如何重新排列 dplyr 的输出答案

【问题标题】：How to rearrange output from dplyr如何重新排列 dplyr 的输出
【发布时间】：2020-09-25 16:51:53
【问题描述】：

当我写下面的代码时

   ddply(milkers, .(dim_cat, lact_cat), function(x) mean(x$milkyield))

我得到以下输出

按库存类别（1 对 2）计算的牛奶产量平均值是正确的。我想最终得到一张更像下面这张的桌子。

实际上，我正在尝试获取每个时间段内的动物数量并计算它们的平均产奶量。问题在于它正在计算所有时间段的动物总数和所有时间段的平均产奶量。

我用来生成此数据的代码如下。

heiferdat <- subset(milkers, lact_cat== 1)
cowdat <- subset(milkers, lact_cat== 2)


ddply(milkers, .(dim_cat), function(x) c(Heifers = sum(milkers$lact_cat==1), H_Milk= mean(heiferdat$milkyield), Cows = sum(milkers$lact_cat==2), C_Milk= mean(cowdat$milkyield)))

我曾预料到，在这段代码中，.(dim_cat) 变量将应用于函数，以限制 sum 和 mean 函数只包括正确时间段内的动物。

我正在寻找有关如何获得每个时间段一行的输出以及每个 lact_cat 类的动物数量和每个 lact_cat 的平均产奶量的建议

谢谢

以下是我正在使用的数据的子集。

dput(milkers[180:200, c(11, 25, 26)]) 
dput(heiferdat[1:20, c(11, 25, 26)])
dput(cowdat[1:20, c(11, 25, 26)])

> dput(milkers[180:200, c(11, 25, 26)]) 
structure(list(milkyield = structure(c(8.42, 38.32, 14.27, 7.68, 
16.59, 17.19, 24.45, 33.47, 36.16, 25.88, 11.61, 18.96, 11.27, 
33.6, 21.57, 20.87, 9.62, 7.93, 21.02, 17.75, 22.01), label = "Milk (L)", class = c("labelled", 
"numeric")), dim_cat = structure(c(5L, 3L, 7L, 7L, 2L, 7L, 2L, 
2L, 2L, 3L, 6L, 6L, 2L, 3L, 6L, 6L, 6L, 6L, 6L, 7L, 6L), .Label = c("<31", 
"31-90", "91-150", "151-210", "211-270", "271-330", ">330"), class = c("labelled", 
"factor"), label = "Days in Milk"), lact_cat = structure(c(2L, 
2L, 1L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L), .Label = c("1", "2"), class = "factor")), row.names = 180:200, class = "data.frame")

> dput(heiferdat[1:20, c(11, 25, 26)]) 
structure(list(milkyield = structure(c(14.27, 17.19, 11.61, 18.96, 
11.27, 21.57, 20.87, 9.62, 7.93, 21.02, 17.75, 22.01, 25.15, 
11.75, 12.6, 15.62, 19.29, 8.85, 15.52, 11.62), label = "Milk (L)", class = c("labelled", 
"numeric")), dim_cat = structure(c(7L, 7L, 6L, 6L, 2L, 6L, 6L, 
6L, 6L, 6L, 7L, 6L, 6L, 6L, 6L, 7L, 6L, 6L, 6L, 6L), .Label = c("<31", 
"31-90", "91-150", "151-210", "211-270", "271-330", ">330"), class = c("labelled", 
"factor"), label = "Days in Milk"), lact_cat = structure(c(1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L), .Label = c("1", "2"), class = "factor")), row.names = c(182L, 
185L, 190L, 191L, 192L, 194L, 195L, 196L, 197L, 198L, 199L, 200L, 
201L, 202L, 203L, 204L, 205L, 206L, 207L, 208L), class = "data.frame")

> dput(cowdat[1:20, c(11, 25, 26)]) 
structure(list(milkyield = structure(c(15.73, 14.56, 16.94, 16.25, 
39.09, 9.79, 8.41, 3.05, 38.89, 11.7, 29.89, 19.73, 18.2, 20.63, 
20.32, 52.99, 10.11, 8.08, 10.84, 33.75), label = "Milk (L)", class = c("labelled", 
"numeric")), dim_cat = structure(c(3L, 6L, 6L, 2L, 3L, 7L, 6L, 
7L, 3L, 7L, 3L, 6L, 3L, 6L, 2L, 2L, 7L, 6L, 7L, 7L), .Label = c("<31", 
"31-90", "91-150", "151-210", "211-270", "271-330", ">330"), class = c("labelled", 
"factor"), label = "Days in Milk"), lact_cat = structure(c(2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L), .Label = c("1", "2"), class = "factor")), row.names = c(NA, 
20L), class = "data.frame")

【问题讨论】：

请添加一些虚拟数据来帮助您！
请创建一个代表 tidyverse.org/help 以帮助您解决此问题。请按照 SO 要求正确编辑您的代码。
ddply 实际上是来自plyr，dplyr 的老祖宗。如果您想使用现代工具（我个人认为您应该这样做，但那是您的决定），您可能希望将其翻译成实际的dplyr:stackoverflow.com/q/25787435/3888000。否则，您可能需要更改问题标题和标签以删除 dplyr 字。

标签： r dplyr mean

【解决方案1】：

根据@DanChaltiel 的建议使用 dplyr。这是一个 dplyr 方法：

library(dplyr)

all_summary = milkers %>%
  group_by(dim_cat, lact_cat) %>%
  summarise(avg = mean(milkyield),
            num = n())

此时您已计算出所有摘要信息。以下代码只是格式化/演示。

heifer_summary = all_summary %>%
  filter(lact_cat == 1) %>%
  select(dim_cat, Heifers = num, H_Milk = avg)
cow_summary = all_summary %>%
  filter(lact_cat == 2) %>%
  select(dim_cat, Cows = num, C_Milk = avg)

arranged_summary = full_join(heifer_summary, cow_summary, by = "dim_cat") %>%
  select(dim_cat, Heifers, H_Milk, Cows, C_Milk) %>%
  arrange(dim_cat)

【讨论】：