Mixing
Ignore Inf value and run a lm Regression
0
these are my variables:
> dput(y)
c(-22.0713165394207, 14.0880914427811, 10.9650636244176, -1.96648890706268,
-5.30593850426708, -7.54651916037787, 3.84914747321197, 4.4986386904214,
1.73067625014435, 2.5585960595839, -2.72766183793304, -3.10167452216202,
2.68853838208521, 1.12662203717498, 1.24951279248057, 3.70075666289518,
-6.11243972144607, -6.91019769671849, 6.46767794752582, 8.84874735514293,
2.95606352319898, 3.23883851668917, -2.61692776879569)
> dput(x)
c(`1` = 0.0520523266234464, `2` = Inf, `3` = 0.0520523266234462,
`4` = 0.0520523266234463, `5` = 0.0520523266234463, `6` = 0.0520523266234461,
`7` = 0.0520523266234463, `8` = 0.0520523266234466, `9` = 0.0520523266234465,
`10` = 0.0520523266234465, `11` = 0.0520523266234465, `12` = 0.0520523266234466,
`13` = 0.0520523266234468, `14` = 0.0520523266234466, `15` = 0.0520523266234467,
`16` = 0.0520523266234464, `17` = 0.0520523266234463, `18` = 0.0520523266234465,
`19` = 0.0520523266234466, `20` = 0.0520523266234463, `21` = 0.0520523266234464,
`22` = 0.0520523266234465, `23` = 0.0520523266234464)
When I run my regression: summary(lm(ex.return ~ ex.return.skew))
I have this error message:
Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
NA/NaN/Inf in 'x'
I tried to delete the positon 2 and then run my regression but I can not do this because data is much bigger. So I am looking for a way to ignore the Inf/Na values and run my regression.
How can I do this?
Any help?
r
asked Nov 20 '18 at 17:33
LauraLaura
35319
add a comment |
0
these are my variables:
> dput(y)
c(-22.0713165394207, 14.0880914427811, 10.9650636244176, -1.96648890706268,
-5.30593850426708, -7.54651916037787, 3.84914747321197, 4.4986386904214,
1.73067625014435, 2.5585960595839, -2.72766183793304, -3.10167452216202,
2.68853838208521, 1.12662203717498, 1.24951279248057, 3.70075666289518,
-6.11243972144607, -6.91019769671849, 6.46767794752582, 8.84874735514293,
2.95606352319898, 3.23883851668917, -2.61692776879569)
> dput(x)
c(`1` = 0.0520523266234464, `2` = Inf, `3` = 0.0520523266234462,
`4` = 0.0520523266234463, `5` = 0.0520523266234463, `6` = 0.0520523266234461,
`7` = 0.0520523266234463, `8` = 0.0520523266234466, `9` = 0.0520523266234465,
`10` = 0.0520523266234465, `11` = 0.0520523266234465, `12` = 0.0520523266234466,
`13` = 0.0520523266234468, `14` = 0.0520523266234466, `15` = 0.0520523266234467,
`16` = 0.0520523266234464, `17` = 0.0520523266234463, `18` = 0.0520523266234465,
`19` = 0.0520523266234466, `20` = 0.0520523266234463, `21` = 0.0520523266234464,
`22` = 0.0520523266234465, `23` = 0.0520523266234464)
When I run my regression: summary(lm(ex.return ~ ex.return.skew))
I have this error message:
Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
NA/NaN/Inf in 'x'
I tried to delete the positon 2 and then run my regression but I can not do this because data is much bigger. So I am looking for a way to ignore the Inf/Na values and run my regression.
How can I do this?
Any help?
r
asked Nov 20 '18 at 17:33
LauraLaura
35319
What do you mean that you cannot delete that observation?
– Julius Vainora
Nov 20 '18 at 17:37
Because, in this specific case It would easy to delete the second position and run the regression. But my data is much bigger than this. It wouldnt be functional to do this.
– Laura
Nov 20 '18 at 17:41
add a comment |
0
0
0
these are my variables:
> dput(y)
c(-22.0713165394207, 14.0880914427811, 10.9650636244176, -1.96648890706268,
-5.30593850426708, -7.54651916037787, 3.84914747321197, 4.4986386904214,
1.73067625014435, 2.5585960595839, -2.72766183793304, -3.10167452216202,
2.68853838208521, 1.12662203717498, 1.24951279248057, 3.70075666289518,
-6.11243972144607, -6.91019769671849, 6.46767794752582, 8.84874735514293,
2.95606352319898, 3.23883851668917, -2.61692776879569)
> dput(x)
c(`1` = 0.0520523266234464, `2` = Inf, `3` = 0.0520523266234462,
`4` = 0.0520523266234463, `5` = 0.0520523266234463, `6` = 0.0520523266234461,
`7` = 0.0520523266234463, `8` = 0.0520523266234466, `9` = 0.0520523266234465,
`10` = 0.0520523266234465, `11` = 0.0520523266234465, `12` = 0.0520523266234466,
`13` = 0.0520523266234468, `14` = 0.0520523266234466, `15` = 0.0520523266234467,
`16` = 0.0520523266234464, `17` = 0.0520523266234463, `18` = 0.0520523266234465,
`19` = 0.0520523266234466, `20` = 0.0520523266234463, `21` = 0.0520523266234464,
`22` = 0.0520523266234465, `23` = 0.0520523266234464)
When I run my regression: summary(lm(ex.return ~ ex.return.skew))
I have this error message:
Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
NA/NaN/Inf in 'x'
I tried to delete the positon 2 and then run my regression but I can not do this because data is much bigger. So I am looking for a way to ignore the Inf/Na values and run my regression.
How can I do this?
Any help?
r
asked Nov 20 '18 at 17:33
LauraLaura
35319
these are my variables:
> dput(y)
c(-22.0713165394207, 14.0880914427811, 10.9650636244176, -1.96648890706268,
-5.30593850426708, -7.54651916037787, 3.84914747321197, 4.4986386904214,
1.73067625014435, 2.5585960595839, -2.72766183793304, -3.10167452216202,
2.68853838208521, 1.12662203717498, 1.24951279248057, 3.70075666289518,
-6.11243972144607, -6.91019769671849, 6.46767794752582, 8.84874735514293,
2.95606352319898, 3.23883851668917, -2.61692776879569)
> dput(x)
c(`1` = 0.0520523266234464, `2` = Inf, `3` = 0.0520523266234462,
`4` = 0.0520523266234463, `5` = 0.0520523266234463, `6` = 0.0520523266234461,
`7` = 0.0520523266234463, `8` = 0.0520523266234466, `9` = 0.0520523266234465,
`10` = 0.0520523266234465, `11` = 0.0520523266234465, `12` = 0.0520523266234466,
`13` = 0.0520523266234468, `14` = 0.0520523266234466, `15` = 0.0520523266234467,
`16` = 0.0520523266234464, `17` = 0.0520523266234463, `18` = 0.0520523266234465,
`19` = 0.0520523266234466, `20` = 0.0520523266234463, `21` = 0.0520523266234464,
`22` = 0.0520523266234465, `23` = 0.0520523266234464)
When I run my regression: summary(lm(ex.return ~ ex.return.skew))
I have this error message:
Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
NA/NaN/Inf in 'x'
I tried to delete the positon 2 and then run my regression but I can not do this because data is much bigger. So I am looking for a way to ignore the Inf/Na values and run my regression.
How can I do this?
Any help?
r
r
asked Nov 20 '18 at 17:33
LauraLaura
35319
asked Nov 20 '18 at 17:33
LauraLaura
35319
asked Nov 20 '18 at 17:33
LauraLaura
35319
asked Nov 20 '18 at 17:33
LauraLaura
35319
asked Nov 20 '18 at 17:33
LauraLaura
35319
35319
What do you mean that you cannot delete that observation?
– Julius Vainora
Nov 20 '18 at 17:37
Because, in this specific case It would easy to delete the second position and run the regression. But my data is much bigger than this. It wouldnt be functional to do this.
– Laura
Nov 20 '18 at 17:41
add a comment |
What do you mean that you cannot delete that observation?
– Julius Vainora
Nov 20 '18 at 17:37
Because, in this specific case It would easy to delete the second position and run the regression. But my data is much bigger than this. It wouldnt be functional to do this.
– Laura
Nov 20 '18 at 17:41
What do you mean that you cannot delete that observation?
– Julius Vainora
Nov 20 '18 at 17:37
What do you mean that you cannot delete that observation?
– Julius Vainora
Nov 20 '18 at 17:37
Because, in this specific case It would easy to delete the second position and run the regression. But my data is much bigger than this. It wouldnt be functional to do this.
– Laura
Nov 20 '18 at 17:41
Because, in this specific case It would easy to delete the second position and run the regression. But my data is much bigger than this. It wouldnt be functional to do this.
– Laura
Nov 20 '18 at 17:41
add a comment |
2 Answers
2
active
oldest
votes
1
We can convert the infinite values to NA and it should work
x[is.infinite(x)] <- NA
summary(lm(y ~ x))
answered Nov 20 '18 at 17:37
akrunakrun
403k13196269
Thanks @akrun but I have NA value for the angular coefficient of my regression. I though it would work, but it didnt. I tried `na.action=na.omit´ but didnt work.
– Laura
Nov 20 '18 at 17:45
@Laura Would it make sense to create a logical index and then ignore those values. i.e.i1 <- is.infinite(x); i2 <- is.infinite(y); i3 <- i1|i2; x1 <- x[!i3]; y1 <- y[!i3]; summary(lm(y1 ~ x1))
– akrun
Nov 20 '18 at 17:58
@akrun this answer doesn’t work because you can’t have NA values in a regression as you get NA coefficients as mentioned. See my solution for the correct answer. The lm function always works best with data.frames rather than separate vectors - see my answer.
– rookie
Nov 21 '18 at 7:52
add a comment |
0
You need to remove the infinite items in both x and y as follows:
summary(lm(ex.return[is.finite(df$ex.return)] ~ ex.return.skew[is.finite(df$ex.return)]))
However, what is better is to put them into a data.frame and add that data.frame in the data argument of lm which filters out the rows of the data.frame.
df <- data.frame(ex.return, ex.return.skew)
summary(lm(ex.return ~ ex.return.skew, df[is.finite(df$ex.return),]))
Note that is.finite() works for NA values as well as -Inf/Inf
is.finite(c(NA, Inf, 10))
[1] FALSE FALSE TRUE
If there is the possibility of Inf/-Inf and NA in any row of any column in your data.frame (i.e. not just in ex.return) then you might need to do something like:
summary(lm(ex.return ~ ex.return.skew, df[is.finite(rowSums(df)),]))
answered Nov 20 '18 at 18:01
rookierookie
863
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
1
We can convert the infinite values to NA and it should work
x[is.infinite(x)] <- NA
summary(lm(y ~ x))
answered Nov 20 '18 at 17:37
akrunakrun
403k13196269
Thanks @akrun but I have NA value for the angular coefficient of my regression. I though it would work, but it didnt. I tried `na.action=na.omit´ but didnt work.
– Laura
Nov 20 '18 at 17:45
@Laura Would it make sense to create a logical index and then ignore those values. i.e.i1 <- is.infinite(x); i2 <- is.infinite(y); i3 <- i1|i2; x1 <- x[!i3]; y1 <- y[!i3]; summary(lm(y1 ~ x1))
– akrun
Nov 20 '18 at 17:58
@akrun this answer doesn’t work because you can’t have NA values in a regression as you get NA coefficients as mentioned. See my solution for the correct answer. The lm function always works best with data.frames rather than separate vectors - see my answer.
– rookie
Nov 21 '18 at 7:52
add a comment |
1
We can convert the infinite values to NA and it should work
x[is.infinite(x)] <- NA
summary(lm(y ~ x))
answered Nov 20 '18 at 17:37
akrunakrun
403k13196269
Thanks @akrun but I have NA value for the angular coefficient of my regression. I though it would work, but it didnt. I tried `na.action=na.omit´ but didnt work.
– Laura
Nov 20 '18 at 17:45
@Laura Would it make sense to create a logical index and then ignore those values. i.e.i1 <- is.infinite(x); i2 <- is.infinite(y); i3 <- i1|i2; x1 <- x[!i3]; y1 <- y[!i3]; summary(lm(y1 ~ x1))
– akrun
Nov 20 '18 at 17:58
@akrun this answer doesn’t work because you can’t have NA values in a regression as you get NA coefficients as mentioned. See my solution for the correct answer. The lm function always works best with data.frames rather than separate vectors - see my answer.
– rookie
Nov 21 '18 at 7:52
add a comment |
1
1
1
We can convert the infinite values to NA and it should work
x[is.infinite(x)] <- NA
summary(lm(y ~ x))
answered Nov 20 '18 at 17:37
akrunakrun
403k13196269
We can convert the infinite values to NA and it should work
x[is.infinite(x)] <- NA
summary(lm(y ~ x))
answered Nov 20 '18 at 17:37
akrunakrun
403k13196269
answered Nov 20 '18 at 17:37
akrunakrun
403k13196269
answered Nov 20 '18 at 17:37
akrunakrun
403k13196269
answered Nov 20 '18 at 17:37
akrunakrun
403k13196269
403k13196269
Thanks @akrun but I have NA value for the angular coefficient of my regression. I though it would work, but it didnt. I tried `na.action=na.omit´ but didnt work.
– Laura
Nov 20 '18 at 17:45
@Laura Would it make sense to create a logical index and then ignore those values. i.e.i1 <- is.infinite(x); i2 <- is.infinite(y); i3 <- i1|i2; x1 <- x[!i3]; y1 <- y[!i3]; summary(lm(y1 ~ x1))
– akrun
Nov 20 '18 at 17:58
@akrun this answer doesn’t work because you can’t have NA values in a regression as you get NA coefficients as mentioned. See my solution for the correct answer. The lm function always works best with data.frames rather than separate vectors - see my answer.
– rookie
Nov 21 '18 at 7:52
add a comment |
Thanks @akrun but I have NA value for the angular coefficient of my regression. I though it would work, but it didnt. I tried `na.action=na.omit´ but didnt work.
– Laura
Nov 20 '18 at 17:45
@Laura Would it make sense to create a logical index and then ignore those values. i.e.i1 <- is.infinite(x); i2 <- is.infinite(y); i3 <- i1|i2; x1 <- x[!i3]; y1 <- y[!i3]; summary(lm(y1 ~ x1))
– akrun
Nov 20 '18 at 17:58
@akrun this answer doesn’t work because you can’t have NA values in a regression as you get NA coefficients as mentioned. See my solution for the correct answer. The lm function always works best with data.frames rather than separate vectors - see my answer.
– rookie
Nov 21 '18 at 7:52
Thanks @akrun but I have NA value for the angular coefficient of my regression. I though it would work, but it didnt. I tried `na.action=na.omit´ but didnt work.
– Laura
Nov 20 '18 at 17:45
Thanks @akrun but I have NA value for the angular coefficient of my regression. I though it would work, but it didnt. I tried `na.action=na.omit´ but didnt work.
– Laura
Nov 20 '18 at 17:45
@Laura Would it make sense to create a logical index and then ignore those values. i.e.
i1 <- is.infinite(x); i2 <- is.infinite(y); i3 <- i1|i2; x1 <- x[!i3]; y1 <- y[!i3]; summary(lm(y1 ~ x1))– akrun
Nov 20 '18 at 17:58
@Laura Would it make sense to create a logical index and then ignore those values. i.e.
i1 <- is.infinite(x); i2 <- is.infinite(y); i3 <- i1|i2; x1 <- x[!i3]; y1 <- y[!i3]; summary(lm(y1 ~ x1))– akrun
Nov 20 '18 at 17:58
@akrun this answer doesn’t work because you can’t have NA values in a regression as you get NA coefficients as mentioned. See my solution for the correct answer. The lm function always works best with data.frames rather than separate vectors - see my answer.
– rookie
Nov 21 '18 at 7:52
@akrun this answer doesn’t work because you can’t have NA values in a regression as you get NA coefficients as mentioned. See my solution for the correct answer. The lm function always works best with data.frames rather than separate vectors - see my answer.
– rookie
Nov 21 '18 at 7:52
add a comment |
0
You need to remove the infinite items in both x and y as follows:
summary(lm(ex.return[is.finite(df$ex.return)] ~ ex.return.skew[is.finite(df$ex.return)]))
However, what is better is to put them into a data.frame and add that data.frame in the data argument of lm which filters out the rows of the data.frame.
df <- data.frame(ex.return, ex.return.skew)
summary(lm(ex.return ~ ex.return.skew, df[is.finite(df$ex.return),]))
Note that is.finite() works for NA values as well as -Inf/Inf
is.finite(c(NA, Inf, 10))
[1] FALSE FALSE TRUE
If there is the possibility of Inf/-Inf and NA in any row of any column in your data.frame (i.e. not just in ex.return) then you might need to do something like:
summary(lm(ex.return ~ ex.return.skew, df[is.finite(rowSums(df)),]))
answered Nov 20 '18 at 18:01
rookierookie
863
add a comment |
0
You need to remove the infinite items in both x and y as follows:
summary(lm(ex.return[is.finite(df$ex.return)] ~ ex.return.skew[is.finite(df$ex.return)]))
However, what is better is to put them into a data.frame and add that data.frame in the data argument of lm which filters out the rows of the data.frame.
df <- data.frame(ex.return, ex.return.skew)
summary(lm(ex.return ~ ex.return.skew, df[is.finite(df$ex.return),]))
Note that is.finite() works for NA values as well as -Inf/Inf
is.finite(c(NA, Inf, 10))
[1] FALSE FALSE TRUE
If there is the possibility of Inf/-Inf and NA in any row of any column in your data.frame (i.e. not just in ex.return) then you might need to do something like:
summary(lm(ex.return ~ ex.return.skew, df[is.finite(rowSums(df)),]))
answered Nov 20 '18 at 18:01
rookierookie
863
add a comment |
0
0
0
You need to remove the infinite items in both x and y as follows:
summary(lm(ex.return[is.finite(df$ex.return)] ~ ex.return.skew[is.finite(df$ex.return)]))
However, what is better is to put them into a data.frame and add that data.frame in the data argument of lm which filters out the rows of the data.frame.
df <- data.frame(ex.return, ex.return.skew)
summary(lm(ex.return ~ ex.return.skew, df[is.finite(df$ex.return),]))
Note that is.finite() works for NA values as well as -Inf/Inf
is.finite(c(NA, Inf, 10))
[1] FALSE FALSE TRUE
If there is the possibility of Inf/-Inf and NA in any row of any column in your data.frame (i.e. not just in ex.return) then you might need to do something like:
summary(lm(ex.return ~ ex.return.skew, df[is.finite(rowSums(df)),]))
answered Nov 20 '18 at 18:01
rookierookie
863
You need to remove the infinite items in both x and y as follows:
summary(lm(ex.return[is.finite(df$ex.return)] ~ ex.return.skew[is.finite(df$ex.return)]))
However, what is better is to put them into a data.frame and add that data.frame in the data argument of lm which filters out the rows of the data.frame.
df <- data.frame(ex.return, ex.return.skew)
summary(lm(ex.return ~ ex.return.skew, df[is.finite(df$ex.return),]))
Note that is.finite() works for NA values as well as -Inf/Inf
is.finite(c(NA, Inf, 10))
[1] FALSE FALSE TRUE
If there is the possibility of Inf/-Inf and NA in any row of any column in your data.frame (i.e. not just in ex.return) then you might need to do something like:
summary(lm(ex.return ~ ex.return.skew, df[is.finite(rowSums(df)),]))
answered Nov 20 '18 at 18:01
rookierookie
863
edited Nov 21 '18 at 13:59
answered Nov 20 '18 at 18:01
rookierookie
863
answered Nov 20 '18 at 18:01
rookierookie
863
answered Nov 20 '18 at 18:01
rookierookie
863
863
add a comment |
add a comment |
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What do you mean that you cannot delete that observation?
– Julius Vainora
Nov 20 '18 at 17:37
Because, in this specific case It would easy to delete the second position and run the regression. But my data is much bigger than this. It wouldnt be functional to do this.
– Laura
Nov 20 '18 at 17:41