将函数应用于data.table的每一行 [英] Applying a function to each row of a data.table
问题描述
我正在寻找一种方法来有效地将一个函数应用到data.table的每一行。让我们考虑下面的数据表:
library(data.table)
library(stringr)
x < - data.table(a = c(1:3,1),b = c('12 13','14 15','16 17','18 19'))
> x
ab
1:1 12 13
2:2 14 15
3:3 16 17
4:1 18 19
$ c $假设我想用空格拆分列 b
的每个元素(因此产生两个原始数据中的每行的行),并连接结果数据表。对于上面的示例,我需要以下结果: a V1
1:1 12
2 :1 13
3:2 14
4:2 15
5:3 16
6:3 17
7:1 18
8:1 19
如果 a
只有唯一值:
x [,list(str_split(b, 1]]),by = a]
除非在原始数据表中有一些相同的行),但是当 x
有很多列和复制列b到结果,这是我想避免的。 / p>
> x [,list(str_split(b,'')[[1]],by = list(a,b)]
ab V1
1:1 12 13 12
2:1 12 13 13
3:2 14 15 14
4:2 14 15 15
5:3 16 17 16
6:3 16 17 17
7:1 18 19 18
8:1 18 19 19
什么是最有效率和惯用的
如何:
<$ p $ p x
ab
1:1 12 13
2:2 14 15
3:3 16 17
4:1 18 19
x [,list(a = rep(a,each = 2),V1 = unlist(strsplit(b,)))
a V1
1:1 12
2:1 13
3:2 14
4:2 15
5:3 16
6:3 17
7:1 18
8:1 19
给出注释的广义解决方案:
$ b b
x [,{s = strsplit(b,); list(a = rep(a,sapply(s,length)),V1 = unlist }]
I looking for a way to efficiently apply a function to each row of data.table. Let's consider the following data table:
library(data.table)
library(stringr)
x <- data.table(a = c(1:3, 1), b = c('12 13', '14 15', '16 17', '18 19'))
> x
a b
1: 1 12 13
2: 2 14 15
3: 3 16 17
4: 1 18 19
Let's say I want to split each element of column b
by space (thus yielding two rows for each row in the original data) and join the resulting data tables. For the example above, I need the following result:
a V1
1: 1 12
2: 1 13
3: 2 14
4: 2 15
5: 3 16
6: 3 17
7: 1 18
8: 1 19
The following would work if column a
has only unique values:
x[, list(str_split(b, ' ')[[1]]), by = a]
The following almost works (unless there are some identical rows in the original data table), but is ugly when x
has many columns and copies column b to the result, which I would like to avoid.
> x[, list(str_split(b, ' ')[[1]]), by = list(a,b)]
a b V1
1: 1 12 13 12
2: 1 12 13 13
3: 2 14 15 14
4: 2 14 15 15
5: 3 16 17 16
6: 3 16 17 17
7: 1 18 19 18
8: 1 18 19 19
What would be the most efficient and idiomatic way to solve this problem?
How about :
x
a b
1: 1 12 13
2: 2 14 15
3: 3 16 17
4: 1 18 19
x[,list(a=rep(a,each=2), V1=unlist(strsplit(b," ")))]
a V1
1: 1 12
2: 1 13
3: 2 14
4: 2 15
5: 3 16
6: 3 17
7: 1 18
8: 1 19
Generalized solution given comment :
x[,{s=strsplit(b," ");list(a=rep(a,sapply(s,length)), V1=unlist(s))}]
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