如何创建连续列（R）

我有处方记录数据，并想知道每个人每年从发行日期到其记录结束时有多少处方。实施例的数据（第5行的每个ID的）：如何创建连续列（R）

 ID Issue_Date index.date other.drugs 
    1: 1 2000-02-08 2011-02-03   1 
    2: 1 2000-04-04 2011-02-03   0 
    3: 1 2000-05-30 2011-02-03   1 
    4: 1 2000-07-25 2011-02-03   1 
    5: 1 2000-08-22 2011-02-03   1 
---          
    1: 2 2007-03-23 2009-04-03   1 
    2: 2 2007-04-04 2009-04-03   1 
    3: 2 2007-04-23 2009-04-03   1 
    4: 2 2007-04-23 2009-04-03   0 
    5: 2 2007-05-21 2009-04-03   1

的other.drugs列是一个指示变量表示在该日期给出的处方是否不在研究兴趣的处方。 index.date是他们进入研究的日期。有超过1000个ID的，只有2个在这里给出。

我想每年在issue.date之后找到每年other.drugs的总和。我分别使用下面的代码计算出这第一年：

dt <- dt[, yearend.1 := Issue_Date[1]+365, by = ID] 
dt <- dt[(Issue_Date<=yearend.1), comorbid.1 := sum(other.drugs), by = ID] 
dt <- dt[, comorbid.1:= comorbid.1[!is.na(comorbid.1)][1], by = ID] 
# the last line copies the value to each cell the ID occupies in the data.table for that column instead of having NA's

这给了以下结果：

 ID Issue_Date index.date other.drugs yearend.1 comorbid.1 
    1: 1 2000-02-08 2011-02-03   1 2001-02-07   8 
    2: 1 2000-04-04 2011-02-03   1 2001-02-07   8 
    3: 1 2000-05-30 2011-02-03   1 2001-02-07   8 
    4: 1 2000-07-25 2011-02-03   1 2001-02-07   8 
    5: 1 2000-08-22 2011-02-03   1 2001-02-07   8 
--- 
    1: 2 2007-03-23 2009-04-03   1 2008-03-22   30 
    2: 2 2007-04-04 2009-04-03   1 2008-03-22   30 
    3: 2 2007-04-23 2009-04-03   1 2008-03-22   30 
    4: 2 2007-04-23 2009-04-03   1 2008-03-22   30 
    5: 2 2007-05-21 2009-04-03   1 2008-03-22   30

解读：ID 1后，他们的第一个issue_date规定在今年8种其他药物和ID 2遵医嘱30.

多年来2-10（有一个最大为11年的记录）我写下面的循环：

years <- seq(730, 3650, 365) 
# number of days in 2-10 years. 
years2 <- seq(2,10,1) 
# numbering the years for column names 
colnames <- paste0("yearend.", years2) 
colnames2 <- paste0("comorbid.", years2) 
# names of columns to be used 

for (i in 1:length(years)) { 
    dt <- dt[, colnames[i] := Issue_Date[1]+years[i], by = ID] 
    dt <- dt[(Issue_Date>=(as.Date(colnames[i], "%d-%m-%Y")) & Issue_Date<(as.Date(colnames[i+1], "%d-%m-%Y"))), 
     colnames2[i] := sum(other.drugs), by = ID] 
    dt <- dt[, colnames2[i]:= colnames2[i][!is.na(colnames2[i])][1], by = ID] 
}

但是应该已经创造了新的栏目有：

 ID Issue_Date index.date other.drugs yearend.1 comorbid.1 yearend.2 comorbid.2 yearend.3 comorbid.3 
    1: 1 2000-02-08 2011-02-03   1 2001-02-07   8 2002-02-07 comorbid.2 2003-02-07 comorbid.3 
    2: 1 2000-04-04 2011-02-03   1 2001-02-07   8 2002-02-07 comorbid.2 2003-02-07 comorbid.3 
    3: 1 2000-05-30 2011-02-03   1 2001-02-07   8 2002-02-07 comorbid.2 2003-02-07 comorbid.3 
    4: 1 2000-07-25 2011-02-03   1 2001-02-07   8 2002-02-07 comorbid.2 2003-02-07 comorbid.3 
    5: 1 2000-08-22 2011-02-03   1 2001-02-07   8 2002-02-07 comorbid.2 2003-02-07 comorbid.3 
---

我想知道什么是我的循环去错了。非常感谢帮助。

来源

2016-03-15 Lb93

无论何时需要在data.table中使用实际来自R中变量的列名称，都需要使用get。因此，你应该重写你的循环这样，

for (i in 1:length(years)) { 
    dt <- dt[, colnames[i] := Issue_Date[1]+years[i], by = ID] 
    dt <- dt[(Issue_Date>=(as.Date(get(colnames[i]), "%d-%m-%Y")) & Issue_Date<(as.Date(get(colnames[i+1]), "%d-%m-%Y"))), 
     colnames2[i] := sum(other.drugs), by = ID] 
    dt <- dt[, colnames2[i]:= get(colnames2[i])[!is.na(get(colnames2[i]))][1], by = ID] 
}

我无法实际测试你的代码，因为它是，因为我有2个问题：

我没有足够的数据，以便我会得到任何东西从你的暂时的情况Issue_Date>...
也许我错过了一些东西，但在你的循环中，你试图使用colnames[i+1]，即yearend.X实际上被创建之前（也许你已经跑了好几次，这就是为什么你不要没有错误？）

我做了这样的事情来测试它，当然的comorbid.2值没有任何意义：

dt 
    ID Issue_Date index.date other.drugs yearend.1 comorbid.1 
1: 1 00-02-08 2011-02-03   1 01-02-07   4 
2: 1 00-04-04 2011-02-03   0 01-02-07   4 
3: 1 00-05-30 2011-02-03   1 01-02-07   4 
4: 1 00-07-25 2011-02-03   1 01-02-07   4 
5: 1 00-08-22 2011-02-03   1 01-02-07   4 
6: 2 07-03-23 2009-04-03   1 08-03-22   4 
7: 2 07-04-04 2009-04-03   1 08-03-22   4 
8: 2 07-04-23 2009-04-03   1 08-03-22   4 
9: 2 07-04-23 2009-04-03   0 08-03-22   4 
10: 2 07-05-21 2009-04-03   1 08-03-22   4 

i <- 1 
dt <- dt[, colnames[i] := Issue_Date[1]+years[i], by = ID] 
dt <- dt[Issue_Date<get(colnames[i]), 
     colnames2[i] := sum(other.drugs), by = ID] 
dt <- dt[, colnames2[i]:= get(colnames2[i])[!is.na(get(colnames2[i]))][1], by = ID] 

dt 
    ID Issue_Date index.date other.drugs yearend.1 comorbid.1 yearend.2 comorbid.2 
1: 1 00-02-08 2011-02-03   1 01-02-07   4 02-02-07   4 
2: 1 00-04-04 2011-02-03   0 01-02-07   4 02-02-07   4 
3: 1 00-05-30 2011-02-03   1 01-02-07   4 02-02-07   4 
4: 1 00-07-25 2011-02-03   1 01-02-07   4 02-02-07   4 
5: 1 00-08-22 2011-02-03   1 01-02-07   4 02-02-07   4 
6: 2 07-03-23 2009-04-03   1 08-03-22   4 09-03-22   4 
7: 2 07-04-04 2009-04-03   1 08-03-22   4 09-03-22   4 
8: 2 07-04-23 2009-04-03   1 08-03-22   4 09-03-22   4 
9: 2 07-04-23 2009-04-03   0 08-03-22   4 09-03-22   4 
10: 2 07-05-21 2009-04-03   1 08-03-22   4 09-03-22   4

希望它能帮助。

来源

2016-03-15 11:29:31 lrnzcig

如何创建连续列（R）

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