傳承ddply分裂的當前值功能

Question

這裏是我想的名字的性別編碼隨着時間的推移一些樣本數據：

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") 

這裏是一個集社會保障數據的最小的，有限的只是那些從1890年和1990年的名字：

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")) 

我定義其子集給出的年一年或範圍的社會保障數據的功能。換句話說，它通過計算男性和女性的出生比例以及這個名字來計算某個特定時間段的名字是男性還是女性。這裏是一個輔助功能沿功能：

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") 
    } 
} 

現在我想要做的就是使用plyr，具體ddply，以子集逐年編碼的數據，並用返回的值合併的每個那件由select_ssa功能。這是我的代碼。

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

當調用select_ssa(year)，這個命令的作品就好了，如果我硬編碼像1890作爲參數傳遞給函數的值。但是，當我試圖通過它year是ddply正在與當前值，我得到一個錯誤信息：

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

如何傳遞的year當前值上ddply？

Answer 1

我認爲你在ddply內部試圖做一個加入讓事情變得複雜。如果我使用dplyr我可能會做更多的事情是這樣的：

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"))) 

需要注意的是0.1.1還是有點挑剔的類型連接列的，所以我不得不將它們轉換。我想我在github上看到了一些活動，表明它在dev版本中是固定的，或者至少是他們正在開發的東西。