问题描述
如果我有数据框架
set.seed(12345)
df = data.frame = rnorm(5),b = rnorm(5))
/ p>
df [6,] = c(5,6)
$ b b
如果我现在在data.table中做同样的操作
library(data.table)
dt = data.table(df)
dt [6,] = c(5,6)
它失败,出现错误。
要扩展@Franks答案,如果在你的您要追加行的特殊情况是:
set.seed(12345)
dt1
microbenchmark(
rbind(dt1, 6)),
rbindlist(list(dt1,list(5,6)))
)
b $ b
我们可以看到:
expr min lq median uq max
rbind (5,6))160.516 166.058 175.089 185.1470 457.735
rbindlist(list(dt1,list(5,6)))130.137 134.037 140.605 149.6365 184.326
如果你想在其他地方插入一行,下面的代码可以工作,但是不漂亮:
rbindlist(list(dt1 [1:3,],list(5,6),dt1 [4:5,]))
或甚至
rbindlist(list(dt1 [1:3, ,as.list(c(5,6)),dt1 [4:5,])
给予:
ab
1:0.5855288 -1.8179560
2:0.7094660 0.6300986
3 :-0.1093033 -0.2761841
4:5.0000000 6.0000000
5:-0.4534972 -0.2841597
6:0.6058875 -0.9193220
如果您正在修改行(这是首选方法),您需要预先定义data.table的大小,即
dt1 set(dt1,i = 6L, j =a,value = 5)#按名称引用列
set(dt1,i = 6L,j = 2L,value = 6)#指按列数字
感谢@Boxuan,我修改了这个答案,以考虑你的建议,这是一个更快,更容易阅读。 >
If I have a data frame
set.seed(12345)
df=data.frame(a=rnorm(5),b=rnorm(5))
I can add a row by e.g.
df[6,] =c(5,6)
If I now do the equivalent in data.table
library(data.table)
dt=data.table(df)
dt[6,]=c(5,6)
It fails with an error. What is the right way to insert a row into a data.table?
To expand on @Franks answer, if in your particular case you are appending a row, it's :
set.seed(12345)
dt1 <- data.table(a=rnorm(5), b=rnorm(5))
The following are equivalent; I find the first easier to read but the second faster:
microbenchmark(
rbind(dt1, list(5, 6)),
rbindlist(list(dt1, list(5, 6)))
)
As we can see:
expr min lq median uq max
rbind(dt1, list(5, 6)) 160.516 166.058 175.089 185.1470 457.735
rbindlist(list(dt1, list(5, 6))) 130.137 134.037 140.605 149.6365 184.326
If you want to insert the row elsewhere, the following will work, but it's not pretty:
rbindlist(list(dt1[1:3, ], list(5, 6), dt1[4:5, ]))
or even
rbindlist(list(dt1[1:3, ], as.list(c(5, 6)), dt1[4:5, ]))
giving:
a b
1: 0.5855288 -1.8179560
2: 0.7094660 0.6300986
3: -0.1093033 -0.2761841
4: 5.0000000 6.0000000
5: -0.4534972 -0.2841597
6: 0.6058875 -0.9193220
If you are modifying a row in place (which is the preferred approach), you will need to define the size of the data.table in advance i.e.
dt1 <- data.table(a=rnorm(6), b=rnorm(6))
set(dt1, i=6L, j="a", value=5) # refer to column by name
set(dt1, i=6L, j=2L, value=6) # refer to column by number
Thanks @Boxuan, I have modified this answer to take account of your suggestion, which is a little faster and easier to read.
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