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r - 根据截止年份添加元素和列？

问题描述

structure(list(`2005` = c(0L, 0L, 0L, 2L, 1L), `2006` = c(0L, 
0L, 0L, 1L, 1L), `2007` = c(1L, 0L, 1L, 0L, 3L), `2008` = c(1L, 
0L, 0L, 4L, 3L), `2009` = c(1L, 0L, 0L, 2L, 3L), `2010` = c(0L, 
0L, 0L, 5L, 0L), `2011` = c(0L, 0L, 0L, 0L, 1L), `2012` = c(0L, 
0L, 0L, 4L, 1L), `2013` = c(1L, 0L, 1L, 0L, 0L), `2014` = c(0L, 
0L, 2L, 0L, 9L), `2015` = c(0L, 0L, 1L, 0L, 2L), `2016` = c(0L, 
0L, 0L, 0L, 0L), Cutoff = c("2011", "2015", "2015", "2005", "2011"
)), .Names = c("2005", "2006", "2007", "2008", "2009", "2010", 
"2011", "2012", "2013", "2014", "2015", "2016", "Cutoff"), row.names = c(NA, 
5L), class = "data.frame")

给定以下数据框。我想在表格中添加 4 列。一列将截止年份之前的元素中的数字相加，一列将截止年份之后的元素中的数字相加。

然后再添加两列，其中一列添加截止前的年/列总数，另一列添加截止后的总年数/列。

截止年份不应包含在相应的行中。

所以决赛桌最终会是这样的：

structure(list(`2005` = c(0L, 0L, 0L, 2L, 1L), `2006` = c(0L, 
0L, 0L, 1L, 1L), `2007` = c(1L, 0L, 1L, 0L, 3L), `2008` = c(1L, 
0L, 0L, 4L, 3L), `2009` = c(1L, 0L, 0L, 2L, 3L), `2010` = c(0L, 
0L, 0L, 5L, 0L), `2011` = c(0L, 0L, 0L, 0L, 1L), `2012` = c(0L, 
0L, 0L, 4L, 1L), `2013` = c(1L, 0L, 1L, 0L, 0L), `2014` = c(0L, 
0L, 2L, 0L, 9L), `2015` = c(0L, 0L, 1L, 0L, 2L), `2016` = c(0L, 
0L, 0L, 0L, 0L), Cutoff = c("2011", "2015", "2015", "2005", "2011"
), Numbers_Before = c(3, 0, 4, 0, 11), Numbers_After = c(1, 0, 
0, 16, 12), Years_Before = c(6, 10, 10, 0, 6), Years_After = c(5, 
1, 1, 11, 5)), .Names = c("2005", "2006", "2007", "2008", "2009", 
"2010", "2011", "2012", "2013", "2014", "2015", "2016", "Cutoff", 
"Numbers_Before", "Numbers_After", "Years_Before", "Years_After"
), row.names = c(NA, 5L), class = "data.frame")

标签： r

解决方案

我发现首先使用melt将表格整理成整齐的格式然后使用一些 data.table 操作来计算年份或截止年份之前和之后的数字更容易。

library(data.table)

dt = setDT(structure(list(`2005` = c(0L, 0L, 0L, 2L, 1L), `2006` = c(0L, 
  0L, 0L, 1L, 1L), `2007` = c(1L, 0L, 1L, 0L, 3L), `2008` = c(1L, 
  0L, 0L, 4L, 3L), `2009` = c(1L, 0L, 0L, 2L, 3L), `2010` = c(0L, 
  0L, 0L, 5L, 0L), `2011` = c(0L, 0L, 0L, 0L, 1L), `2012` = c(0L, 
  0L, 0L, 4L, 1L), `2013` = c(1L, 0L, 1L, 0L, 0L), `2014` = c(0L, 
  0L, 2L, 0L, 9L), `2015` = c(0L, 0L, 1L, 0L, 2L), `2016` = c(0L, 
  0L, 0L, 0L, 0L), Cutoff = c("2011", "2015", "2015", "2005", "2011"
  )), .Names = c("2005", "2006", "2007", "2008", "2009", "2010", 
  "2011", "2012", "2013", "2014", "2015", "2016", "Cutoff"), row.names = c(NA, 
  5L), class = "data.frame"))

dt[, row := rownames(dt)]
dt2 = melt(dt, id.vars = c('Cutoff', 'row'), variable.name = 'Year', variable.factor = F)

dt2[, Numbers_Before := ifelse(Year < Cutoff, value, 0)] 
dt2[, Numbers_After := ifelse(Year > Cutoff, value, 0)]
dt2[, Years_Before := ifelse(Year < Cutoff, 1, 0)]
dt2[, Years_After := ifelse(Year > Cutoff, 1, 0)]

dt3 = dt2[, .(Numbers_Before = sum(Numbers_Before), Numbers_After = sum(Numbers_After), 
              Years_Before = sum(Years_Before), Years_After = sum(Years_After)), by = row]

dt = merge(dt,dt3, by = 'row')

> dt
   row 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 Cutoff Numbers_Before Numbers_After Years_Before Years_After
1:   1    0    0    1    1    1    0    0    0    1    0    0    0   2011              3             1            6           5
2:   2    0    0    0    0    0    0    0    0    0    0    0    0   2015              0             0           10           1
3:   3    0    0    1    0    0    0    0    0    1    2    1    0   2015              4             0           10           1
4:   4    2    1    0    4    2    5    0    4    0    0    0    0   2005              0            16            0          11
5:   5    1    1    3    3    3    0    1    1    0    9    2    0   2011             11            12            6           5

编辑：这里使用更聪明的数据表语法和 dcast 而不是 ifelses：

dt[, row := rownames(dt)]
dt2 = melt(dt, id.vars = c('Cutoff', 'row'), variable.name = 'Year', variable.factor = F)
dt2 = dt2[Year != Cutoff][, .(Numbers = sum(value), Years = .N), by = .(row, Year > Cutoff, Cutoff)]
dt2 = dcast(dt2, row + Cutoff ~ Year, value.var = c('Numbers', 'Years'), fill = 0)
dt = merge(dt, dt2, by = c('row', 'Cutoff'))

> dt
   row Cutoff 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 Numbers_FALSE
1:   1   2011    0    0    1    1    1    0    0    0    1    0    0    0             3
2:   2   2015    0    0    0    0    0    0    0    0    0    0    0    0             0
3:   3   2015    0    0    1    0    0    0    0    0    1    2    1    0             4
4:   4   2005    2    1    0    4    2    5    0    4    0    0    0    0             0
5:   5   2011    1    1    3    3    3    0    1    1    0    9    2    0            11
   Numbers_TRUE Years_FALSE Years_TRUE
1:            1           6          5
2:            0          10          1
3:            0          10          1
4:           16           0         11
5:           12           6          5

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