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我想将以下示例数据汇总到一个新的数据框中，如下所示：

人口、样本量 (N)、完成百分比 (%)

样本大小是每个总体的所有记录的计数。我可以使用 table 命令或 tapply 来做到这一点。完成百分比是带有“结束日期”的记录的百分比（所有没有“结束日期”的记录都被假定为未完成。这就是我迷路的地方！

样本数据

 sample <- structure(list(Population = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 
    2L, 2L, 3L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 
    1L, 2L, 2L, 3L, 3L, 3L, 3L, 1L, 1L, 3L, 3L, 3L, 3L), .Label = c("Glommen", 
    "Kaseberga", "Steninge"), class = "factor"), Start_Date = structure(c(16032, 
    16032, 16032, 16032, 16032, 16036, 16036, 16036, 16037, 16038, 
    16038, 16039, 16039, 16039, 16039, 16039, 16039, 16041, 16041, 
    16041, 16041, 16041, 16041, 16044, 16044, 16045, 16045, 16045, 
    16045, 16048, 16048, 16048, 16048, 16048, 16048), class = "Date"), 
        End_Date = structure(c(NA, 16037, NA, NA, 16036, 16043, 16040, 
        16041, 16042, 16042, 16042, 16043, 16043, 16043, 16043, 16043, 
        16043, 16045, 16045, 16045, 16045, 16045, NA, 16048, 16048, 
        16049, 16049, NA, NA, 16052, 16052, 16052, 16052, 16052, 
        16052), class = "Date")), .Names = c("Population", "Start_Date", 
    "End_Date"), row.names = c(NA, 35L), class = "data.frame")
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2 回答 2

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您可以使用拆分/应用/组合来执行此操作：

spl = split(sample, sample$Population)
new.rows = lapply(spl, function(x) data.frame(Population=x$Population[1],
                                              SampleSize=nrow(x),
                                              PctComplete=sum(!is.na(x$End_Date))/nrow(x)))
combined = do.call(rbind, new.rows)
combined

#           Population SampleSize PctComplete
# Glommen      Glommen         13   0.6923077
# Kaseberga  Kaseberga          7   1.0000000
# Steninge    Steninge         15   0.8666667

一个警告：sample是一个基本函数的名称，所以你应该为你的数据框选择一个不同的名称。

于 2013-12-19T20:50:26.567 回答
2

使用该plyr软件包很容易：

library(plyr)
ddply(sample, .(Population), summarize, 
      Sample_Size = length(End_Date),
      Percent_Completed = mean(!is.na(End_Date)) * 100)

#   Population Sample_Size Percent_Completed
# 1    Glommen          13          69.23077
# 2  Kaseberga           7         100.00000
# 3   Steninge          15          86.66667
于 2013-12-19T20:52:11.160 回答