如何使用 plyr/dplyr 更快地从现有变量创建上限和下限年龄变量

Question

我有一些 'untidy' 格式的数据 - 变量名称中嵌入了 'age'。使用 dplyr，我想创建一个 'tidy' 格式的数据集，其中的键是数据区、年份和年龄组，并且年龄组中的低年龄和高年龄是单独的变量。

一切都很好，除了最后一步花费的时间比我希望的要长得多。有没有更快的方法做到这一点仍然是 'readable'?

完整的可复制示例（使用 repmis 提取文件）

require(repmis)
require(stringr)
require(tidyr)
require(plyr)
require(dplyr)


persons <- source_DropboxData(
    file="persons.csv",
    key="vcz7qngb44vbynq"
    ) %>% 
    tbl_df() %>%
    select(datazone, year,
           contains("hspeop")
    )

names(persons) <- names(persons) %>%str_replace_all( "GR.hspeop", "count_both_")
persons <- persons %>% gather(age_group, count, -datazone, -year)

persons <- persons %>% mutate(gender="both", age_group=str_replace_all(age_group, "count_both_", ""))
persons$age_group <- persons$age_group %>% revalue(
    c(
        "1619" = "16_19",
        "2024" = "20_24",
        "2529" = "25_29", 
        "3034" = "30_34",
        "3539" = "35_39",
        "4044" = "40_44",
        "4549" = "45_49",
        "5054" = "50_54",
        "5559" = "55_59",
        "6064" = "60_64",
        "6569" = "65_69",
        "7074" = "70_74",
        "7579" = "75_79",
        "8084" = "80_84",
        "85over" = "85_100"
    )
)

# deal with "" separately as revalue can't cope
persons$age_group[nchar(persons$age_group)==0] <- "all"

persons_by_age <- persons %>% filter(grepl("_", age_group)) # this is how to filter by contents of age_group

persons_by_age <- persons_by_age %>% 
    group_by(age_group) %>% 
    mutate(
        lower_age = str_split(age_group, "_")[[1]][1] %>% as.numeric(),
        upper_age = str_split(age_group, "_")[[1]][2] %>% as.numeric()
    )

显然我在 mutate 中创建了同一个对象两次，因此速度可能会加倍。我还认为 group_by 意味着每个年龄组只需要完成一次操作，但它似乎对每一行都这样做。例如，按年龄组汇总、变异、然后加入是否是一种更快的方法？

编辑

上面的代码已经创建了输出，但比我希望的要慢得多。

最终输出的几个例子：

> persons_by_age
Source: local data frame [5,854,500 x 7]

    datazone year age_group count gender lower_age upper_age
1  S01000001 1996         0     8   both         0         0
2  S01000002 1996         0     4   both         0         0
3  S01000003 1996         0    18   both         0         0
4  S01000004 1996         0     4   both         0         0
5  S01000005 1996         0    17   both         0         0
6  S01000006 1996         0     1   both         0         0
7  S01000007 1996         0     9   both         0         0
8  S01000008 1996         0    10   both         0         0
9  S01000009 1996         0     8   both         0         0
10 S01000010 1996         0     9   both         0         0
..       ...  ...       ...   ...    ...       ...       ...
> persons_by_age %>% filter(year==2000 & gender=="male" & lower_age > 30)
Source: local data frame [71,555 x 7]

    datazone year age_group count gender lower_age upper_age
1  S01000001 2000     35_39    34   male        35        39
2  S01000002 2000     35_39    41   male        35        39
3  S01000003 2000     35_39    61   male        35        39
4  S01000004 2000     35_39    43   male        35        39
5  S01000005 2000     35_39    43   male        35        39
6  S01000006 2000     35_39    24   male        35        39
7  S01000007 2000     35_39    34   male        35        39
8  S01000008 2000     35_39    23   male        35        39
9  S01000009 2000     35_39    30   male        35        39
10 S01000010 2000     35_39    37   male        35        39
..       ...  ...       ...   ...    ...       ...       ...
> persons_by_age %>% filter(year==2000 & gender=="female" & lower_age > 30)
Source: local data frame [71,555 x 7]

    datazone year age_group count gender lower_age upper_age
1  S01000001 2000     35_39    37 female        35        39
2  S01000002 2000     35_39    30 female        35        39
3  S01000003 2000     35_39    58 female        35        39
4  S01000004 2000     35_39    46 female        35        39
5  S01000005 2000     35_39    28 female        35        39
6  S01000006 2000     35_39    29 female        35        39
7  S01000007 2000     35_39    33 female        35        39
8  S01000008 2000     35_39    25 female        35        39
9  S01000009 2000     35_39    36 female        35        39
10 S01000010 2000     35_39    38 female        35        39
..       ...  ...       ...   ...    ...       ...       ...

Answer 1

你可以试试这个：

persons_by_age<-persons_by_age %>% 
  group_by(age_group) %>% 
  do(cbind(.,matrix(rep(unlist(strsplit(as.character(.[1,3]), "_")),nrow(.)),ncol=2,byrow=TRUE)))

. 允许您访问 group_by

中的组

对于每个组，age_group 列的第一行 (.[1,3]) 被拆分，lower 和 upper 被做成一个向量，然后重复尽可能多的行在群里。

然后将得到的矩阵绑定到组中。几秒钟后运行。

separate 正如@jazzurro 所建议的那样要容易得多： separate(persons_by_age, age_group, c("lower", "upper"), sep = "_",remove=FALSE)

如何使用 plyr/dplyr 更快地从现有变量创建上限和下限年龄变量

How to make creating upper and lower age variables from existing variable much faster using plyr/dplyr

r

stringr

dplyr

编辑