I have the following vector:
我有以下向量:
tmp3 <- c("1500 2", "1500 1", "1510 2", "1510 1", "1520 2", "1520 1", "1530 2",
"1530 1", "1540 2", "1540 1")
I would like to just retain the second number in each of the atoms of this vector, so it would read:
我想在这个向量的每个原子中保留第二个数字,所以它会读到:
c(2,1,2,1,2,1,2,1,2,1)
8 个解决方案
#1
32
There's probably a better way, but here are two approaches with strsplit()
:
可能有更好的方法,但这里有两种方法:strsplit():
as.numeric(data.frame(strsplit(tmp3, " "))[2,])
as.numeric(lapply(strsplit(tmp3," "), function(x) x[2]))
The as.numeric() may not be necessary if you can use characters...
如果您可以使用字符,则可能不需要数字()。
#2
21
One could use read.table
on textConnection
:
一个可以用阅读。在textConnection表:
X <- read.table(textConnection(tmp3))
then
然后
> str(X)
'data.frame': 10 obs. of 2 variables:
$ V1: int 1500 1500 1510 1510 1520 1520 1530 1530 1540 1540
$ V2: int 2 1 2 1 2 1 2 1 2 1
so X$V2
is what you need.
所以X$V2是你需要的。
#3
9
It depends a little bit on how closely your actual data matches the example data you've given. I you're just trying to get everything after the space, you can use gsub
:
这取决于你的实际数据与你给出的示例数据有多接近。我你只是想把所有的东西都做完,你可以用gsub:
gsub(".+\\s+", "", tmp3)
[1] "2" "1" "2" "1" "2" "1" "2" "1" "2" "1"
If you're trying to implement a rule more complicated than "take everything after the space", you'll need a more complicated regular expresion.
如果你想要实现一个比“把所有东西都放进去”更复杂的规则,你需要一个更复杂的正则表达式。
#4
9
What I think is the most elegant way to do this
我认为这是最优雅的方式!
> res <- sapply(strsplit(tmp3, " "), "[[", 2)
If you need it to be an integer
如果你需要它是一个整数。
> storage.mode(res) <- "integer"
#5
5
substr(x = tmp3, start = 6, stop = 6)
So long as your strings are always the same length, this should do the trick.
只要你的字符串长度相同,这就行了。
(And, of course, you don't have to specify the argument names - substr(tmp3, 6, 6)
works fine, too)
(当然,您不必指定参数名——substr(tmp3, 6, 6)也可以正常工作)
#6
4
This should do it:
这应该这样做:
library(plyr)
ldply(strsplit(tmp3, split = " "))[[2]]
If you need a numeric vector, use
如果您需要一个数字矢量,请使用。
as.numeric(ldply(strsplit(tmp3, split = " "))[[2]])
#7
1
Another option is scan()
. To get the second value, we can use a logical subset.
另一个选择是扫描()。为了获得第二个值,我们可以使用一个逻辑子集。
scan(text = tmp3)[c(FALSE, TRUE)]
# [1] 2 1 2 1 2 1 2 1 2 1
#8
-1
An easier way to split 1 column into 2 columns via data.table
一种更简单的方法,将1列分成2列,通过数据表。
require(data.table)
data_ex = data.table( a = paste( sample(1:3, size=10, replace=TRUE),"-separate", sep="" ))
data_ex[, number:= unlist( strsplit(x=a, split="-") )[[1]], by=a]
data_ex[, word:= unlist( strsplit(x=a, split="-") )[[2]], by=a ]
#1
32
There's probably a better way, but here are two approaches with strsplit()
:
可能有更好的方法,但这里有两种方法:strsplit():
as.numeric(data.frame(strsplit(tmp3, " "))[2,])
as.numeric(lapply(strsplit(tmp3," "), function(x) x[2]))
The as.numeric() may not be necessary if you can use characters...
如果您可以使用字符,则可能不需要数字()。
#2
21
One could use read.table
on textConnection
:
一个可以用阅读。在textConnection表:
X <- read.table(textConnection(tmp3))
then
然后
> str(X)
'data.frame': 10 obs. of 2 variables:
$ V1: int 1500 1500 1510 1510 1520 1520 1530 1530 1540 1540
$ V2: int 2 1 2 1 2 1 2 1 2 1
so X$V2
is what you need.
所以X$V2是你需要的。
#3
9
It depends a little bit on how closely your actual data matches the example data you've given. I you're just trying to get everything after the space, you can use gsub
:
这取决于你的实际数据与你给出的示例数据有多接近。我你只是想把所有的东西都做完,你可以用gsub:
gsub(".+\\s+", "", tmp3)
[1] "2" "1" "2" "1" "2" "1" "2" "1" "2" "1"
If you're trying to implement a rule more complicated than "take everything after the space", you'll need a more complicated regular expresion.
如果你想要实现一个比“把所有东西都放进去”更复杂的规则,你需要一个更复杂的正则表达式。
#4
9
What I think is the most elegant way to do this
我认为这是最优雅的方式!
> res <- sapply(strsplit(tmp3, " "), "[[", 2)
If you need it to be an integer
如果你需要它是一个整数。
> storage.mode(res) <- "integer"
#5
5
substr(x = tmp3, start = 6, stop = 6)
So long as your strings are always the same length, this should do the trick.
只要你的字符串长度相同,这就行了。
(And, of course, you don't have to specify the argument names - substr(tmp3, 6, 6)
works fine, too)
(当然,您不必指定参数名——substr(tmp3, 6, 6)也可以正常工作)
#6
4
This should do it:
这应该这样做:
library(plyr)
ldply(strsplit(tmp3, split = " "))[[2]]
If you need a numeric vector, use
如果您需要一个数字矢量,请使用。
as.numeric(ldply(strsplit(tmp3, split = " "))[[2]])
#7
1
Another option is scan()
. To get the second value, we can use a logical subset.
另一个选择是扫描()。为了获得第二个值,我们可以使用一个逻辑子集。
scan(text = tmp3)[c(FALSE, TRUE)]
# [1] 2 1 2 1 2 1 2 1 2 1
#8
-1
An easier way to split 1 column into 2 columns via data.table
一种更简单的方法,将1列分成2列,通过数据表。
require(data.table)
data_ex = data.table( a = paste( sample(1:3, size=10, replace=TRUE),"-separate", sep="" ))
data_ex[, number:= unlist( strsplit(x=a, split="-") )[[1]], by=a]
data_ex[, word:= unlist( strsplit(x=a, split="-") )[[2]], by=a ]