如何仅从整数值中减去中值

问题描述 投票:0回答:1

我有这个数据集

    df=structure(list(Dt = structure(1:39, .Label = c("2018-02-20 00:00:00.000", 
"2018-02-21 00:00:00.000", "2018-02-22 00:00:00.000", "2018-02-23 00:00:00.000", 
"2018-02-24 00:00:00.000", "2018-02-25 00:00:00.000", "2018-02-26 00:00:00.000", 
"2018-02-27 00:00:00.000", "2018-02-28 00:00:00.000", "2018-03-01 00:00:00.000", 
"2018-03-02 00:00:00.000", "2018-03-03 00:00:00.000", "2018-03-04 00:00:00.000", 
"2018-03-05 00:00:00.000", "2018-03-06 00:00:00.000", "2018-03-07 00:00:00.000", 
"2018-03-08 00:00:00.000", "2018-03-09 00:00:00.000", "2018-03-10 00:00:00.000", 
"2018-03-11 00:00:00.000", "2018-03-12 00:00:00.000", "2018-03-13 00:00:00.000", 
"2018-03-14 00:00:00.000", "2018-03-15 00:00:00.000", "2018-03-16 00:00:00.000", 
"2018-03-17 00:00:00.000", "2018-03-18 00:00:00.000", "2018-03-19 00:00:00.000", 
"2018-03-20 00:00:00.000", "2018-03-21 00:00:00.000", "2018-03-22 00:00:00.000", 
"2018-03-23 00:00:00.000", "2018-03-24 00:00:00.000", "2018-03-25 00:00:00.000", 
"2018-03-26 00:00:00.000", "2018-03-27 00:00:00.000", "2018-03-28 00:00:00.000", 
"2018-03-29 00:00:00.000", "2018-03-30 00:00:00.000"), class = "factor"), 
    ItemRelation = c(158043L, 158043L, 158043L, 158043L, 158043L, 
    158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 
    158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 
    158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 
    158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 
    158043L, 158043L, 158043L, 158043L, 158043L, 158043L), stuff = c(200L, 
    0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 3600L, 0L, 0L, 0L, 0L, 
    700L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1000L, 
    2600L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 700L), num = c(1459L, 
    1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 
    1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 
    1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 
    1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 
    1459L, 1459L), year = c(2018L, 2018L, 2018L, 2018L, 2018L, 
    2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 
    2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 
    2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 
    2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L), action = c(0L, 
    0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 
    0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 
    0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L)), .Names = c("Dt", "ItemRelation", 
"stuff", "num", "year", "action"), class = "data.frame", row.names = c(NA, 
-39L))

动作列只有两个值0和1.我必须通过1个动作类别的东西计算中位数,然后使用零动作类别的东西计算中位数,使用一个类别之前的最后五个整数值。我只取最后5个观察结果,有必要在零行动类别中取最后5个观察值,但在我们的例子中只有整数值这是

200
3600
700
1000
2600

然后从一个类别的中位数中减去零类别的中位数。

零作业类别中的东西观察数量可以在0-10之间变化。如果我们有10个零类别的整数值,我们采用最后五个。如果只有1,2,3,4,5个整数值,我们减去实数整数值的中位数。如果我们只有0而没有整数,那么我们只需要取0。

怎么做?

edit expected output

Dt                  ItemRelation    DocumentNum DocumentYear    value
2018-03-30 00:00:00.000 158043        1459       2018          -300

*-300=(700-median(   200,
    3600,
    700,
    1000,
    2600)

note, if for 1 category of action by stuff there is one value(in our case it is 700) we don't calculate median, we work only with this value

如果有两个值,我们按东西计算1个类别的中位数

r dplyr plyr lapply
1个回答
1
投票
df.0 <- df %>% filter(action == 0 & stuff != 0) %>% arrange(Dt) %>% top_n(5)
df.1 <- df %>% filter(action==1 & stuff!=0)

new.df <- rbind(df.0,df.1)


View(
  df %>% select (everything()) %>%  group_by(ItemRelation, num, year) %>%
    summarise(
      median.1 = median(stuff[action == 1 & stuff != 0], na.rm = T),
      median.0 = median(stuff[action == 0 &
                                stuff != 0], na.rm = T)
    ) %>%
    mutate(
      value = median.1 - median.0,
      DocumentNum = num,
      DocumentYear = year
    ) %>%
    select(ItemRelation, DocumentNum, DocumentYear, value)

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