将ddply的当前值传递给函数

Question

Here is some sample data for which I want to encode the gender of the names over time:

names_to_encode <- structure(list(names = structure(c(2L, 2L, 1L, 1L, 3L, 3L), .Label = c("jane", "john", "madison"), class = "factor"), year = c(1890, 1990, 1890, 1990, 1890, 2012)), .Names = c("names", "year"), row.names = c(NA, -6L), class = "data.frame")

Here is a minimal set of the Social Security data, limited to just those names from 1890 and 1990:

ssa_demo <- structure(list(name = c("jane", "jane", "john", "john", "madison", "madison"), year = c(1890L, 1990L, 1890L, 1990L, 1890L, 1990L), female = c(372, 771, 56, 81, 0, 1407), male = c(0, 8, 8502, 29066, 14, 145)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -6L), .Names = c("name", "year", "female", "male"))

I've defined a function which subsets the Social Security data given a year or range of years. In other words, it calculates whether a name was male or female over a given time period by figuring out the proportion of male and female births with that name. Here is the function along with a helper function:

require(plyr)
require(dplyr)

select_ssa <- function(years) {

  # If we get only one year (1890) convert it to a range of years (1890-1890)
  if (length(years) == 1) years <- c(years, years)

  # Calculate the male and female proportions for the given range of years
  ssa_select <- ssa_demo %.%
    filter(year >= years[1], year <= years[2]) %.%
    group_by(name) %.%
    summarise(female = sum(female),
              male = sum(male)) %.%
    mutate(proportion_male = round((male / (male + female)), digits = 4),
           proportion_female = round((female / (male + female)), digits = 4)) %.%
    mutate(gender = sapply(proportion_female, male_or_female))

  return(ssa_select)
}

# Helper function to determine whether a name is male or female in a given year
male_or_female <- function(proportion_female) {
  if (proportion_female > 0.5) {
    return("female")
  } else if(proportion_female == 0.5000) {
    return("either")
  } else {
    return("male")
  }
}

Now what I want to do is use plyr, specifically ddply , to subset the data to be encoded by year, and merge each of those pieces with the value returned by the select_ssa function. This is the code I have.

ddply(names_to_encode, .(year), merge, y = select_ssa(year), by.x = "names", by.y = "name", all.x = TRUE)

When calling select_ssa(year) , this command works just fine if I hard code a value like 1890 as the argument to the function. But when I try to pass it the current value for year that ddply is working with, I get an error message:

Error in filter_impl(.data, dots(...), environment()) : 
  (list) object cannot be coerced to type 'integer'

How can I pass the current value of year on to ddply ?

Answer 1

I think you're making things too complicated by trying to do a join inside ddply . If I were to use dplyr I would probably do something more like this:

names_to_encode <- structure(list(name = structure(c(2L, 2L, 1L, 1L, 3L, 3L), .Label = c("jane", "john", "madison"), class = "factor"), year = c(1890, 1990, 1890, 1990, 1890, 2012)), .Names = c("name", "year"), row.names = c(NA, -6L), class = "data.frame")

ssa_demo <- structure(list(name = c("jane", "jane", "john", "john", "madison", "madison"), year = c(1890L, 1990L, 1890L, 1990L, 1890L, 1990L), female = c(372, 771, 56, 81, 0, 1407), male = c(0, 8, 8502, 29066, 14, 145)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -6L), .Names = c("name", "year", "female", "male"))

names_to_encode$name <- as.character(names_to_encode$name)
names_to_encode$year <- as.integer(names_to_encode$year)

tmp <- left_join(ssa_demo,names_to_encode) %.%
        group_by(year,name) %.%
        summarise(female = sum(female),
              male = sum(male)) %.%
        mutate(proportion_male = round((male / (male + female)), digits = 4),
           proportion_female = round((female / (male + female)), digits = 4)) %.%
        mutate(gender = ifelse(proportion_female == 0.5,"either",
                        ifelse(proportion_female > 0.5,"female","male")))

Note that 0.1.1 is still a little finicky about the types of join columns, so I had to convert them. I think I saw some activity on github that suggested that was either fixed in the dev version, or at least something they're working on.