如何对行进行有条件的计数器？R
我得到了这个公司的数据集，我已经“完成了面板”，所以只要数量变量（销售、工资）为0，公司就关闭了。NA表示我已经完成了该小组，这意味着所有公司都有相同的年份，但NA表示该公司之前（或之后）不存在 我想为公司的第一次倒闭做一个柜台 因此，我的数据如下所示：如何对行进行有条件的计数器？R,r,database,dataframe,dplyr,tidyverse,R,Database,Dataframe,Dplyr,Tidyverse,我得到了这个公司的数据集，我已经“完成了面板”，所以只要数量变量（销售、工资）为0，公司就关闭了。NA表示我已经完成了该小组，这意味着所有公司都有相同的年份，但NA表示该公司之前（或之后）不存在 我想为公司的第一次倒闭做一个柜台 因此，我的数据如下所示： Year Firm sales wages 2014 A 12 4 2015 A 8 3 2016 A 0 0 2017
Year Firm sales wages
2014 A 12 4
2015 A 8 3
2016 A 0 0
2017 A NA NA
2018 A NA NA
2014 B NA NA
2015 B 8 3
2016 B 4 2
2017 B 9 5
2018 B 8 6
2014 C 9 5
2015 C 7 6
2016 C 0 0
2017 C 0 0
2018 C 0 0
Year Firm sales wages Closure
2014 A 12 4 0
2015 A 8 3 0
2016 A 0 0 1
2017 A NA NA 2 # After the closure in 2016 it doesn't appear on the original dataset anymore
2018 A NA NA 3 # Same here
2014 B NA NA 0 # Here the firm has not been created yet
2015 B NA NA 0 # Here too
2016 B 4 2 0
2017 B 9 5 0
2018 B 8 6 0
2014 C 9 5 0
2015 C 7 6 0
2016 C 0 0 1
2017 C 0 0 2 #After the closure it continues appearing because the firm has some debts or some pending
2018 C 0 0 3 #Here the same, still appears bc it still have obligations
Year Firm sales wages
2014 A 12 4
2015 A 8 3
2016 A 0 0
2017 A NA NA
2018 A NA NA
2014 B NA NA
2015 B 8 3
2016 B 4 2
2017 B 9 5
2018 B 8 6
2014 C 9 5
2015 C 7 6
2016 C 0 0
2017 C 0 0
2018 C 0 0
Year Firm sales wages Closure
2014 A 12 4 0
2015 A 8 3 0
2016 A 0 0 1
2017 A NA NA 2 # After the closure in 2016 it doesn't appear on the original dataset anymore
2018 A NA NA 3 # Same here
2014 B NA NA 0 # Here the firm has not been created yet
2015 B NA NA 0 # Here too
2016 B 4 2 0
2017 B 9 5 0
2018 B 8 6 0
2014 C 9 5 0
2015 C 7 6 0
2016 C 0 0 1
2017 C 0 0 2 #After the closure it continues appearing because the firm has some debts or some pending
2018 C 0 0 3 #Here the same, still appears bc it still have obligations
library(dplyr)
library(tidyr)
df1 %>%
group_by(Firm) %>%
mutate(Closure = replace_na(cumsum(lead(is.na(sales) &
is.na(wages), default = TRUE)|(sales == 0 & wages == 0)), 0)) %>%
ungroup
# A tibble: 15 x 5
# Year Firm sales wages Closure
# <int> <chr> <int> <int> <dbl>
# 1 2014 A 12 4 0
# 2 2015 A 8 3 0
# 3 2016 A 0 0 1
# 4 2017 A NA NA 2
# 5 2018 A NA NA 3
# 6 2014 B NA NA 0
# 7 2015 B 8 3 0
# 8 2016 B 4 2 0
# 9 2017 B 9 5 0
#10 2018 B 8 6 0
#11 2014 C 9 5 0
#12 2015 C 7 6 0
#13 2016 C 0 0 1
#14 2017 C 0 0 2
#15 2018 C 0 0 3
#一个tible:15 x 5
#年公司销售工资关闭
#
#1 2014 A 12 4 0
#2 2015 A 8 3 0
#2016年3月0日1
#4 2017 A不适用2
#5 2018 A不适用3
#2014年6月B日NA 0
#7 2015 B 8 3 0
#8 2016 B 4 2 0
#9 2017 B 9 5 0
#10 2018 B 8 6 0
#11 2014 C 9 5 0
#12 2015 C 7 6 0
#13 2016 c0 0 1
#14 2017 C 0 2
#15 2018 C 0 3
df1是的，它起作用了。但是我忘了假设销售=0，工资=0所以现在我想我还需要两杯可乐。我会付账的，但现在我要再问一次，请帮帮我。@JorgeParedes我刚刚看过你编辑的评论。你能不能作为一个新问题发表。谢谢你，现在就发了。