How to label a integer column based on a conditional statement in R
I want to label this vector C into either [1] HY HY HY HY LY HY (2 levels) or vice versa [1] LY LY LY LY HY LY (2 levels) based on a condition I specified using factor.
For example,
C <- c(1, 1, 1, 1, 2, 1)
ifelse(50 > 100,
factor(C, labels = c('HY','LY')),
factor(C, labels = c('LY','HY')))
This prints out [1] 1 rather than my expectation.
But factor(C, labels = c('LY','HY')) works fine. Why is that?
Then I did a test by taking out factor, but it still doesn't give me 'LY' 'HY'.
ifelse(50 > 100,
c('HY','LY'),
c('LY','HY'))
[1] "LY"
Another option I can think of is to change the vector into characters like this and then change it to factor. In either case, it should give me a vector, not a value.
ifelse(50 > 100,
ifelse(C==1, 'HY', 'LY'),
ifelse(C==1, 'LY', 'HY'))
[1] "LY"
if-statement label
asked Nov 15 '18 at 3:16
Golden JiangGolden Jiang
4819
add a comment |
I want to label this vector C into either [1] HY HY HY HY LY HY (2 levels) or vice versa [1] LY LY LY LY HY LY (2 levels) based on a condition I specified using factor.
For example,
C <- c(1, 1, 1, 1, 2, 1)
ifelse(50 > 100,
factor(C, labels = c('HY','LY')),
factor(C, labels = c('LY','HY')))
This prints out [1] 1 rather than my expectation.
But factor(C, labels = c('LY','HY')) works fine. Why is that?
Then I did a test by taking out factor, but it still doesn't give me 'LY' 'HY'.
ifelse(50 > 100,
c('HY','LY'),
c('LY','HY'))
[1] "LY"
Another option I can think of is to change the vector into characters like this and then change it to factor. In either case, it should give me a vector, not a value.
ifelse(50 > 100,
ifelse(C==1, 'HY', 'LY'),
ifelse(C==1, 'LY', 'HY'))
[1] "LY"
if-statement label
asked Nov 15 '18 at 3:16
Golden JiangGolden Jiang
4819
add a comment |
I want to label this vector C into either [1] HY HY HY HY LY HY (2 levels) or vice versa [1] LY LY LY LY HY LY (2 levels) based on a condition I specified using factor.
For example,
C <- c(1, 1, 1, 1, 2, 1)
ifelse(50 > 100,
factor(C, labels = c('HY','LY')),
factor(C, labels = c('LY','HY')))
This prints out [1] 1 rather than my expectation.
But factor(C, labels = c('LY','HY')) works fine. Why is that?
Then I did a test by taking out factor, but it still doesn't give me 'LY' 'HY'.
ifelse(50 > 100,
c('HY','LY'),
c('LY','HY'))
[1] "LY"
Another option I can think of is to change the vector into characters like this and then change it to factor. In either case, it should give me a vector, not a value.
ifelse(50 > 100,
ifelse(C==1, 'HY', 'LY'),
ifelse(C==1, 'LY', 'HY'))
[1] "LY"
if-statement label
asked Nov 15 '18 at 3:16
Golden JiangGolden Jiang
4819
I want to label this vector C into either [1] HY HY HY HY LY HY (2 levels) or vice versa [1] LY LY LY LY HY LY (2 levels) based on a condition I specified using factor.
For example,
C <- c(1, 1, 1, 1, 2, 1)
ifelse(50 > 100,
factor(C, labels = c('HY','LY')),
factor(C, labels = c('LY','HY')))
This prints out [1] 1 rather than my expectation.
But factor(C, labels = c('LY','HY')) works fine. Why is that?
Then I did a test by taking out factor, but it still doesn't give me 'LY' 'HY'.
ifelse(50 > 100,
c('HY','LY'),
c('LY','HY'))
[1] "LY"
Another option I can think of is to change the vector into characters like this and then change it to factor. In either case, it should give me a vector, not a value.
ifelse(50 > 100,
ifelse(C==1, 'HY', 'LY'),
ifelse(C==1, 'LY', 'HY'))
[1] "LY"
if-statement label
if-statement label
asked Nov 15 '18 at 3:16
Golden JiangGolden Jiang
4819
asked Nov 15 '18 at 3:16
Golden JiangGolden Jiang
4819
edited Nov 15 '18 at 4:41
Golden Jiang
asked Nov 15 '18 at 3:16
Golden JiangGolden Jiang
4819
asked Nov 15 '18 at 3:16
Golden JiangGolden Jiang
4819
asked Nov 15 '18 at 3:16
Golden JiangGolden Jiang
4819
4819
add a comment |
add a comment |
1 Answer
1
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oldest
votes
So it turns out ifelse is like a for loop that accepts a vector input and prints out a vector of the same length, i.e. 50>100 prints out FALSE of the length 1 and so as the output.
When I use if instead of ifelse, the problem solved.
if(50>100){
ifelse(C==1, 'HY', 'LY')
} else {
ifelse(C==1, 'LY', 'HY')
}
[1] "LY" "LY" "LY" "LY" "HY" "LY"
answered Nov 16 '18 at 4:41
Golden JiangGolden Jiang
4819
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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So it turns out ifelse is like a for loop that accepts a vector input and prints out a vector of the same length, i.e. 50>100 prints out FALSE of the length 1 and so as the output.
When I use if instead of ifelse, the problem solved.
if(50>100){
ifelse(C==1, 'HY', 'LY')
} else {
ifelse(C==1, 'LY', 'HY')
}
[1] "LY" "LY" "LY" "LY" "HY" "LY"
answered Nov 16 '18 at 4:41
Golden JiangGolden Jiang
4819
add a comment |
So it turns out ifelse is like a for loop that accepts a vector input and prints out a vector of the same length, i.e. 50>100 prints out FALSE of the length 1 and so as the output.
When I use if instead of ifelse, the problem solved.
if(50>100){
ifelse(C==1, 'HY', 'LY')
} else {
ifelse(C==1, 'LY', 'HY')
}
[1] "LY" "LY" "LY" "LY" "HY" "LY"
answered Nov 16 '18 at 4:41
Golden JiangGolden Jiang
4819
add a comment |
So it turns out ifelse is like a for loop that accepts a vector input and prints out a vector of the same length, i.e. 50>100 prints out FALSE of the length 1 and so as the output.
When I use if instead of ifelse, the problem solved.
if(50>100){
ifelse(C==1, 'HY', 'LY')
} else {
ifelse(C==1, 'LY', 'HY')
}
[1] "LY" "LY" "LY" "LY" "HY" "LY"
answered Nov 16 '18 at 4:41
Golden JiangGolden Jiang
4819
So it turns out ifelse is like a for loop that accepts a vector input and prints out a vector of the same length, i.e. 50>100 prints out FALSE of the length 1 and so as the output.
When I use if instead of ifelse, the problem solved.
if(50>100){
ifelse(C==1, 'HY', 'LY')
} else {
ifelse(C==1, 'LY', 'HY')
}
[1] "LY" "LY" "LY" "LY" "HY" "LY"
answered Nov 16 '18 at 4:41
Golden JiangGolden Jiang
4819
answered Nov 16 '18 at 4:41
Golden JiangGolden Jiang
4819
answered Nov 16 '18 at 4:41
Golden JiangGolden Jiang
4819
answered Nov 16 '18 at 4:41
Golden JiangGolden Jiang
4819
4819
add a comment |
add a comment |
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