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
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
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
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
edited Nov 15 '18 at 4:41
Golden Jiang
asked Nov 15 '18 at 3:16
Golden JiangGolden Jiang
4819
4819
add a comment |
add a comment |
1 Answer
1
active
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"
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
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"
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"
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"
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
4819
add a comment |
add a comment |
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