如何将“补丁”应用到数据。

For lack of a better word, how do I apply a "patch" to a R data.frame? Suppose I have a master database with firm and outlet columns and an ownership shares variable that is 1 or 0 in this example, but could be any percentage.

如果没有更好的词，我如何将“补丁”应用到R数据。假设我有一个具有公司和outlet列的主数据库，以及在这个例子中是1或0的所有权共享变量，但是可以是任何百分比。

// master
     firm outlet shares.pre
1    five      1          0
2     one      1          1
3     red      1          0
4  yellow      1          0
5    five      2          0
6     one      2          0
// many more

I want to let firm "one" sell outlet "1" to firm "red", which transaction I have in another data.frame

我想让我在另一个data.frame中的“one”sell outlet“1”to“firm”red”

// delta
  firm outlet shares.delta
1  one      1         -1
2  red      1          1

What is the most efficient way in R to apply this "patch" or transaction to my master database? The end result should look like this:

在R中，将这个“补丁”或事务应用到主数据库的最有效的方法是什么?最终结果应该是这样的:

// preferably master, NOT a copy
     firm outlet shares.post
1    five      1          0
2     one      1          0  <--- was 1
3     red      1          1  <--- was 0
4  yellow      1          0
5    five      2          0
6     one      2          0
// many more

I am not particular about keeping the suffixes pre, post or delta. If they were all named shares that would be fine too, I simply want to "add" these data frames.

我并不是特别要保留前缀、post或delta。如果它们都被命名为共享，那也没问题，我只想“添加”这些数据帧。

UPDATE: my current approach is this

更新:我目前的方法是这样的

update <- (master$firm %in% delta$firm) & (master$outlet %in% delta$outlet)
master[update,]$shares <- master[update,]$shares + delta$shares

Yes, I'm aware it does a vector scan to creat the Boolean update vector, and that the subsetting is also not very efficient. But the thing I don't like about it most is that I have to write out the matching columns.

是的，我知道它会进行向量扫描来创建布尔更新向量，而且子设置也不是很有效。但我最不喜欢的是我必须写出匹配的列。

2 个解决方案

#1

Another way using data.table. Assuming you've loaded both your data in df1 and df2 data.frames,

另一种方式使用data.table。假设您已经在df1和df2数据中加载了数据。

require(data.table)
dt1 <- data.table(df1)
dt2 <- data.table(df2)
setkey(dt1, firm, outlet)
setkey(dt2, firm, outlet)
dt1 <- dt2[dt1]
dt1[is.na(dt1)] <- 0
dt1[, shares.post := shares.delta + shares.pre]

#      firm outlet shares.delta shares.pre shares.post
# 1:   five      1            0          0           0
# 2:   five      2            0          0           0
# 3:    one      1           -1          1           0
# 4:    one      2            0          0           0
# 5:    red      1            1          0           1
# 6: yellow      1            0          0           0

#2

I'd give a more precise answer if you had provided a reproducible example, but here's one way:

如果你能提供一个可重复的例子，我会给出一个更精确的答案，但有一个方法:

Call your first data.frame dat and your second chg
打电话给你的第一个数据

Then you could merge the two:

然后你可以合并这两个:

dat <- merge(dat,chg)

And just subtract:

就减:

dat$shares <- with(dat, shares.pre + shares.delta )

#1