Re: [R] function problem: multi selection in one argument
> Rui Barradas
> on Tue, 25 Jan 2022 14:22:47 + writes:
> Hello,
> Here are 3 functions that do what the question asks for. The first 2 are
> tidyverse solutions, the last one a base R function.
> I don't understand why the group_by and mutate, that's why I've included
> a 2nd tidyverse function. And the base R function follows the same lines
> of outputing the table only without data transformation.
> And the test dataset is not iris, it only has one factor variable. It
> doesn't make sense to table a continuous variable such as Petal.Width.
> The data set mtcars has several numeric variables that are in fact
> categorical ("vs", "am") and one, "cyl", that can be tabled. The
> examples below use mtcars.
> f3 <- function(data, ...){
> groups <- unlist(list(...))
> temp <- data %>%
> select(all_of({{groups}})) %>%
> group_by(across(all_of({{groups}}))) %>%
> mutate(numbering = row_number(), max = n())
> temp %>%
> select(all_of({{groups}})) %>%
> table()
> }
> f4 <- function(data, ...){
> groups <- unlist(list(...))
> data %>%
> select(all_of({{groups}})) %>%
> table()
> }
> f5 <- function(data, ...){
> temp <- lapply(list(...), \(col) data[[col]])
> table(setNames(temp, list(...)))
> }
> f3(mtcars, "am", "cyl")
> f4(mtcars, "am", "cyl")
> f5(mtcars, "am", "cyl")
> Hope this helps,
> Rui Barradas
Thank you, Rui!
Note that your base R solution can be vastly simplified :
> f6 <- function(data, ...) table(data[, unlist(list(...))])
> f6(mtcars, "am", "cyl")
cyl
am 4 6 8
0 3 4 12
1 8 3 2
>
If you started measuring carefully, I'm sure this would not only
be the shortest but also by far the fastest solution ...
Best,
Martin
--
Martin Maechler
ETH Zurich and R Core Team
> Às 00:14 de 25/01/2022, Kai Yang via R-help escreveu:
>> Hello Team,
>> I can run the function below:
>>
>> library(tidyverse)
>>
>> f2 <- function(indata, subgrp1){
>> indata0 <- indata
>> temp<- indata0 %>% select({{subgrp1}}) %>% arrange({{subgrp1}}) %>%
>> group_by({{subgrp1}}) %>%
>> mutate(numbering =row_number(), max=max(numbering))
>> view(temp)
>> f_table <- table(temp$Species)
>> view(f_table)
>> return(f_table)
>> }
>> f2(iris, Species)
>>
>> You can see the second argument I use Species only, and it works fine.
>> But If I say, I want the 2nd argument = Petal.Width, Species , how
should I write the argument? I did try f2(iris, c(Petal.Width, Species)), but I
got error message:
>> Error: arrange() failed at implicit mutate() step.
>> * Problem with `mutate()` column `..1`.
>> i `..1 = c(Petal.Width, Species)`.
>> i `..1` must be size 150 or 1, not 300.
>>
>> I'm not sure how to fix the problem either in function or can fix it
when using the function.
>> Thank you,
>> Kai
>> [[alternative HTML version deleted]]
>>
>> __
>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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> PLEASE do read the posting guide
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> and provide commented, minimal, self-contained, reproducible code.
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Re: [R] function problem: multi selection in one argument
Hello,
Here are 3 functions that do what the question asks for. The first 2 are
tidyverse solutions, the last one a base R function.
I don't understand why the group_by and mutate, that's why I've included
a 2nd tidyverse function. And the base R function follows the same lines
of outputing the table only without data transformation.
And the test dataset is not iris, it only has one factor variable. It
doesn't make sense to table a continuous variable such as Petal.Width.
The data set mtcars has several numeric variables that are in fact
categorical ("vs", "am") and one, "cyl", that can be tabled. The
examples below use mtcars.
f3 <- function(data, ...){
groups <- unlist(list(...))
temp <- data %>%
select(all_of({{groups}})) %>%
group_by(across(all_of({{groups}}))) %>%
mutate(numbering = row_number(), max = n())
temp %>%
select(all_of({{groups}})) %>%
table()
}
f4 <- function(data, ...){
groups <- unlist(list(...))
data %>%
select(all_of({{groups}})) %>%
table()
}
f5 <- function(data, ...){
temp <- lapply(list(...), \(col) data[[col]])
table(setNames(temp, list(...)))
}
f3(mtcars, "am", "cyl")
f4(mtcars, "am", "cyl")
f5(mtcars, "am", "cyl")
Hope this helps,
Rui Barradas
Às 00:14 de 25/01/2022, Kai Yang via R-help escreveu:
Hello Team,
I can run the function below:
library(tidyverse)
f2 <- function(indata, subgrp1){
indata0 <- indata
temp <- indata0 %>% select({{subgrp1}}) %>% arrange({{subgrp1}}) %>%
group_by({{subgrp1}}) %>%
mutate(numbering =row_number(), max=max(numbering))
view(temp)
f_table <- table(temp$Species)
view(f_table)
return(f_table)
}
f2(iris, Species)
You can see the second argument I use Species only, and it works fine.
But If I say, I want the 2nd argument = Petal.Width, Species , how should I
write the argument? I did try f2(iris, c(Petal.Width, Species)), but I got
error message:
Error: arrange() failed at implicit mutate() step.
* Problem with `mutate()` column `..1`.
i `..1 = c(Petal.Width, Species)`.
i `..1` must be size 150 or 1, not 300.
I'm not sure how to fix the problem either in function or can fix it when using
the function.
Thank you,
Kai
[[alternative HTML version deleted]]
__
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] function problem: multi selection in one argument
Hallo
You should explain better what do you want as many people here do not use
tidyverse functions.
I am not sure what the function should do.
table(iris$Species)
give the same result and whatever you do in tidyverse part it always finalise
in table(...$Species)
Cheers
Petr
> -Original Message-
> From: R-help On Behalf Of Kai Yang via R-help
> Sent: Tuesday, January 25, 2022 1:14 AM
> To: R-help Mailing List
> Subject: [R] function problem: multi selection in one argument
>
> Hello Team,
> I can run the function below:
>
> library(tidyverse)
>
> f2 <- function(indata, subgrp1){
> indata0 <- indata
> temp<- indata0 %>% select({{subgrp1}}) %>% arrange({{subgrp1}}) %>%
> group_by({{subgrp1}}) %>%
> mutate(numbering =row_number(), max=max(numbering))
> view(temp)
> f_table <- table(temp$Species)
> view(f_table)
> return(f_table)
> }
> f2(iris, Species)
>
> You can see the second argument I use Species only, and it works fine. But
> If I
> say, I want the 2nd argument = Petal.Width, Species , how should I write the
> argument? I did try f2(iris, c(Petal.Width, Species)), but I got error
> message:
> Error: arrange() failed at implicit mutate() step.
> * Problem with `mutate()` column `..1`.
> i `..1 = c(Petal.Width, Species)`.
> i `..1` must be size 150 or 1, not 300.
>
> I'm not sure how to fix the problem either in function or can fix it when
> using the
> function.
> Thank you,
> Kai
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
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[R] function problem: multi selection in one argument
Hello Team,
I can run the function below:
library(tidyverse)
f2 <- function(indata, subgrp1){
indata0 <- indata
temp <- indata0 %>% select({{subgrp1}}) %>% arrange({{subgrp1}}) %>%
group_by({{subgrp1}}) %>%
mutate(numbering =row_number(), max=max(numbering))
view(temp)
f_table <- table(temp$Species)
view(f_table)
return(f_table)
}
f2(iris, Species)
You can see the second argument I use Species only, and it works fine.
But If I say, I want the 2nd argument = Petal.Width, Species , how should I
write the argument? I did try f2(iris, c(Petal.Width, Species)), but I got
error message:
Error: arrange() failed at implicit mutate() step.
* Problem with `mutate()` column `..1`.
i `..1 = c(Petal.Width, Species)`.
i `..1` must be size 150 or 1, not 300.
I'm not sure how to fix the problem either in function or can fix it when using
the function.
Thank you,
Kai
[[alternative HTML version deleted]]
__
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[R] function problem
I have a series of columns that need to be evaluated in various tables. I
need to apply a function in the following manner somers2(name1,name2). name1
is a vector of x inputs for the function which correspond to a vector of y
inputs in name2.
y-rep(c(3,4,5,8),6)
z-rep(c(23,24,25,26,27,28),4)
name1-sprintf(Pred_pres_%s_indpdt[,%s,,],x,y)
name2-sprintf(population[,%s],z)
rank-function(i,j){
for(i in name1){
for(j in name2){
somers2(i,j)
}}}
Thanks
Robert
--
View this message in context:
http://www.nabble.com/function-problem-tp24759253p24759253.html
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Re: [R] function problem
Hi,
On Jul 31, 2009, at 12:25 PM, RR! wrote:
I have a series of columns that need to be evaluated in various
tables. I
need to apply a function in the following manner
somers2(name1,name2). name1
is a vector of x inputs for the function which correspond to a
vector of y
inputs in name2.
y-rep(c(3,4,5,8),6)
z-rep(c(23,24,25,26,27,28),4)
name1-sprintf(Pred_pres_%s_indpdt[,%s,,],x,y)
name2-sprintf(population[,%s],z)
rank-function(i,j){
for(i in name1){
for(j in name2){
somers2(i,j)
}}}
You're missing a value for x, so I'm finding it hard to understand
what you're trying to do, exactly. From your double for loop, it seems
like you just want to compute the function over all vals in X, and
each x against all Y.
Does this help?
R x - LETTERS[1:3]
R y - LETTERS[10:15]
R all.vals - expand.grid(x,y, stringsAsFactors=F)
R all.vals[1:3,]
Var1 Var2
1AJ
2BJ
3CJ
R ans - lapply(seq(nrow(all.vals)), function(i) paste(all.vals[i,1],
all.vals[i,2], sep='(*)'))
R ans[1:3]
[[1]]
[1] A(*)J
[[2]]
[1] B(*)J
[[3]]
[1] C(*)J
-steve
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
__
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Re: [R] function problem
I am sorry I should have stated the whole problem:
I appreciate all the help I have received from this list and also not being
flamed because I am new to R. Many of my problems are in automation so far.
I am trying to create multiple objects that are outputs of functions. Here
are the tasks:
aGAM-somers2(Pred_pres_a_indpdt[,3,,],population[,23])
aGBM-somers2(Pred_pres_a_indpdt[,4,,],population[,23])
cGLM-somers2(Pred_pres_a_indpdt[,5,,],population[,23])
cRF-somers2(Pred_pres_a_indpdt[,8,,],population[,23])
bGAM-somers2(Pred_pres_b_indpdt[,3,,],population[,24])
bGBM-somers2(Pred_pres_b_indpdt[,4,,],population[,24])
bGLM-somers2(Pred_pres_b_indpdt[,5,,],population[,24])
bRF-somers2(Pred_pres_b_indpdt[,8,,],population[,24])
...and so on through the letter f.
so the variables (x,y,z in this example) are as follows:
xGAM-somers2(Pred_pres_x_indpdt[,y,,],population[,z])
the GAM, GBM, GLM, and RF are stored in a column y in Pred_pres_x_indpdt.
x is denoting a species, which corresponds to column z in population, so
a:f == 23:28
~~~
This may be pushing it, but it would be great if I could get all of these in
a matrix of dim[1,96]
Here is what I have so far:
x - c(a,b,c,d,e,f)
nums - c(23:28)
rank-function(x){
for(y in nums){
xGAM-somers2(Pred_pres_x_indpdt[,3,,],population[,y])
xGBM-somers2(Pred_pres_x_indpdt[,4,,],population[,y])
xGLM-somers2(Pred_pres_x_indpdt[,5,,],population[,y])
xRF-somers2(Pred_pres_x_indpdt[,8,,],population[,y])
rank(x)
}}
#for each species, there is 16 columns of output, so a total of 96 columns
is needed
x_matrix-matrix(c(xGAM[1:2],Evaluation.results.TSS[[1]][1,3:4],xGBM[1:2],Evaluation.results.TSS[[1]][2,3:4],
xGLM[1:2],Evaluation.results.TSS[[1]][3,3:4],xRF[1:2],Evaluation.results.TSS[[1]][4,3:4]),
nrow=1,ncol=16,
dimnames=list(c(),
c(x_gam_AUC, x_gam_Dxy,x_gam_TSS,x_gam_Cutoff,x_gbm_AUC,
x_gbm_Dxy,x_gbm_TSS,x_gbm_Cutoff,
x_glm_AUC, x_glm_Dxy,x_glm_TSS,x_glm_Cutoff,x_rf_AUC,
x_rf_Dxy,x_rf_TSS,x_rf_Cutoff)))
Thanks again for helping me out
Steve Lianoglou-6 wrote:
Hi,
On Jul 31, 2009, at 12:25 PM, RR! wrote:
I have a series of columns that need to be evaluated in various
tables. I
need to apply a function in the following manner
somers2(name1,name2). name1
is a vector of x inputs for the function which correspond to a
vector of y
inputs in name2.
y-rep(c(3,4,5,8),6)
z-rep(c(23,24,25,26,27,28),4)
name1-sprintf(Pred_pres_%s_indpdt[,%s,,],x,y)
name2-sprintf(population[,%s],z)
rank-function(i,j){
for(i in name1){
for(j in name2){
somers2(i,j)
}}}
You're missing a value for x, so I'm finding it hard to understand
what you're trying to do, exactly. From your double for loop, it seems
like you just want to compute the function over all vals in X, and
each x against all Y.
Does this help?
R x - LETTERS[1:3]
R y - LETTERS[10:15]
R all.vals - expand.grid(x,y, stringsAsFactors=F)
R all.vals[1:3,]
Var1 Var2
1AJ
2BJ
3CJ
R ans - lapply(seq(nrow(all.vals)), function(i) paste(all.vals[i,1],
all.vals[i,2], sep='(*)'))
R ans[1:3]
[[1]]
[1] A(*)J
[[2]]
[1] B(*)J
[[3]]
[1] C(*)J
-steve
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
__
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and provide commented, minimal, self-contained, reproducible code.
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