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以下是一些示例数据，我想随着时间的推移对其名称的性别进行编码：

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")

这是一组最小的社会保障数据，仅限于 1890 年和 1990 年的那些名字：

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"))

我已经定义了一个函数，它对给定年份或年份范围的社会保障数据进行子集化。换句话说，它通过计算出使用该名字的男性和女性出生的比例来计算给定时间段内一个名字是男性还是女性。这是该函数和一个辅助函数：

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")
  }
}

现在我想要做的是使用 plyr，特别是ddply，对要按年份编码的数据进行子集化，并将每个片段与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)

调用时select_ssa(year)，如果我硬编码一个值1890作为函数的参数，这个命令就可以正常工作。但是当我尝试将当前值传递给它year时ddply，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 on 的当前值传递year给ddply？

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1 回答 1

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我认为您尝试在内部进行 join 会使事情变得过于复杂ddply。如果我要使用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 对连接列的类型仍然有些挑剔，所以我不得不对其进行转换。我想我在 github 上看到了一些活动，表明这些活动要么已在开发版本中修复，要么至少是他们正在研究的东西。

于 2014-02-21T15:51:14.390 回答