Mixing
Diffrents between gini, information gain and sum of square of errors in rpart R
1
I made a short code in R to check how split criterias work. I got unexpected results, all of them choose the same value to split. Can someone explain it? Here is the code:
set.seed(1)
y <- sample(c(1, 0), 10000, replace = T)
x <- seq(1, 10000)
data <- data.frame(x, y)
library(rpart)
rpart(y~x,data = data,parms=list(split="gini"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
rpart(y~x,data = data,parms=list(split="information"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
rpart(y~x,data = data,method = "anova",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
r split rpart
asked Nov 21 '18 at 12:10
RobertRobert
83
add a comment |
1
I made a short code in R to check how split criterias work. I got unexpected results, all of them choose the same value to split. Can someone explain it? Here is the code:
set.seed(1)
y <- sample(c(1, 0), 10000, replace = T)
x <- seq(1, 10000)
data <- data.frame(x, y)
library(rpart)
rpart(y~x,data = data,parms=list(split="gini"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
rpart(y~x,data = data,parms=list(split="information"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
rpart(y~x,data = data,method = "anova",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
r split rpart
asked Nov 21 '18 at 12:10
RobertRobert
83
add a comment |
1
1
1
1
I made a short code in R to check how split criterias work. I got unexpected results, all of them choose the same value to split. Can someone explain it? Here is the code:
set.seed(1)
y <- sample(c(1, 0), 10000, replace = T)
x <- seq(1, 10000)
data <- data.frame(x, y)
library(rpart)
rpart(y~x,data = data,parms=list(split="gini"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
rpart(y~x,data = data,parms=list(split="information"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
rpart(y~x,data = data,method = "anova",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
r split rpart
asked Nov 21 '18 at 12:10
RobertRobert
83
I made a short code in R to check how split criterias work. I got unexpected results, all of them choose the same value to split. Can someone explain it? Here is the code:
set.seed(1)
y <- sample(c(1, 0), 10000, replace = T)
x <- seq(1, 10000)
data <- data.frame(x, y)
library(rpart)
rpart(y~x,data = data,parms=list(split="gini"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
rpart(y~x,data = data,parms=list(split="information"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
rpart(y~x,data = data,method = "anova",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
r split rpart
r split rpart
asked Nov 21 '18 at 12:10
RobertRobert
83
asked Nov 21 '18 at 12:10
RobertRobert
83
edited Nov 21 '18 at 12:50
Robert
asked Nov 21 '18 at 12:10
RobertRobert
83
asked Nov 21 '18 at 12:10
RobertRobert
83
asked Nov 21 '18 at 12:10
RobertRobert
83
83
add a comment |
add a comment |
1 Answer
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In my case only the last rpart command did split something:
> set.seed(1)
> y <- sample(c(1, 0), 1000, replace = T)
> x <- seq(1, 1000)
> data <- data.frame(x, y)
> library(rpart)
No split with split="gini":
> rpart(y~x,data = data,parms=list(split="gini"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
n= 1000
node), split, n, loss, yval, (yprob)
* denotes terminal node
1) root 1000 480 1 (0.4800000 0.5200000) *
No split with split="information":
> rpart(y~x,data = data,parms=list(split="information"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
n= 1000
node), split, n, loss, yval, (yprob)
* denotes terminal node
1) root 1000 480 1 (0.4800000 0.5200000) *
There is a single split with split="anova":
> rpart(y~x,data = data,method = "anova",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
n= 1000
node), split, n, deviance, yval
* denotes terminal node
1) root 1000 249.6000 0.5200000
2) x< 841.5 841 210.1831 0.5089180 *
3) x>=841.5 159 38.7673 0.5786164 *
As regards to why the split points can be in the same position, a couple of extract from the rpart documentation:
- Gini measure vs. Information impurity (page 6): "For the two class problem the measures differ only slightly, and will nearly always choose the same split point."
- Gini measure vs. [ANalysis Of] Variances (page 41): "... for the two class case the Gini splitting rule reduces to 2p(1 − p), which is the variance of a node."
So it seems like in the case of two class problem the different measures may produce similar split points.
answered Nov 21 '18 at 12:33
HeikkiHeikki
1,2871018
You're right, (I added this control param to avoid it however it anyway happen sometimes - idk why), change the data size to 10 000 (its 1000) and it should happen anymore
– Robert
Nov 21 '18 at 12:49
should not* happen anymore
– Robert
Nov 21 '18 at 12:57
@Robert, edited the answer in order to try to explain why split points may occur at same positions.
– Heikki
Nov 21 '18 at 14:01
Thanks a lot! Its good to know it.
– Robert
Nov 21 '18 at 14:24
add a comment |
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In my case only the last rpart command did split something:
> set.seed(1)
> y <- sample(c(1, 0), 1000, replace = T)
> x <- seq(1, 1000)
> data <- data.frame(x, y)
> library(rpart)
No split with split="gini":
> rpart(y~x,data = data,parms=list(split="gini"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
n= 1000
node), split, n, loss, yval, (yprob)
* denotes terminal node
1) root 1000 480 1 (0.4800000 0.5200000) *
No split with split="information":
> rpart(y~x,data = data,parms=list(split="information"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
n= 1000
node), split, n, loss, yval, (yprob)
* denotes terminal node
1) root 1000 480 1 (0.4800000 0.5200000) *
There is a single split with split="anova":
> rpart(y~x,data = data,method = "anova",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
n= 1000
node), split, n, deviance, yval
* denotes terminal node
1) root 1000 249.6000 0.5200000
2) x< 841.5 841 210.1831 0.5089180 *
3) x>=841.5 159 38.7673 0.5786164 *
As regards to why the split points can be in the same position, a couple of extract from the rpart documentation:
- Gini measure vs. Information impurity (page 6): "For the two class problem the measures differ only slightly, and will nearly always choose the same split point."
- Gini measure vs. [ANalysis Of] Variances (page 41): "... for the two class case the Gini splitting rule reduces to 2p(1 − p), which is the variance of a node."
So it seems like in the case of two class problem the different measures may produce similar split points.
answered Nov 21 '18 at 12:33
HeikkiHeikki
1,2871018
You're right, (I added this control param to avoid it however it anyway happen sometimes - idk why), change the data size to 10 000 (its 1000) and it should happen anymore
– Robert
Nov 21 '18 at 12:49
should not* happen anymore
– Robert
Nov 21 '18 at 12:57
@Robert, edited the answer in order to try to explain why split points may occur at same positions.
– Heikki
Nov 21 '18 at 14:01
Thanks a lot! Its good to know it.
– Robert
Nov 21 '18 at 14:24
add a comment |
0
In my case only the last rpart command did split something:
> set.seed(1)
> y <- sample(c(1, 0), 1000, replace = T)
> x <- seq(1, 1000)
> data <- data.frame(x, y)
> library(rpart)
No split with split="gini":
> rpart(y~x,data = data,parms=list(split="gini"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
n= 1000
node), split, n, loss, yval, (yprob)
* denotes terminal node
1) root 1000 480 1 (0.4800000 0.5200000) *
No split with split="information":
> rpart(y~x,data = data,parms=list(split="information"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
n= 1000
node), split, n, loss, yval, (yprob)
* denotes terminal node
1) root 1000 480 1 (0.4800000 0.5200000) *
There is a single split with split="anova":
> rpart(y~x,data = data,method = "anova",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
n= 1000
node), split, n, deviance, yval
* denotes terminal node
1) root 1000 249.6000 0.5200000
2) x< 841.5 841 210.1831 0.5089180 *
3) x>=841.5 159 38.7673 0.5786164 *
As regards to why the split points can be in the same position, a couple of extract from the rpart documentation:
- Gini measure vs. Information impurity (page 6): "For the two class problem the measures differ only slightly, and will nearly always choose the same split point."
- Gini measure vs. [ANalysis Of] Variances (page 41): "... for the two class case the Gini splitting rule reduces to 2p(1 − p), which is the variance of a node."
So it seems like in the case of two class problem the different measures may produce similar split points.
answered Nov 21 '18 at 12:33
HeikkiHeikki
1,2871018
You're right, (I added this control param to avoid it however it anyway happen sometimes - idk why), change the data size to 10 000 (its 1000) and it should happen anymore
– Robert
Nov 21 '18 at 12:49
should not* happen anymore
– Robert
Nov 21 '18 at 12:57
@Robert, edited the answer in order to try to explain why split points may occur at same positions.
– Heikki
Nov 21 '18 at 14:01
Thanks a lot! Its good to know it.
– Robert
Nov 21 '18 at 14:24
add a comment |
0
0
0
In my case only the last rpart command did split something:
> set.seed(1)
> y <- sample(c(1, 0), 1000, replace = T)
> x <- seq(1, 1000)
> data <- data.frame(x, y)
> library(rpart)
No split with split="gini":
> rpart(y~x,data = data,parms=list(split="gini"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
n= 1000
node), split, n, loss, yval, (yprob)
* denotes terminal node
1) root 1000 480 1 (0.4800000 0.5200000) *
No split with split="information":
> rpart(y~x,data = data,parms=list(split="information"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
n= 1000
node), split, n, loss, yval, (yprob)
* denotes terminal node
1) root 1000 480 1 (0.4800000 0.5200000) *
There is a single split with split="anova":
> rpart(y~x,data = data,method = "anova",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
n= 1000
node), split, n, deviance, yval
* denotes terminal node
1) root 1000 249.6000 0.5200000
2) x< 841.5 841 210.1831 0.5089180 *
3) x>=841.5 159 38.7673 0.5786164 *
As regards to why the split points can be in the same position, a couple of extract from the rpart documentation:
- Gini measure vs. Information impurity (page 6): "For the two class problem the measures differ only slightly, and will nearly always choose the same split point."
- Gini measure vs. [ANalysis Of] Variances (page 41): "... for the two class case the Gini splitting rule reduces to 2p(1 − p), which is the variance of a node."
So it seems like in the case of two class problem the different measures may produce similar split points.
answered Nov 21 '18 at 12:33
HeikkiHeikki
1,2871018
In my case only the last rpart command did split something:
> set.seed(1)
> y <- sample(c(1, 0), 1000, replace = T)
> x <- seq(1, 1000)
> data <- data.frame(x, y)
> library(rpart)
No split with split="gini":
> rpart(y~x,data = data,parms=list(split="gini"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
n= 1000
node), split, n, loss, yval, (yprob)
* denotes terminal node
1) root 1000 480 1 (0.4800000 0.5200000) *
No split with split="information":
> rpart(y~x,data = data,parms=list(split="information"),method = "class",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
n= 1000
node), split, n, loss, yval, (yprob)
* denotes terminal node
1) root 1000 480 1 (0.4800000 0.5200000) *
There is a single split with split="anova":
> rpart(y~x,data = data,method = "anova",control = list(maxdepth = 1,cp=0.0001,minsplit=1))
n= 1000
node), split, n, deviance, yval
* denotes terminal node
1) root 1000 249.6000 0.5200000
2) x< 841.5 841 210.1831 0.5089180 *
3) x>=841.5 159 38.7673 0.5786164 *
As regards to why the split points can be in the same position, a couple of extract from the rpart documentation:
- Gini measure vs. Information impurity (page 6): "For the two class problem the measures differ only slightly, and will nearly always choose the same split point."
- Gini measure vs. [ANalysis Of] Variances (page 41): "... for the two class case the Gini splitting rule reduces to 2p(1 − p), which is the variance of a node."
So it seems like in the case of two class problem the different measures may produce similar split points.
answered Nov 21 '18 at 12:33
HeikkiHeikki
1,2871018
edited Nov 21 '18 at 14:00
answered Nov 21 '18 at 12:33
HeikkiHeikki
1,2871018
answered Nov 21 '18 at 12:33
HeikkiHeikki
1,2871018
answered Nov 21 '18 at 12:33
HeikkiHeikki
1,2871018
1,2871018
You're right, (I added this control param to avoid it however it anyway happen sometimes - idk why), change the data size to 10 000 (its 1000) and it should happen anymore
– Robert
Nov 21 '18 at 12:49
should not* happen anymore
– Robert
Nov 21 '18 at 12:57
@Robert, edited the answer in order to try to explain why split points may occur at same positions.
– Heikki
Nov 21 '18 at 14:01
Thanks a lot! Its good to know it.
– Robert
Nov 21 '18 at 14:24
add a comment |
You're right, (I added this control param to avoid it however it anyway happen sometimes - idk why), change the data size to 10 000 (its 1000) and it should happen anymore
– Robert
Nov 21 '18 at 12:49
should not* happen anymore
– Robert
Nov 21 '18 at 12:57
@Robert, edited the answer in order to try to explain why split points may occur at same positions.
– Heikki
Nov 21 '18 at 14:01
Thanks a lot! Its good to know it.
– Robert
Nov 21 '18 at 14:24
You're right, (I added this control param to avoid it however it anyway happen sometimes - idk why), change the data size to 10 000 (its 1000) and it should happen anymore
– Robert
Nov 21 '18 at 12:49
You're right, (I added this control param to avoid it however it anyway happen sometimes - idk why), change the data size to 10 000 (its 1000) and it should happen anymore
– Robert
Nov 21 '18 at 12:49
should not* happen anymore
– Robert
Nov 21 '18 at 12:57
should not* happen anymore
– Robert
Nov 21 '18 at 12:57
@Robert, edited the answer in order to try to explain why split points may occur at same positions.
– Heikki
Nov 21 '18 at 14:01
@Robert, edited the answer in order to try to explain why split points may occur at same positions.
– Heikki
Nov 21 '18 at 14:01
Thanks a lot! Its good to know it.
– Robert
Nov 21 '18 at 14:24
Thanks a lot! Its good to know it.
– Robert
Nov 21 '18 at 14:24
add a comment |
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