我有一个包含名称、日期和几个分类列的数据集。让我们说
data <- data.table(name = c('Anne', 'Ben', 'Cal', 'Anne', 'Ben', 'Cal', 'Anne', 'Ben', 'Ben', 'Ben', 'Cal'),
period = c(1,1,1,1,1,1,2,2,2,3,3),
category = c("A","A","A","B","B","B","A","B","A","B","A"))
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看起来像这样:
name period category
Anne 1 A
Ben 1 A
Cal 1 A
Anne 1 B
Ben 1 B
Cal 1 B
Anne 2 A
Ben 2 B
Ben 2 A
Ben 3 A
Cal 3 B
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我想计算,对于每个时期,对于我的每组分类变量,过去时期存在多少个名字。输出应如下所示:
period category recurrence_count
2 A 2 # due to Anne and Ben being on A, period 1
2 B 1 # due to Ben being on B, period 1
3 A 1 # due to Ben being on A, period 2
3 B 0 # no match from B, period 2
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我知道 data.table 中的 .I 和 .GRP 运算符,但我不知道如何在语句的 j 条目中编写“下一组”的概念。我想像这样的事情可能是一条合理的路径,但我无法弄清楚正确的语法:
data[, .(recurrence_count = length(intersect(name, name[last(.GRP)]))), by = .(category, period)]
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另一种data.table选择。对于可以有前一个句点 ( period != 1) 的行,创建这样一个变量 ( prev_period := period - 1)。
将原始数据与具有“prev_period”值的子集 ( data[data[!is.na(prev_period)]) 连接起来。连接“类别”、“期间 = prev_period”和“名称”。
在生成的数据集中,对于每个“周期”和“类别”( ),计算原始数据 ( ) 中与前一个周期 ( ) 匹配的by = .(period = i.period, category)名称数量。x.namelength(na.omit(x.name))
data[period != 1, prev_period := period - 1]
data[data[!is.na(prev_period)], on = c("category", period = "prev_period", "name"),
.(category, i.period, x.name)][
, .(n = length(na.omit(x.name))), by = .(period = i.period, category)]
# period category n
# 1: 2 A 2
# 2: 2 B 1
# 3: 3 B 1
# 4: 3 A 0
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