Question

以下是一些示例数据，我想对其中的名称性别进行编码：

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")

以下是社会保障数据的最小集合，仅限于1890年和1990年的名称：

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"))

我已经定义了一个功能，该功能在给定一年或一年的范围内对社会保障数据进行子集。换句话说，它通过计算具有该名称的男性和女性出生比例来计算在给定时间段内姓名是男性还是女性。这是函数和辅助函数：

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")
  }
}

现在我要做的是使用plyr，特别是ddply来对要按年编码的数据进行子集化，并将每个部分与select_ssa函数返回的值合并。这是我的代码。

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

当调用select_ssa(year)时，如果我将类似1890的值硬编码为函数的参数，则此命令可以正常工作。但是当我尝试传递year正在使用的ddply的当前值时，我收到一条错误消息：

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

如何将year的当前值传递给ddply？

Answer 1

我认为你试图在ddply内进行联接会让事情变得太复杂。如果我使用dplyr，我可能会做更多这样的事情：

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")))

请注意，0.1.1对于连接列的类型仍然有点挑剔，所以我不得不转换它们。我想我在github上看到了一些活动，这些活动建议在dev版本中修复，或者至少是他们正在处理的东西。

将ddply的当前值传递给函数

1 个答案: