根据其他变量的组合计算不同行中变量值的差异'；R中不同行的值
我在根据其他变量的组合计算不同行中变量值的差异'；R中不同行的值,r,data.table,panel-data,R,Data.table,Panel Data,我在数据表中有id和period标识的不平衡面板数据。共有8278个观测值和230个变量 我想知道我的数据中的公司（由id标识）从计划进入市场（plan\u entry==“yes”，包含NAs）到实际进入市场需要多长时间（enter\u market==“yes”，包含noNAs） 因此，如果在期间==4一家公司计划进入市场，并且在期间==9最终进入市场，我想生成例如进入时间==5。数据的结构大致如下，并且已经包含了所需的输出变量。请注意，公司可能会进入一个市场，而无需说明之前的任何计划。也可
数据表中有id
和period
标识的不平衡面板数据。共有8278个观测值和230个变量
我想知道我的数据中的公司（由id
标识）从计划进入市场（plan\u entry==“yes”
，包含NA
s）到实际进入市场需要多长时间（enter\u market==“yes”
，包含noNA
s）
因此，如果在期间==4
一家公司计划进入市场，并且在期间==9
最终进入市场，我想生成例如进入时间==5
。数据的结构大致如下，并且已经包含了所需的输出变量。请注意，公司可能会进入一个市场，而无需说明之前的任何计划。也可能是他们计划进入一个新市场，并在同一时期进入一个新市场。在这两种情况下，我都希望time\u to\u entry==0
。如果一家公司从未进入任何市场，则该值也应为0
示例性数据和预期结果
library(data.table)
desired_output <-
data.table(id = as.factor(c(rep("C001", 3), "C002", rep("C003", 5), rep("C004", 2), rep("C005", 7))),
period = as.factor(c(1, 2, 3, 2, 1, 4, 5, 6, 10, 3, 4, 2, 3, 4, 7, 8, 9, 10)),
plan_entry = as.factor(c(rep(NA, 2), "yes", "no", NA, rep("no", 2), rep("yes", 4), rep(NA, 2), rep("yes", 4), "no")),
enter_market = as.factor(c(rep("no", 3), "yes", rep("no", 5), rep("yes", 2), rep("no", 5), rep("yes", 2))),
time_to_entry = c(rep(0, 10), 1, rep(0, 5), 5, 1))
desired_output
# id period plan_entry enter_market time_to_entry
# 1: C001 1 <NA> no 0
# 2: C001 2 <NA> no 0
# 3: C001 3 yes no 0
# 4: C002 2 no yes 0
# 5: C003 1 <NA> no 0
# 6: C003 4 no no 0
# 7: C003 5 no no 0
# 8: C003 6 yes no 0
# 9: C003 10 yes no 0
#10: C004 3 yes yes 0 ! there might be cases
# where companies enter a market without stating any plans to do so in previous periods
#11: C004 4 yes yes 1
#12: C005 2 <NA> no 0
#13: C005 3 <NA> no 0
#14: C005 4 yes no 0
#15: C005 7 yes no 0
#16: C005 8 yes no 0
#17: C005 9 yes yes 5
#18: C005 10 no yes 1
dt$plan_条目==“是”的所有期间
dt$plan\u entry\u period这里有一个解决方案，可以重现您的所需结果
dt <-
data.table(id = as.factor(c(rep("C001", 3), "C002", rep("C003", 5), rep("C004", 2), rep("C005", 7))),
period = as.numeric(c(1, 2, 3, 2, 1, 4, 5, 6, 10, 3, 4, 2, 3, 4, 7, 8, 9, 10)),
plan_entry = as.factor(c(rep(NA, 2), "yes", "no", NA, rep("no", 2), rep("yes", 4), rep(NA, 2), rep("yes", 4), "no")),
enter_market = as.factor(c(rep("no", 3), "yes", rep("no", 5), rep("yes", 2), rep("no", 5), rep("yes", 2))))
dt[, time_to_entry_with_plan := period - min(period),
by = .(id,plan_entry)]
dt[, time_to_entry_without_plan := period - min(period),
by = .(id,enter_market)]
dt[, time_to_entry:=fcase(enter_market == "yes" & plan_entry == "yes", time_to_entry_with_plan,
enter_market == "yes" & plan_entry == "no", time_to_entry_without_plan,
default = 0)]
dt这里有一个使用非等联接的选项：
#find the previous latest enter_market before current row
DT[enter_market=="yes", prev_entry :=
fcoalesce(.SD[.SD, on=.(id, period<period), mult="last", x.period], 0L)
]
#non-equi join to find the first plan_entry before current enter_market but after previous latest enter_market
DT[enter_market=="yes", plan_period :=
DT[plan_entry=="yes"][.SD, on=.(id, period>=prev_entry, period<period), mult="first", x.period]
]
#calculate time_to_entry and set NAs to 0
DT[, time_to_entry := fcoalesce(period - plan_period, 0L)]
DT
#在当前行之前查找上一个最新进入的市场
DT[输入市场==“是”，上一个输入：=
fcoalesce（.SD[.SD，on=。（id，period=prev_entry，period非常感谢这种方法，Peace！我喜欢它，因为它看起来有点简单。但是，当我运行它时，它并没有完全重现所需的_输出。我收到的不是time_to_entry
最后一个单元格中的1，而是0。在整个数据集上运行它时，问题似乎总是不同在计算当前期间（其中enter\u market==“yes”
和第一个期间（其中plan\u entry==“yes”
）之间，而不是计算下一个期间（其中plan\u entry==“yes”
）。您确定吗？我在time\u to\u entry的最后一个单元格中得到1而不是0。我编辑答案以显示结果（请参见最后一列time\u to\u entry）。太好了，chinsoon12！非常感谢，这似乎工作得很好。您能解释一下为什么“.”例如inon=。（id，period表达式“（）”是list（）（from？data.table
）的简写别名。x.
用于引用x[i，on=keys]联接中右表中的列吗
和i
是左边的表格。从数据中读取所有的渐晕图。表格是一个很好的开始。太好了，非常感谢！
# fill in first entry_period for each observation by company
library(zoo) # for na.locf()
dt <- as.data.table(dt)
dt[, entry_period := na.locf(entry_period, na.rm = FALSE, fromLast = FALSE), by = id]
dt[, entry_period := na.locf(entry_period, na.rm = FALSE, fromLast = TRUE), by = id]
dt$time_to_entry <-
ifelse(
dt$plan_entry == "yes",
dt$entry_period - dt$plan_entry_period,
NA)
# check variable
summary(dt$time_to_entry)
dt
# id period plan_entry enter_market min_period min_entry_period entry_period plan_entry_period time_to_entry
# 1: C001 1 <NA> no 1 1 NA NA NA
# 2: C001 2 <NA> no 1 1 NA NA NA
# 3: C001 3 yes no 1 1 NA 3 NA
# 4: C002 2 no yes 2 2 NA NA 0
# 5: C003 1 <NA> no 1 1 NA NA NA
# 6: C003 4 no no 1 1 NA NA 0
# 7: C003 5 no no 1 1 NA NA 0
# 8: C003 6 yes no 1 1 NA 6 NA
# 9: C003 10 yes no 1 1 NA 10 NA
#10: C004 3 yes yes 3 3 NA 3 NA
#11: C004 4 yes yes 3 3 NA 4 NA
#12: C005 2 <NA> no 2 2 9 NA NA
#13: C005 3 <NA> no 2 2 9 NA NA
#14: C005 4 yes no 2 2 9 4 5
#15: C005 7 yes no 2 2 9 7 2
#16: C005 8 yes no 2 2 9 8 1
#17: C005 9 yes yes 2 9 9 9 0
#18: C005 10 no yes 2 9 9 NA 0
dt <-
data.table(id = as.factor(c(rep("C001", 3), "C002", rep("C003", 5), rep("C004", 2), rep("C005", 7))),
period = as.numeric(c(1, 2, 3, 2, 1, 4, 5, 6, 10, 3, 4, 2, 3, 4, 7, 8, 9, 10)),
plan_entry = as.factor(c(rep(NA, 2), "yes", "no", NA, rep("no", 2), rep("yes", 4), rep(NA, 2), rep("yes", 4), "no")),
enter_market = as.factor(c(rep("no", 3), "yes", rep("no", 5), rep("yes", 2), rep("no", 5), rep("yes", 2))))
dt[, time_to_entry_with_plan := period - min(period),
by = .(id,plan_entry)]
dt[, time_to_entry_without_plan := period - min(period),
by = .(id,enter_market)]
dt[, time_to_entry:=fcase(enter_market == "yes" & plan_entry == "yes", time_to_entry_with_plan,
enter_market == "yes" & plan_entry == "no", time_to_entry_without_plan,
default = 0)]
id period plan_entry enter_market time_to_entry_with_plan time_to_entry_without_plan time_to_entry
1: C001 1 <NA> no 0 0 0
2: C001 2 <NA> no 1 1 0
3: C001 3 yes no 0 2 0
4: C002 2 no yes 0 0 0
5: C003 1 <NA> no 0 0 0
6: C003 4 no no 0 3 0
7: C003 5 no no 1 4 0
8: C003 6 yes no 0 5 0
9: C003 10 yes no 4 9 0
10: C004 3 yes yes 0 0 0
11: C004 4 yes yes 1 1 1
12: C005 2 <NA> no 0 0 0
13: C005 3 <NA> no 1 1 0
14: C005 4 yes no 0 2 0
15: C005 7 yes no 3 5 0
16: C005 8 yes no 4 6 0
17: C005 9 yes yes 5 0 5
18: C005 10 no yes 0 1 1
#find the previous latest enter_market before current row
DT[enter_market=="yes", prev_entry :=
fcoalesce(.SD[.SD, on=.(id, period<period), mult="last", x.period], 0L)
]
#non-equi join to find the first plan_entry before current enter_market but after previous latest enter_market
DT[enter_market=="yes", plan_period :=
DT[plan_entry=="yes"][.SD, on=.(id, period>=prev_entry, period<period), mult="first", x.period]
]
#calculate time_to_entry and set NAs to 0
DT[, time_to_entry := fcoalesce(period - plan_period, 0L)]
DT
DT <-
data.table(id = c(rep("C001", 3), "C002", rep("C003", 5), rep("C004", 2), rep("C005", 7)),
period = as.integer(c(1, 2, 3, 2, 1, 4, 5, 6, 10, 3, 4, 2, 3, 4, 7, 8, 9, 10)),
plan_entry = c(rep(NA, 2), "yes", "no", NA, rep("no", 2), rep("yes", 4), rep(NA, 2), rep("yes", 4), "no"),
enter_market = c(rep("no", 3), "yes", rep("no", 5), rep("yes", 2), rep("no", 5), rep("yes", 2)))