15 19529
#> 5 Ind1 2012 Area1 20 18284
#> 6 Ind1 2012 Area1 25 17478
<- data.frame(indicator = rep(c("Ind1","Ind2","Ind3","Ind4"), each = 19 * 2 * 5),
deaths period = rep(2012:2016, each = 19 * 2),
region = rep(rep(c("Area1","Area2"),each=19), times = 5),
ageband = rep(c(0,5,10,15,20,25,30,35,40,45,50,
55,60,65,70,75,80,85,90),times = 10),
dths = sample(200, 19 * 2 * 5 * 4, replace=TRUE))
head(deaths)
#> indicator period region ageband dths
#> 1 Ind1 2012 Area1 0 83
#> 2 Ind1 2012 Area1 5 24
#> 3 Ind1 2012 Area1 10 65
#> 4 Ind1 2012 Area1 15 121
#> 5 Ind1 2012 Area1 20 145
#> 6 Ind1 2012 Area1 25 92
Our data contains records for 4 different indicators, 5 time periods and 2 geographies so let’s calculate a DSR for each combination - that’s 40 separate DSRs from a single execution of the phe_dsr function……
First we’ll need to join our datasets to create the input data.frame for the function and specify the grouping sets:
<- left_join(pops,deaths, by = c("indicator","period","region","ageband")) %>%
df group_by(indicator, period, region)
It is important that your data meets the following criteria in order for the phe_dsr function to work so it is wise to check this before we move on.
1. Each grouping set within your data must contain an equal number of records.
The phe_dsr function has built in error handling to check this requirement, or you can check your data manually using code like this:
# check equal number of records in each grouping set - eyeball check
summarise(df,n=n())
#> `summarise()` has grouped output by 'indicator', 'period'. You can override
#> using the `.groups` argument.
#> # A tibble: 40 × 4
#> # Groups: indicator, period [20]
#> indicator period region n
#> <chr> <int> <chr> <int>
#> 1 Ind1 2012 Area1 19
#> 2 Ind1 2012 Area2 19
#> 3 Ind1 2013 Area1 19
#> 4 Ind1 2013 Area2 19
#> 5 Ind1 2014 Area1 19
#> 6 Ind1 2014 Area2 19
#> 7 Ind1 2015 Area1 19
#> 8 Ind1 2015 Area2 19
#> 9 Ind1 2016 Area1 19
#> 10 Ind1 2016 Area2 19
#> # ℹ 30 more rows
# or alternatively the following should return TRUE
n_distinct(select(ungroup(summarise(df,n=n())),n)) == 1
#> `summarise()` has grouped output by 'indicator', 'period'. You can override
#> using the `.groups` argument.
#> [1] TRUE
2. If you are supplying your standard population in vector format (stdpoptype=“vector”) then this vector must also contain the same number of records as each grouping set within your data.
In this example we’re going to use the default esp2013 vector that is provided with the PHEindicatormethods package for our standard population - it contains 19 ordered values representing the 5-year age bands 0-4, 5-9, 10-14….85-89, 90+.
The phe_dsr function has built in error handling to check this requirement, or you can check your data manually using code like this:
# check standard population has same number of records as in each grouping set of data in check 1 above - eyeball check
length(esp2013)
#> [1] 19
# or alternatively the following should return TRUE
pull(slice(select(ungroup(summarise(df,n=n())),n),1)) == length(esp2013)
#> `summarise()` has grouped output by 'indicator', 'period'. You can override
#> using the `.groups` argument.
#> [1] TRUE
3. If you are supplying your standard population in vector format (stdpoptype=“vector”) then the standard population and your data (for each grouping set) must be sorted in the same standardisation category order because the function will join these by position. This would normally mean sorting both the standard population vector and the records for each group within your data by age band from youngest to oldest.
The phe_dsr function does not have any built in error handling to check this requirement as the function does not require the standardisation category labels to be provided in your data. It is therefore the responsibility of the function user to ensure this requirement is met. If the standardisation category labels are included with your data (as in our example) then the following code can be used to check the requirement manually:
# check data is ordered by required agebands from youngest to oldest
all(df$ageband == rep(c(0,5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90)))
#> [1] TRUE
CAUTION: Failure to ensure that this 3rd requirement is met could result in apparent successful execution of the phe_dsr function even though the data may have been incorrectly standardised. This is demonstrated in the sample code below:
<- phe_dsr(df,dths,pop)
right
<- df %>%
wrong arrange(desc(ageband)) %>%
phe_dsr(dths,pop)
# the following statement shows that execution of the phe_dsr function on the deliberately-incorrectly sorted data frame produces different (and incorrect) results.
identical(right,wrong)
#> [1] FALSE
Now we’re ready to calculate the DSRs using our correctly ordered df data frame.
By default the function will apply 95% confidence, a 100,000 multiplier and will output 3 data fields against each grouping set:
It will also output 3 metadata fields as an audit showing which argument parameters were passed:
phe_dsr(df, dths, pop)
#> # A tibble: 40 × 11
#> # Groups: indicator, period, region [40]
#> indicator period region total_count total_pop value lowercl uppercl
#> <chr> <int> <chr> <int> <int> <dbl> <dbl> <dbl>
#> 1 Ind1 2012 Area1 1897 289520 742. 706. 778.
#> 2 Ind1 2012 Area2 1860 299654 623. 593. 654.
#> 3 Ind1 2013 Area1 1924 285860 686. 654. 719.
#> 4 Ind1 2013 Area2 2081 282576 834. 795. 874.
#> 5 Ind1 2014 Area1 1801 284966 734. 699. 771.
#> 6 Ind1 2014 Area2 1341 277988 560. 529. 592.
#> 7 Ind1 2015 Area1 2165 267246 892. 853. 933.
#> 8 Ind1 2015 Area2 1762 276244 664. 632. 698.
#> 9 Ind1 2016 Area1 2110 283514 714. 682. 747.
#> 10 Ind1 2016 Area2 1417 280720 553. 522. 585.
#> # ℹ 30 more rows
#> # ℹ 3 more variables: confidence <chr>, statistic <chr>, method <chr>
Alternatively, we can just output the data fields by specifying the ‘type’ argument value as ‘standard’:
phe_dsr(df, dths, pop, type = "standard", confidence = 99.8, multiplier = 10000)
#> # A tibble: 40 × 8
#> # Groups: indicator, period, region [40]
#> indicator period region total_count total_pop value lowercl uppercl
#> <chr> <int> <chr> <int> <int> <dbl> <dbl> <dbl>
#> 1 Ind1 2012 Area1 1897 289520 74.2 68.7 79.9
#> 2 Ind1 2012 Area2 1860 299654 62.3 57.6 67.2
#> 3 Ind1 2013 Area1 1924 285860 68.6 63.6 73.8
#> 4 Ind1 2013 Area2 2081 282576 83.4 77.4 89.7
#> 5 Ind1 2014 Area1 1801 284966 73.4 67.9 79.2
#> 6 Ind1 2014 Area2 1341 277988 56.0 51.2 61.1
#> 7 Ind1 2015 Area1 2165 267246 89.2 83.0 95.6
#> 8 Ind1 2015 Area2 1762 276244 66.4 61.4 71.8
#> 9 Ind1 2016 Area1 2110 283514 71.4 66.4 76.7
#> 10 Ind1 2016 Area2 1417 280720 55.3 50.5 60.3
#> # ℹ 30 more rows
In some cases you may wish to standardise against a different population to the default esp2013 one provided - such as the 1976 European Standard Population or an age and sex standardised population. There are two ways to specify an alternative standard population:
In the example below, the 1976 European Standard Population (which has 18 age groups) is provided as a vector and then referenced in the function call. To ensure the function works we must also ensure that our data has been broken down into these same 18 age bands (for the purposes of this example I’ve just combined the 85-90 and 90+ age band data into a single 85+ age band from the data.frame we used earlier).
The phe_dsr function can then be executed using a user-defined standard population:
<- c(8000, 7000, 7000, 7000, 7000, 7000, 7000, 7000, 7000, 7000, 7000, 6000, 5000, 4000, 3000, 2000, 1000, 1000)
esp1976
<- df
df18 $dths[df18$ageband == 85] <- df18$dths[df18$ageband == 85] + df18$dths[df18$ageband == 90]
df18$pop[df18$ageband == 85] <- df18$pop[df18$ageband == 85] + df18$pop[df18$ageband == 90]
df18<- filter(df18,ageband != 90)
df18
phe_dsr(df18,dths,pop,stdpop = esp1976)
#> # A tibble: 40 × 11
#> # Groups: indicator, period, region [40]
#> indicator period region total_count total_pop value lowercl uppercl
#> <chr> <int> <chr> <int> <int> <dbl> <dbl> <dbl>
#> 1 Ind1 2012 Area1 1897 289520 741. 704. 780.
#> 2 Ind1 2012 Area2 1860 299654 621. 589. 653.
#> 3 Ind1 2013 Area1 1924 285860 624. 594. 656.
#> 4 Ind1 2013 Area2 2081 282576 832. 792. 873.
#> 5 Ind1 2014 Area1 1801 284966 810. 769. 853.
#> 6 Ind1 2014 Area2 1341 277988 580. 547. 615.
#> 7 Ind1 2015 Area1 2165 267246 903. 863. 945.
#> 8 Ind1 2015 Area2 1762 276244 659. 625. 694.
#> 9 Ind1 2016 Area1 2110 283514 751. 715. 789.
#> 10 Ind1 2016 Area2 1417 280720 551. 519. 584.
#> # ℹ 30 more rows
#> # ℹ 3 more variables: confidence <chr>, statistic <chr>, method <chr>
In the example below, the esp2013 standard population is appended to our data frame prior to calling the phe_dsr function. The field name can then be specified in the function call. If stdpop is specified as a field name we must also tell the function this by specifying stdpoptype = “field” as below:
<- df %>%
df_with_stdpop mutate(spop = esp2013)
names(df_with_stdpop)
#> [1] "indicator" "period" "region" "ageband" "pop" "dths"
#> [7] "spop"
phe_dsr(df_with_stdpop, dths, pop, stdpop = spop, stdpoptype = "field")
#> # A tibble: 40 × 11
#> # Groups: indicator, period, region [40]
#> indicator period region total_count total_pop value lowercl uppercl
#> <chr> <int> <chr> <int> <int> <dbl> <dbl> <dbl>
#> 1 Ind1 2012 Area1 1897 289520 742. 706. 778.
#> 2 Ind1 2012 Area2 1860 299654 623. 593. 654.
#> 3 Ind1 2013 Area1 1924 285860 686. 654. 719.
#> 4 Ind1 2013 Area2 2081 282576 834. 795. 874.
#> 5 Ind1 2014 Area1 1801 284966 734. 699. 771.
#> 6 Ind1 2014 Area2 1341 277988 560. 529. 592.
#> 7 Ind1 2015 Area1 2165 267246 892. 853. 933.
#> 8 Ind1 2015 Area2 1762 276244 664. 632. 698.
#> 9 Ind1 2016 Area1 2110 283514 714. 682. 747.
#> 10 Ind1 2016 Area2 1417 280720 553. 522. 585.
#> # ℹ 30 more rows
#> # ℹ 3 more variables: confidence <chr>, statistic <chr>, method <chr>
This would be a fairly common scenario - maybe you have Local Authority data and there are no deaths in some of the younger age groups for some of the smaller areas. From PHEindicatormethods version 1.1.0 onwards, the phe_dsr function can handle this scenario and will automatically assign a zero death count where a death count is missing or recorded as NA.
Let’s fudge a couple of data frames to represent this. In this example, there are no deaths in the 10-14, 15-20 and 20-14 age bands:
<- data.frame(ageband = c( 0, 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90),
pops2 pop = c(30,35,35,35,40,40,45,50,50,50,60,60,70,75,70,60,20,20,15))
<- data.frame(ageband = c(0,5,25,30,35,40,45,50,55,60,65,70,75,80,85,90),
deaths2 dths = c(1,1, 1, 1, 3, 3, 3, 3,10,10,10,10, 8, 8, 8, 8))
If we join these data frames to produce the input data frame required for the phe_dsr function then we get NA values in the Deaths column. From PHEindicatormethods version 1.1.0 onwards, the phe_dsr function will return the correct DSR, assuming zero deaths in the age groups with no deaths recorded. If you are using an earlier version of PHEindicatormethods then an error will be returned. See what you get…..
<- left_join(pops2, deaths2, by="ageband")
df2 phe_dsr(df2, dths, pop)
#> total_count total_pop value lowercl uppercl confidence statistic
#> 1 88 860 8283.016 6570.819 10289.65 95% dsr per 100000
#> method
#> 1 Dobson
For earlier versions of PHEindicatormethods, the NA values must be replaced with zeros before executing the function:
<- df2 %>%
df3 mutate(dths = replace(dths, which(is.na(dths)), 0))
phe_dsr(df3, dths, pop)
#> total_count total_pop value lowercl uppercl confidence statistic
#> 1 88 860 8283.016 6570.819 10289.65 95% dsr per 100000
#> method
#> 1 Dobson
These binaries (installable software) and packages are in development.
They may not be fully stable and should be used with caution. We make no claims about them.