加入表格时添加“空”行

Question

我正在开发一个在 R 中监控工人生产力的应用程序。对于每个工人，我都有他们完成的工作，我想看看他们一年中每周完成了多少工作（所以第 1 到 52 周）。问题是某些工人在某些周内没有任何工作记录（例如，工人 1 可能工作了第 1 到 48 周，但不是第 49 到 52 周）。

源数据是每个工作顶部的表格，其中观察是工人，格式如下：

UniqueID  Date     mapper
10000001 3/3/2015  person1

从这里开始，我将日期分解为年、月、日字段，然后使用lubridate 来获取周数。这适用于每个源数据文件（其中有 10 个）。

然后我按用户和周对每个表进行分组，并计算每个工人完成的工作数量，请参见此处的代码：

library(tidyverse)
library(readxl)
library(rlang)
library(pryr)
library(lubridate)
library(extrafont)
loadfonts(device = "win")


current_week <- week(Sys.Date())
feature_sheets <- excel_sheets("./typesMerged.xlsx")
list_sheets <- lapply(feature_sheets, function(x) read_excel(path = "./typesMerged.xlsx",col_types = c("numeric", "date", "text"), sheet = x))

users = list('person1', 'person2', 'person3', 'person4', 'person5', 'person6')


createTibble <- function(i, yr, wk){
  is_missing <- missing(wk)
  list_sheets[[i]]%>%
    filter(LASTUSER %in% users ) %>% 
    mutate(year = as.numeric(format(DATEMODIFI, format = "%Y")),
           month = as.numeric(format(DATEMODIFI, format = "%m")),
           week = week(DATEMODIFI),
           day = as.numeric(format(DATEMODIFI, format = "%d")))%>% 
    select(-DATEMODIFI) %>%
    filter(year == yr) %>%
    filter(if(is_missing) TRUE else week == wk) 
}




tableNames <- c('dpd', 'fuse', 'ohprimary', 'ohsecondary', 'poles', 'pv', 'switch','transformers', 'ugprimary', 'ugsecondary' )
features <- vector('list', length(tableNames))


for (feature in seq_along(features)){

  features[[feature]] <- createTibble(i = feature, yr = 2019)


}

# use for iteratively creating tibbles/df if needed
# for (feature in seq_along(features)){
#   assign(features[feature], value = createTibble(feature))
# }


countFeatures <- function(x){
  x %>%
    group_by(LASTUSER, week) %>%
    summarize(n = n())
}


featureCountTibbles <-map(.x = features,.f = countFeatures)

featureCountsByUser <- featureCountTibbles %>% reduce(full_join, by = c("LASTUSER", "week")) %>% 
  rename(mapper = LASTUSER) %>% 
  rename(dpd = n.x) %>%
  rename(fuse = n.y) %>% 
  rename(ohprimary = n.x.x) %>% 
  rename(ohsecondary = n.y.y) %>% 
  rename(poles = n.x.x.x) %>% 
  rename(pv = n.y.y.y) %>% 
  rename(switch = n.x.x.x.x) %>% 
  rename(transformers = n.y.y.y.y) %>% 
  rename(ugprimary = n.x.x.x.x.x) %>% 
  rename(ugsecondary = n.y.y.y.y.y) %>% 
  replace_na(replace = list(dpd = 0, fuse = 0, ohprimary = 0, ohsecondary = 0, poles = 0, pv = 0, switch = 0, transformers = 0, ugprimary = 0, ugsecondary = 0))

这导致一个看起来像这样的表，这里的问题是 person1 在第 5 周没有工作，所以他没有一行：

mapper  week  dpd  fuse  etc.
person1  1    10    50   ...
person1  2     0    50   ...
person1  3    10     0   ...
person1  4    10    50   ...
person1  6    10    50   ...
person2  1    10    50   ...
person2  2    50    50   ...
person2  3    10     0   ...
person2  4    10    50   ...
person2  5    10    50   ***
person2  6    10    50   ...

这就是我希望它在几周没有工作的地方看到的样子：

mapper  week  dpd  fuse  etc.
person1  1    10    50   ...
person1  2     0    50   ...
person1  3    10     0   ...
person1  4    10    50   ...
person1  5     0     0   ***
person1  6    10    50   ...
person2  1    10    50   ...
person2  2    50    50   ...
person2  3    10     0   ...
person2  4    10    50   ...
person2  5    10    50   ...
person2  6    10    50   ...

如何才能添加这些空行？这听起来很简单，但我真的想不出在 R 中做这件事的方法（对它来说还是有点新的）。如果这是 SQL，我会在开始时添加一个新字段 week，用 1 - 52 填充它（这将添加我需要的所有行），然后使用我在几周内拥有的实际数据更新表。

已实施的解决方案：

featureCountsByUser <- featureCountTibbles %>% reduce(full_join, by = c("LASTUSER", "week")) %>%
  rename(mapper = LASTUSER) %>%
  rename(dpd = n.x) %>%
  rename(fuse = n.y) %>% 
  rename(ohprimary = n.x.x) %>% 
  rename(ohsecondary = n.y.y) %>% 
  rename(poles = n.x.x.x) %>% 
  rename(pv = n.y.y.y) %>% 
  rename(switch = n.x.x.x.x) %>% 
  rename(transformers = n.y.y.y.y) %>% 
  rename(ugprimary = n.x.x.x.x.x) %>% 
  rename(ugsecondary = n.y.y.y.y.y) %>% 
  ungroup() %>% 
  complete(mapper,week) %>% 
  group_by(mapper,week) %>% 
  replace_na(replace = list(dpd = 0, fuse = 0, ohprimary = 0, ohsecondary = 0, poles = 0, pv = 0, switch = 0, transformers = 0, ugprimary = 0, ugsecondary = 0)) 

complete() 是答案；但与文档相反，它不尊重我的分组，并且由于我计算完成工作的方式的性质，我无法在原始数据上complete()，所以这是我的工作。谢谢大家！

Answer 1

使用您发布的非您想要的结果数据：

library(tidyverse)
x <- c("Worker  Week  dpd  fuse ", 
  "person1  1    10    5   ",
  "person1  2     0    5   ",
  "person1  3    10        ",
  "person1  4    10    5   ",
  "person1  6    10    5   ",
  "person2  1    10    5   ",
  "person2  2    50    5   ",
  "person2  3    10        ",
  "person2  4    10    5   ",
  "person2  5    10    5   ",
  "person2  6    10    5   ") %>%
  read_table()


x %>% complete(Worker, Week)

应该给：

# A tibble: 12 x 4
   Worker   Week   dpd  fuse
   <chr>   <dbl> <dbl> <dbl>
 1 person1     1    10     5
 2 person1     2     0     5
 3 person1     3    10    NA
 4 person1     4    10     5
 5 person1     5    NA    NA
 6 person1     6    10     5
 7 person2     1    10     5
 8 person2     2    50     5
 9 person2     3    10    NA
10 person2     4    10     5
11 person2     5    10     5
12 person2     6    10     5

complete() 有填写缺失数据的选项，@aosmith 提供的以上参考链接。用 0 填充 NA 应该不是问题。