R 从面板中删除行，同时保持面板平衡 [英] R remove rows from panel while keeping the panel balanced

问题描述

有没有一种优雅的方法来平衡不平衡的面板数据集?我想从一个不平衡的面板开始(即，有些人缺少一些数据)并以一个平衡的面板结束(即，所有的人都没有丢失数据).下面是一些示例代码.正确的最终结果是保留对 'Frank' 和 'Edward' 的所有观察，并删除对 'Tony' 的所有观察，因为他有一些缺失的数据.谢谢.

Is there an elegant way to balance an unbalanced panel data set? I would like to start with an unbalanced panel (ie, some individuals are missing some data) and end up with a balanced panel (ie, all individuals are missing no data). Below is some sample code. The correct end result is for all observations on 'Frank' and 'Edward' to remain and for all observations on 'Tony' to be removed since he has some missing data. Thank you.

unbal <- data.frame(PERSON=c(rep('Frank',5),rep('Tony',5),rep('Edward',5)), YEAR=c(2001,2002,2003,2004,2005,2001,2002,2003,2004,2005,2001,2002,2003,2004,2005), Y=c(21,22,23,24,25,5,6,NA,7,8,31,32,33,34,35), X=c(1:15))
unbal

推荐答案

平衡面板的一种方式是去除数据不完整的个体，另一种方式是填入一个值，比如NA或 0 表示缺失的观察值.对于第一种方法，您可以使用 complete.cases 来查找其中没有 NA 的行.然后你可以找到所有的PERSON，至少有一个缺失的案例.

One way to balance a panel is to remove individuals with incomplete data, another way is to fill in a value, such as NA or 0 for the missing observations. For the first approach, you can use complete.cases to find rows that have no NA in them. Then you can find all the PERSON with at least one missing case.

missing.at.least.one <- unique(unbal$PERSON[!complete.cases(unbal)])
unbal[!(unbal$PERSON %in% missing.at.least.one),]
#    PERSON YEAR  Y  X
# 1   Frank 2001 21  1
# 2   Frank 2002 22  2
# 3   Frank 2003 23  3
# 4   Frank 2004 24  4
# 5   Frank 2005 25  5
# 11 Edward 2001 31 11
# 12 Edward 2002 32 12
# 13 Edward 2003 33 13
# 14 Edward 2004 34 14
# 15 Edward 2005 35 15

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