R 加注NAs NOCB（下一次观察向后进行）

我的数据是从KDB数据库查询返回的，以列表的形式出现 如果我是str（myData）

我面临的问题是，前20个左右的回报是NaN，输出中可能还有更多NaN

如果我们的列表是这样的，我想用第一行中没有NaN的结果替换NaN

[1]     [2]           [3]    [4]   [5]     [6]    [7]
09:30   Nan           Nan    Nan   Nan     Nan    Nan 
09:31   Nan           Nan    Nan   Nan     Nan    Nan 
09:32   2013:01:01    ABC    100   50      101    35 
09:33   2013:01:01    ABC    100   50      102    60
09:34   Nan           Nan    Nan   Nan     Nan    Nan 
09:35   2013:01:01    ABC    99   40      101    50


This would come out as follows:
[1]     [2]           [3]    [4]   [5]     [6]    [7]
09:30   2013:01:01    ABC    100   50      101    35 
09:31   2013:01:0     ABC    100   50      101    35 
09:32   2013:01:01    ABC    100   50      101    35 
09:33   2013:01:01    ABC    100   50      102    60
09:34   2013:01:01    ABC    99   40       101    50 
09:35   2013:01:01    ABC    99   40       101    50

也就是说，Nan'a被同一列（列表）中的下一个有效条目替换。

您可以使用和：

---

d使用
zoo:：na.locf

## if your data is in a data frame you can do it like this 
your_data = zoo::na.locf(your_data, fromLast = TRUE)

## if your data is a list, do this
your_data = lapply(your_data, zoo::na.locf, fromLast = TRUE)

## if your data is in a data frame you can do it like this 
your_data = zoo::na.locf(your_data, fromLast = TRUE)

## if your data is a list, do this
your_data = lapply(your_data, zoo::na.locf, fromLast = TRUE)