Mixing
How to recode multiple-conditionals to -1/0/+1?
I have my data like:
Name A B C
apple 1 -1 0
banana 2 -2 1
pear -3 0 1
I want to replace all positive value to -1, all negative values to +1, but remain the 0 to be 0, how do I achieve that?
r recode
edited Nov 21 '18 at 8:40
smci
14.9k674104
asked Nov 21 '18 at 8:32
Lennon Lee Lennon Lee
498
add a comment |
I have my data like:
Name A B C
apple 1 -1 0
banana 2 -2 1
pear -3 0 1
I want to replace all positive value to -1, all negative values to +1, but remain the 0 to be 0, how do I achieve that?
r recode
edited Nov 21 '18 at 8:40
smci
14.9k674104
asked Nov 21 '18 at 8:32
Lennon Lee Lennon Lee
498
add a comment |
I have my data like:
Name A B C
apple 1 -1 0
banana 2 -2 1
pear -3 0 1
I want to replace all positive value to -1, all negative values to +1, but remain the 0 to be 0, how do I achieve that?
r recode
edited Nov 21 '18 at 8:40
smci
14.9k674104
asked Nov 21 '18 at 8:32
Lennon Lee Lennon Lee
498
I have my data like:
Name A B C
apple 1 -1 0
banana 2 -2 1
pear -3 0 1
I want to replace all positive value to -1, all negative values to +1, but remain the 0 to be 0, how do I achieve that?
r recode
r recode
edited Nov 21 '18 at 8:40
smci
14.9k674104
asked Nov 21 '18 at 8:32
Lennon Lee Lennon Lee
498
edited Nov 21 '18 at 8:40
smci
14.9k674104
asked Nov 21 '18 at 8:32
Lennon Lee Lennon Lee
498
edited Nov 21 '18 at 8:40
smci
14.9k674104
edited Nov 21 '18 at 8:40
smci
14.9k674104
edited Nov 21 '18 at 8:40
smci
14.9k674104
14.9k674104
asked Nov 21 '18 at 8:32
Lennon Lee Lennon Lee
498
asked Nov 21 '18 at 8:32
Lennon Lee Lennon Lee
498
asked Nov 21 '18 at 8:32
Lennon Lee Lennon Lee
498
498
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
We could use sign to get the sign of all the elements of the dataframe and then reverse it by multiplying it to -1.
df[-1] <- sign(df[-1]) * -1
df
# Name A B C
#1 apple -1 1 0
#2 banana -1 1 -1
#3 pear 1 0 -1
From ?sign
sign returns a vector with the signs of the corresponding elements of x (the sign of a real number is 1, 0, or -1 if the number is positive, zero, or negative, respectively).
data
df <- structure(list(Name = structure(1:3, .Label = c("apple", "banana",
"pear"), class = "factor"), A = c(-1, -1, 1), B = c(1, 1, 0),
C = c(0, -1, -1)), .Names = c("Name", "A", "B", "C"), row.names = c(NA,
-3L), class = "data.frame")
answered Nov 21 '18 at 8:36
Ronak ShahRonak Shah
38.2k104161
1
why not simplify to-sign(df[-1])?
– Andre Elrico
Nov 21 '18 at 9:11
@AndreElrico yes, that would give the same result as well.
– Ronak Shah
Nov 21 '18 at 9:16
add a comment |
By no means this is as good as the top voted, but here is an alternative:
ans <- df1[,-1] / -abs(df1[,-1])
ans[sapply(ans,is.nan)] <- 0
#> ans
# A B C
#1 -1 1 0
#2 -1 1 -1
#3 1 0 -1
answered Nov 21 '18 at 9:08
Andre ElricoAndre Elrico
5,72011029
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
We could use sign to get the sign of all the elements of the dataframe and then reverse it by multiplying it to -1.
df[-1] <- sign(df[-1]) * -1
df
# Name A B C
#1 apple -1 1 0
#2 banana -1 1 -1
#3 pear 1 0 -1
From ?sign
sign returns a vector with the signs of the corresponding elements of x (the sign of a real number is 1, 0, or -1 if the number is positive, zero, or negative, respectively).
data
df <- structure(list(Name = structure(1:3, .Label = c("apple", "banana",
"pear"), class = "factor"), A = c(-1, -1, 1), B = c(1, 1, 0),
C = c(0, -1, -1)), .Names = c("Name", "A", "B", "C"), row.names = c(NA,
-3L), class = "data.frame")
answered Nov 21 '18 at 8:36
Ronak ShahRonak Shah
38.2k104161
1
why not simplify to-sign(df[-1])?
– Andre Elrico
Nov 21 '18 at 9:11
@AndreElrico yes, that would give the same result as well.
– Ronak Shah
Nov 21 '18 at 9:16
add a comment |
We could use sign to get the sign of all the elements of the dataframe and then reverse it by multiplying it to -1.
df[-1] <- sign(df[-1]) * -1
df
# Name A B C
#1 apple -1 1 0
#2 banana -1 1 -1
#3 pear 1 0 -1
From ?sign
sign returns a vector with the signs of the corresponding elements of x (the sign of a real number is 1, 0, or -1 if the number is positive, zero, or negative, respectively).
data
df <- structure(list(Name = structure(1:3, .Label = c("apple", "banana",
"pear"), class = "factor"), A = c(-1, -1, 1), B = c(1, 1, 0),
C = c(0, -1, -1)), .Names = c("Name", "A", "B", "C"), row.names = c(NA,
-3L), class = "data.frame")
answered Nov 21 '18 at 8:36
Ronak ShahRonak Shah
38.2k104161
1
why not simplify to-sign(df[-1])?
– Andre Elrico
Nov 21 '18 at 9:11
@AndreElrico yes, that would give the same result as well.
– Ronak Shah
Nov 21 '18 at 9:16
add a comment |
We could use sign to get the sign of all the elements of the dataframe and then reverse it by multiplying it to -1.
df[-1] <- sign(df[-1]) * -1
df
# Name A B C
#1 apple -1 1 0
#2 banana -1 1 -1
#3 pear 1 0 -1
From ?sign
sign returns a vector with the signs of the corresponding elements of x (the sign of a real number is 1, 0, or -1 if the number is positive, zero, or negative, respectively).
data
df <- structure(list(Name = structure(1:3, .Label = c("apple", "banana",
"pear"), class = "factor"), A = c(-1, -1, 1), B = c(1, 1, 0),
C = c(0, -1, -1)), .Names = c("Name", "A", "B", "C"), row.names = c(NA,
-3L), class = "data.frame")
answered Nov 21 '18 at 8:36
Ronak ShahRonak Shah
38.2k104161
We could use sign to get the sign of all the elements of the dataframe and then reverse it by multiplying it to -1.
df[-1] <- sign(df[-1]) * -1
df
# Name A B C
#1 apple -1 1 0
#2 banana -1 1 -1
#3 pear 1 0 -1
From ?sign
sign returns a vector with the signs of the corresponding elements of x (the sign of a real number is 1, 0, or -1 if the number is positive, zero, or negative, respectively).
data
df <- structure(list(Name = structure(1:3, .Label = c("apple", "banana",
"pear"), class = "factor"), A = c(-1, -1, 1), B = c(1, 1, 0),
C = c(0, -1, -1)), .Names = c("Name", "A", "B", "C"), row.names = c(NA,
-3L), class = "data.frame")
answered Nov 21 '18 at 8:36
Ronak ShahRonak Shah
38.2k104161
answered Nov 21 '18 at 8:36
Ronak ShahRonak Shah
38.2k104161
answered Nov 21 '18 at 8:36
Ronak ShahRonak Shah
38.2k104161
answered Nov 21 '18 at 8:36
Ronak ShahRonak Shah
38.2k104161
38.2k104161
1
why not simplify to-sign(df[-1])?
– Andre Elrico
Nov 21 '18 at 9:11
@AndreElrico yes, that would give the same result as well.
– Ronak Shah
Nov 21 '18 at 9:16
add a comment |
1
why not simplify to-sign(df[-1])?
– Andre Elrico
Nov 21 '18 at 9:11
@AndreElrico yes, that would give the same result as well.
– Ronak Shah
Nov 21 '18 at 9:16
1
1
why not simplify to
-sign(df[-1])?– Andre Elrico
Nov 21 '18 at 9:11
why not simplify to
-sign(df[-1])?– Andre Elrico
Nov 21 '18 at 9:11
@AndreElrico yes, that would give the same result as well.
– Ronak Shah
Nov 21 '18 at 9:16
@AndreElrico yes, that would give the same result as well.
– Ronak Shah
Nov 21 '18 at 9:16
add a comment |
By no means this is as good as the top voted, but here is an alternative:
ans <- df1[,-1] / -abs(df1[,-1])
ans[sapply(ans,is.nan)] <- 0
#> ans
# A B C
#1 -1 1 0
#2 -1 1 -1
#3 1 0 -1
answered Nov 21 '18 at 9:08
Andre ElricoAndre Elrico
5,72011029
add a comment |
By no means this is as good as the top voted, but here is an alternative:
ans <- df1[,-1] / -abs(df1[,-1])
ans[sapply(ans,is.nan)] <- 0
#> ans
# A B C
#1 -1 1 0
#2 -1 1 -1
#3 1 0 -1
answered Nov 21 '18 at 9:08
Andre ElricoAndre Elrico
5,72011029
add a comment |
By no means this is as good as the top voted, but here is an alternative:
ans <- df1[,-1] / -abs(df1[,-1])
ans[sapply(ans,is.nan)] <- 0
#> ans
# A B C
#1 -1 1 0
#2 -1 1 -1
#3 1 0 -1
answered Nov 21 '18 at 9:08
Andre ElricoAndre Elrico
5,72011029
By no means this is as good as the top voted, but here is an alternative:
ans <- df1[,-1] / -abs(df1[,-1])
ans[sapply(ans,is.nan)] <- 0
#> ans
# A B C
#1 -1 1 0
#2 -1 1 -1
#3 1 0 -1
answered Nov 21 '18 at 9:08
Andre ElricoAndre Elrico
5,72011029
answered Nov 21 '18 at 9:08
Andre ElricoAndre Elrico
5,72011029
answered Nov 21 '18 at 9:08
Andre ElricoAndre Elrico
5,72011029
answered Nov 21 '18 at 9:08
Andre ElricoAndre Elrico
5,72011029
5,72011029
add a comment |
add a comment |
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