Question

我的每小时/每日库存退货数据类似于以下内容：

  set.seed(1)
  dt <- data.table(stock = c(rep("a",24),rep("b",24),rep("c",24),rep("d",24)),
  hour = rep(1:24,4), day1 = sample(-5:5,96,replace = TRUE), 
  day2 = sample(-10:-1,96,replace = TRUE), day3 = sample(0:10,96,replace = TRUE),
  day4 = 0)

看起来像：

  stock hour    day1    day2    day3    day4
  a      1      -3       -6      1       0
  a      2      -1       -6      10      0
  a      3       1       -2      3       0
  ...                   
  d      22      4       -5      1       0
  d      23      3       -3      3       0
  d      24      3       -7      1       0

我想每天计算以下内容：

有多少股票显示所有小时的负回报
有多少股票显示所有时段的正回报
24小时内有多少股票显示出正反复收益
24小时内有多少股票显示至少一个零
所有24小时内有多少股票显示为零

输出应该类似于：

  counts                            day1    day2    day3    day4
  stocks_where_all_hours_negative   0       4       0       0
  stocks_where_all_hours_positive   0       0       4       0
  stocks_where_mixed_pos_and_neg    4       0       0       0
  stocks_where_at_least_one_zero    4       0       2       4
  stocks_where_all_zero             0       0       0       4

非常感谢。

Answer 1

mdt <- melt(
  dt,
  id.vars = c('stock', 'hour'),
  measure.vars = c('day1', 'day2', 'day3', 'day4'),
  variable.name = 'day',
  value.name = 'position'
)
counts <- mdt[ ,.(stocks_where_all_hours_negative = all(position < 0),
        stocks_where_all_hours_positive = all(position > 0),
        stocks_where_mixed_pos_and_neg = any(position < 0) & any(position > 0),
        stocks_where_at_least_one_zero = any(position != 0),
        stocks_where_all_zero = all(position == 0)),
     by = c('day',
            'stock')][ ,.(stocks_where_all_hours_negative = sum(stocks_where_all_hours_negative),
                          stocks_where_all_hours_positive = sum(stocks_where_all_hours_positive),
                          stocks_where_mixed_pos_and_neg = sum(stocks_where_mixed_pos_and_neg),
                          stocks_where_at_least_one_zero = sum(stocks_where_at_least_one_zero),
                          stocks_where_all_zero = sum(stocks_where_all_zero)),
                       by = 'day']
t(counts)

首先将数据重新整形为“长”格式，以便您可以按天分组。 any和all函数将检查每个向量的逻辑并返回您设置的方案。

Answer 2

由于您所做的摘要非常专业，您可以编写自己的函数：

summarize_stocks <- function(stock, day, return_names = FALSE){
  if(!return_names) return( c(length(unique(stock[as.logical(ave(day, stock, FUN = function(x) all(x<0)))])),
                              length(unique(stock[as.logical(ave(day, stock, FUN = function(x) all(x>0)))])),
                              length(unique(stock[as.logical(ave(day, stock, FUN = function(x) any(x>0) & any(x<0)))])),
                              length(unique(stock[as.logical(ave(day, stock, FUN = function(x) any(x==0)))])),
                              length(unique(stock[as.logical(ave(day, stock, FUN = function(x) all(x==0)))]))
                              ))
  else return(c("stocks_where_all_hours_negative", "stocks_where_all_hours_positive", "stocks_where_mixed_pos_and_neg",
                  "stocks_where_at_least_one_zero", "stocks_where_all_zero"))
}

res <- dt[, lapply(.SD, summarize_stocks, stock = stock), .SDcols = c("day1","day2", "day3", "day4")]
res[, counts := summarize_stocks(return_names = TRUE)]

res
#       day1 day2 day3 day4                          counts
#1:    0    4    0    0 stocks_where_all_hours_negative
#2:    0    0    2    0 stocks_where_all_hours_positive
#3:    4    0    0    0  stocks_where_mixed_pos_and_neg
#4:    4    0    2    4  stocks_where_at_least_one_zero
#5:    0    0    0    4           stocks_where_all_zero

或者，如果您想获得幻想，可以将summarize_stocks更改为：

summarize_stocks <- function(stock, day, return_names = FALSE){
  funs <- list(function(x) all(x<0),
               function(x) all(x>0),
               function(x) any(x>0) & any(x<0),
               function(x) any(x==0),
               function(x) all(x==0))

  if(!return_names) return( vapply(funs, function(f) length(unique(stock[as.logical(ave(day, stock, FUN = function(x) f(x)))])), numeric(1L)))

  else return(c("stocks_where_all_hours_negative", "stocks_where_all_hours_positive", "stocks_where_mixed_pos_and_neg",
                  "stocks_where_at_least_one_zero", "stocks_where_all_zero"))
}

使用data.table计算满足条件的分组观察值

2 个答案: