邏輯條件下的計數頻率

這裏是R代碼。邏輯條件下的計數頻率

illdate<-c("2009-05-26","2010-12-03","2011-06-05","2010-08-10","2009-05-14","2010-01-08","2012-08-07","2010-11-13","2013-06-06","2010-08-26") 

Treated<-c("T","T","N","T","N","N","T","T","T","N") 

Reason<-c("Finish","Finish","Dead","Dead","Cured","Cured","Finish","Finish","Dead","Cured") 

Enddate<-c("2009-12-12","2011-05-08","2011-02-28","2010-10-30","2011-10-08","2011-03-03","2013-05-05","2011-05-06","2014-08-08","2012-07-07") 

number<-c(1:10) 

mydata<-data.frame(number,illdate,Treated,Reason,Enddate)

而且我想算適合的邏輯條件的案件數量：

1:illdate between "2010-01-01" and "2010-12-31" 

2:Treated= "T" 

3:Reason= "Finish" & "Cured" 

4:Enddate between "2011-01-01" and "2012-12-31"

請告訴我如何寫R.代碼非常感謝你。

來源

2017-08-04 Binnnnn

你嘗試過什麼方法到現在爲止？這個論壇不是通用代碼生成的地方。 – zwep

只要執行'mydata％>％filter（之間（illdate，「2010-01-01」，「」2010-12-31「），處理==」T「，原因％在％c（」完成「，」（Enddate，「2011-01-01」，「2012-12-31」））'更改爲'as.Date（「2010-01-01」）'如果需要 – akrun

@akrun您應該添加作爲回答 – Sotos

我們可以做

res <- mydata %>% 
     mutate_at(vars(matches("date")), as.Date) %>% 
     filter(between(illdate, as.Date("2010-01-01"), as.Date("2010-12-31")), 
      Treated == "T", 
      Reason %in% c("Finish", "Cured"), 
      between(Enddate, as.Date("2011-01-01"), as.Date("2012-12-31"))) 

dim(res) 
#[1] 2 5

來源

2017-08-04 09:01:30 akrun

它的工作。你這麼多。:-) – Binnnnn

@ WuBin謝謝你的評論，你也可以點擊[這裏]（https://stackoverflow.com/help/someone-answers） – akrun

這裏是另一種方式，也許更少美麗：

nrow(subset(mydata,"2010-01-01" < as.Date(illdate) & as.Date(illdate) < "2010-12-31" & 
    Treated == "T" & Reason %in% c("Finish", "Cured") & 
    "2011-01-01" < as.Date(Enddate) & as.Date(Enddate) < "2012-12-31"))

來源

2017-08-04 09:05:53

它的工作，非常感謝。 - ） – Binnnnn

邏輯條件下的計數頻率

回答

相關問題