Fail to extract the gender information from the first name in R

[YuJiang]

Hello, I attempted to extract gender information from the first name using gender package in R. I tried both 'ssa' and 'genderize' for argument method.

Here is my demo sample code.

unique_id <- seq(0:6)
first_name <- c("annie j", "Juan", "Richard", "Aj",
                "Dana", "annie j", "liyuan")

demo1 <- as.data.frame(cbind(unique_id, first_name))

For ssa, it uses names based from the U.S. Social Security Administration baby name data. Therefore, if the name does not include in ssa, it will return the error as shown below.

demo1$gender <- gender(demo1$first_name, method="ssa")$gender

I know this is because 'annie j' does not include in the name dataset, ssa. Any suggestions or advice to fix it?
Really appreciate your help and reply.

[technocrat]

library(gender)

(d <- data.frame(
  unique_id = 0:6,
  first_name = c("annie j", "Juan", "Richard", "Aj",
                "Dana", "annie j", "liyuan")))
#>   unique_id first_name
#> 1         0    annie j
#> 2         1       Juan
#> 3         2    Richard
#> 4         3         Aj
#> 5         4       Dana
#> 6         5    annie j
#> 7         6     liyuan

gender(d$first_name, method="ssa")$gender
#> [1] "male"   "female" "male"   "male"

gender(d$first_name, method="ssa")[c("name","gender")]
#> # A tibble: 4 × 2
#>   name    gender
#>   <chr>   <chr> 
#> 1 Aj      male  
#> 2 Dana    female
#> 3 Juan    male  
#> 4 Richard male

^{Created on 2023-03-24 with reprex v2.0.2}

[YuJiang]

Hello technocrat. Thanks for your reply but it does not return the entire list of names. How can I fix this problem?

[nirgrahamuk]

To have a data.frame of all names you wanted to test, and their gender results (and gaps where gender was not detected) you would perform a Left Join with your original data.frame on the left and the shorter gender results data.frame on the right.

dplyr package has a function that helps you do this; the left_join() function

[technocrat]

f(x=y

x is d
y the the object to be created, which isn't clear from the description
f is the function to transform x to y and may be composite

Often the most difficult part is clearly having in mind the content of y. In the example below, I assume a data frame with variables unique_id, first-name and gender. Because ssa does not recognize space separated first names gender() will return only single-word names, so f should in addition to single word names should trim multiple word names to a single word to recheck. Single word names with no match should be included with gender of NA. I compose f to illustrate this stepwise for clarity.

library(gender)

(d <- data.frame(
  unique_id = 0:9,
  first_name = c("mary ann", "billy bob", "norma rae",
                "jim bob","Juan", "Richard", "Aj",
                 "Dana", "annie j", "liyuan")))
#>    unique_id first_name
#> 1          0   mary ann
#> 2          1  billy bob
#> 3          2  norma rae
#> 4          3    jim bob
#> 5          4       Juan
#> 6          5    Richard
#> 7          6         Aj
#> 8          7       Dana
#> 9          8    annie j
#> 10         9     liyuan

(keepers <- gender(d$first_name, method="ssa"))
#> # A tibble: 4 × 6
#>   name    proportion_male proportion_female gender year_min year_max
#>   <chr>             <dbl>             <dbl> <chr>     <dbl>    <dbl>
#> 1 Aj                0.988            0.0119 male       1932     2012
#> 2 Dana              0.202            0.798  female     1932     2012
#> 3 Juan              0.992            0.0084 male       1932     2012
#> 4 Richard           0.996            0.0037 male       1932     2012

(unfound <- d[-which(d$first_name %in% keepers$name),])
#>    unique_id first_name
#> 1          0   mary ann
#> 2          1  billy bob
#> 3          2  norma rae
#> 4          3    jim bob
#> 9          8    annie j
#> 10         9     liyuan

unfound$first_name <- sub(" .*$","",unfound$first_name)

(one_name <- gender(unfound$first_name, method="ssa"))
#> # A tibble: 5 × 6
#>   name  proportion_male proportion_female gender year_min year_max
#>   <chr>           <dbl>             <dbl> <chr>     <dbl>    <dbl>
#> 1 annie          0.0053            0.995  female     1932     2012
#> 2 billy          0.988             0.0119 male       1932     2012
#> 3 jim            0.997             0.0034 male       1932     2012
#> 4 mary           0.0036            0.996  female     1932     2012
#> 5 norma          0.0051            0.995  female     1932     2012

(lost <- setdiff(unfound$first_name,one_name$name))
#> [1] "liyuan"
  
(combined <- rbind(keepers,one_name))
#> # A tibble: 9 × 6
#>   name    proportion_male proportion_female gender year_min year_max
#>   <chr>             <dbl>             <dbl> <chr>     <dbl>    <dbl>
#> 1 Aj               0.988             0.0119 male       1932     2012
#> 2 Dana             0.202             0.798  female     1932     2012
#> 3 Juan             0.992             0.0084 male       1932     2012
#> 4 Richard          0.996             0.0037 male       1932     2012
#> 5 annie            0.0053            0.995  female     1932     2012
#> 6 billy            0.988             0.0119 male       1932     2012
#> 7 jim              0.997             0.0034 male       1932     2012
#> 8 mary             0.0036            0.996  female     1932     2012
#> 9 norma            0.0051            0.995  female     1932     2012
combined$name <- casefold(combined$name)
combined <- combined[c(1,4)]
combined <- rbind(combined,c(lost,NA))
colnames(combined)[1] <- "first_name"
result <- dplyr::left_join(d,combined)
#> Joining with `by = join_by(first_name)`
result$gender <- combined$gender
result
#>    unique_id first_name gender
#> 1          0   mary ann   male
#> 2          1  billy bob female
#> 3          2  norma rae   male
#> 4          3    jim bob   male
#> 5          4       Juan female
#> 6          5    Richard   male
#> 7          6         Aj   male
#> 8          7       Dana female
#> 9          8    annie j female
#> 10         9     liyuan   <NA>