I need to use the qchisq function on a column of a sparklyr data frame.
The problem is that it seems that qchisq function is not implemented in Spark. If I am reading the error message below correctly, sparklyr tried execute a function called "QCHISQ", however this doesn't exist neither in Hive SQL, nor in Spark.
In general, is there a way to run arbitrary functions that are not implemented in Hive or Spark, with sparklyr? I know about spark_apply, but haven't figured out how to configure it yet.
> mydf = data.frame(beta=runif(100, -5, 5), pval = runif(100, 0.001, 0.1))
> mydf_tbl = copy_to(con, mydf)
> mydf_tbl
# Source: table<mydf> [?? x 2]
# Database: spark_connection
beta pval
<dbl> <dbl>
1 3.42 0.0913
2 -1.72 0.0629
3 0.515 0.0335
4 -3.12 0.0717
5 -2.12 0.0253
6 1.36 0.00640
7 -3.33 0.0896
8 1.36 0.0235
9 0.619 0.0414
10 4.73 0.0416
> mydf_tbl %>% mutate(se = sqrt(beta^2/qchisq(pval)))
Error: org.apache.spark.sql.AnalysisException: Undefined function: 'QCHISQ'.
This function is neither a registered temporary function nor a permanent function registered in the database 'default'.; line 1 pos 49
As you noted you can use spark_apply:
mydf_tbl %>%
spark_apply(function(df)
dplyr::mutate(df, se = sqrt(beta^2/qchisq(pval, df = 12))))
# # Source: table<sparklyr_tmp_14bd5feacf5> [?? x 3]
# # Database: spark_connection
# beta pval X3
# <dbl> <dbl> <dbl>
# 1 1.66 0.0763 0.686
# 2 0.153 0.0872 0.0623
# 3 2.96 0.0485 1.30
# 4 4.86 0.0349 2.22
# 5 -1.82 0.0712 0.760
# 6 2.34 0.0295 1.10
# 7 3.54 0.0297 1.65
# 8 4.57 0.0784 1.88
# 9 4.94 0.0394 2.23
# 10 -0.610 0.0906 0.246
# # ... with more rows
but fair warning - it is embarrassingly slow. Unfortunately you don't have alternative here, short of writing your own Scala / Java extensions.
In the end I've used an horrible hack, which for this case works fine.
Another solution would have been to write a User Defined Function (UDF), but sparklyr doesn't support it yet: https://github.com/rstudio/sparklyr/issues/1052
This is the hack I've used. In short, I precompute a qchisq table, upload it as a sparklyr object, then join. If I compare this with results calculated on a local data frame, I get a correlation of r=0.99999990902236146617.
#' #param n: number of significant digits to use
> check_precomputed_strategy = function(n) {
chisq = data.frame(pval=seq(0, 1, 1/(10**(n)))) %>%
mutate(qval=qchisq(pval, df=1, lower.tail = FALSE)) %>%
mutate(pval_s = as.character(round(as.integer(pval*10**n),0)))
chisq %>% head %>% print
chisq_tbl = copy_to(con, chisq, overwrite=T)
mydf = data.frame(beta=runif(100, -5, 5), pval = runif(100, 0.001, 0.1)) %>%
mutate(se1 = sqrt(beta^2/qchisq(pval, df=1, lower.tail = FALSE)))
mydf_tbl = copy_to(con, mydf)
mydf_tbl.up = mydf_tbl %>%
mutate(pval_s=as.character(round(as.integer(pval*10**n),0))) %>%
left_join(chisq_tbl, by="pval_s") %>%
mutate(se=sqrt(beta^2 / qval)) %>%
collect %>%
filter(!duplicated(beta))
mydf_tbl.up %>% head %>% print
mydf_tbl.up %>% filter(complete.cases(.)) %>% nrow %>% print
mydf_tbl.up %>% filter(complete.cases(.)) %>% select(se, se1) %>% cor
}
> check_precomputed_strategy(4)
pval qval pval_s
1 0.00000000000000000000000 Inf 0
2 0.00010000000000000000479 15.136705226623396570 1
3 0.00020000000000000000958 13.831083619091122827 2
4 0.00030000000000000002793 13.070394140069462097 3
5 0.00040000000000000001917 12.532193305401813532 4
6 0.00050000000000000001041 12.115665146397173402 5
# A tibble: 6 x 8
beta pval.x se1 myvar pval_s pval.y qval se
<dbl> <dbl> <dbl> <dbl> <chr> <dbl> <dbl> <dbl>
1 3.42 0.0913 2.03 1. 912 0.0912 2.85 2.03
2 -1.72 0.0629 0.927 1. 628 0.0628 3.46 0.927
3 0.515 0.0335 0.242 1. 335 0.0335 4.52 0.242
4 -3.12 0.0717 1.73 1. 716 0.0716 3.25 1.73
5 -2.12 0.0253 0.947 1. 253 0.0253 5.00 0.946
6 1.36 0.00640 0.498 1. 63 0.00630 7.46 0.497
[1] 100
se se1
se 1.00000000000000000000 0.99999990902236146617
se1 0.99999990902236146617 1.00000000000000000000
Related
Good Morning,
i am using the "epiR" packages to assess test accuracy.
https://search.r-project.org/CRAN/refmans/epiR/html/epi.tests.html
## Generate a data set listing test results and true disease status:
dis <- c(rep(1, times = 744), rep(0, times = 842))
tes <- c(rep(1, times = 670), rep(0, times = 74),
rep(1, times = 202), rep(0, times = 640))
dat.df02 <- data.frame(dis, tes)
tmp.df02 <- dat.df02 %>%
mutate(dis = factor(dis, levels = c(1,0), labels = c("Dis+","Dis-"))) %>%
mutate(tes = factor(tes, levels = c(1,0), labels = c("Test+","Test-"))) %>%
group_by(tes, dis) %>%
summarise(n = n())
tmp.df02
## View the data in conventional 2 by 2 table format:
pivot_wider(tmp.df02, id_cols = c(tes), names_from = dis, values_from = n)
rval.tes02 <- epi.tests(tmp.df02, method = "exact", digits = 2,
conf.level = 0.95)
summary(rval.tes02)
The data type is listed as "epi.test". I would like to export the summary statistics to a table (i.e. gtsummary or flextable).
As summary is a function of base R, I am struggling to do this. Can anyone help? Thank you very much
The epi.tests function has been edited so it writes the results out to a data frame (instead of a list). This will simplify export to gtsummary or flextable. epiR version 2.0.50 to be uploaded to CRAN shortly.
This was not quite as straight forward as I expected.
It appears that summary() when applied to an object x of class epi.tests simply prints x$details. x$details is a list of data.frames with statistic names as row names. That last bit makes things slightly more complicated than they would otherwise have been.
A potential tidyverse solution is
library(tidyverse)
lapply(
names(rval.tes02$detail),
function(x) {
as_tibble(rval.tes02$detail[[x]]) %>%
add_column(statistic=x, .before=1)
}
) %>%
bind_rows()
# A tibble: 18 × 4
statistic est lower upper
<chr> <dbl> <dbl> <dbl>
1 ap 0.550 0.525 0.574
2 tp 0.469 0.444 0.494
3 se 0.901 0.877 0.921
4 sp 0.760 0.730 0.789
5 diag.ac 0.826 0.806 0.844
6 diag.or 28.7 21.5 38.2
7 nndx 1.51 1.41 1.65
8 youden 0.661 0.607 0.710
9 pv.pos 0.768 0.739 0.796
10 pv.neg 0.896 0.872 0.918
11 lr.pos 3.75 3.32 4.24
12 lr.neg 0.131 0.105 0.163
13 p.rout 0.450 0.426 0.475
14 p.rin 0.550 0.525 0.574
15 p.tpdn 0.240 0.211 0.270
16 p.tndp 0.0995 0.0789 0.123
17 p.dntp 0.232 0.204 0.261
18 p.dptn 0.104 0.0823 0.128
Which is a tibble containing the same information as summary(rval.tes02), which you should be able to pass on to gtsummary or flextable. Unusually, the broom package doesn't have a tidy() verb for epi.tests objects.
I have a database which looks like this but with much more rows and columns.
Several variables (x,y,z) measured at different time (1,2,3).
df <-
tibble(
x1 = rnorm(10),
x2 = rnorm(10),
x3 = rnorm(10),
y1 = rnorm(10),
y2 = rnorm(10),
y3 = rnorm(10),
z1 = rnorm(10),
z2 = rnorm(10),
z3 = rnorm(10),
)
I am trying to create dummies variables from the variables with the same suffix (measured at the same time) like this:
df <- df %>%
mutate(var1= ifelse(x1>0 & (y1<0.5 |z1<0.5),0,1)) %>%
mutate(var2= ifelse(x2>0 & (y2<0.5 |z2<0.5),0,1)) %>%
mutate(var3= ifelse(x3>0 & (y1<0.5 |z3<0.5),0,1))
I am used to coding in SAS or Stata, so I would like to use a function or a loop because I have many more variables in my database.
But I think I don't have the right approach in R to deal with this.
Thank you very much for your help !
{dplyover} makes this kind of operation easy (disclaimer: I'm the maintainer), given that your desired output contains a typo:
I think you want to use all variables with the same digit (1, 2, 3 and so on) in each calculation:
df <- df %>%
mutate(var1= ifelse(x1>0 & (y1<0.5 |z1<0.5),0,1)) %>%
mutate(var2= ifelse(x2>0 & (y2<0.5 |z2<0.5),0,1)) %>%
mutate(var3= ifelse(x3>0 & (y3<0.5 |z3<0.5),0,1))
If that is the case we can use dplyover::over to apply the same function over a vector. Here we construct the vector with extract_names("[0-9]{1}$") which gets us all ending numbers of our variable names here: c(1,2,3). We can then construct the variable names using a special syntax: .("x{.x}"). Here .x evaluates to the first number in our vector so it would return the object name x1 (not a string!) which we can use inside the function argument of over.
library(dplyr)
library(dplyover) # Only on GitHub: https://github.com/TimTeaFan/dplyover
df %>%
mutate(over(cut_names("^[a-z]{1}"),
~ ifelse(.("x{.x}") > 0 & (.("y{.x}") < 0.5 | .("z{.x}") < 0.5), 0, 1),
.names = "var{x}"
))
#> # A tibble: 10 x 12
#> x1 x2 x3 y1 y2 y3 z1 z2 z3 var1
#> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 0.690 0.550 0.911 0.203 -0.111 0.530 -2.09 0.189 0.147 0
#> 2 -0.238 1.32 -0.145 0.744 1.05 -0.448 2.05 -1.04 1.50 1
#> 3 0.888 0.898 -1.46 -1.87 -1.14 1.59 1.91 -0.155 1.46 0
#> 4 -2.78 -1.34 -0.486 -0.0674 0.246 0.141 0.154 1.08 -0.319 1
#> 5 -1.20 0.835 1.28 -1.32 -0.674 0.115 0.362 1.06 0.515 1
#> 6 0.622 -0.713 0.0525 1.79 -0.427 0.819 -1.53 -0.885 0.00237 0
#> 7 -2.54 0.0197 0.942 0.230 -1.37 -1.02 -1.55 -0.721 -1.06 1
#> 8 -0.434 1.97 -0.274 0.848 -0.482 -0.422 0.197 0.497 -0.600 1
#> 9 -0.316 -0.219 0.467 -1.97 -0.718 -0.442 -1.39 -0.877 1.52 1
#> 10 -1.03 0.226 2.04 0.432 -1.02 -0.535 0.954 -1.11 0.804 1
#> # ... with 2 more variables: var2 <dbl>, var3 <dbl>
Alternatively we can use dplyr::across and use cur_column(), get() and gsub() to alter the name of the column on the fly. To name the new variables correctly we use gsub() in the .names argument of across and wrap it in curly braces {} to evaluate the expression.
library(dplyr)
df %>%
mutate(across(starts_with("x"),
~ {
cur_c <- dplyr::cur_column()
ifelse(.x > 0 & (get(gsub("x","y", cur_c)) < 0.5 | get(gsub("x","z", cur_c)) < 0.5), 0, 1)
},
.names = '{gsub("x", "var", .col)}'
))
#> # A tibble: 10 x 12
#> x1 x2 x3 y1 y2 y3 z1 z2 z3 var1
#> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 -0.423 -1.42 -1.15 -1.54 1.92 -0.511 -0.739 0.501 0.451 1
#> 2 -0.358 0.164 0.971 -1.61 1.96 -0.675 -0.0188 -1.88 1.63 1
#> 3 -0.453 -0.758 -0.258 -0.449 -0.795 -0.362 -1.81 -0.780 -1.90 1
#> 4 0.855 0.335 -1.36 0.796 -0.674 -1.37 -1.42 -1.03 -0.560 0
#> 5 0.436 -0.0487 -0.639 0.352 -0.325 -0.893 -0.746 0.0548 -0.394 0
#> 6 -0.228 -0.240 -0.854 -0.197 0.884 0.118 -0.0713 1.09 -0.0289 1
#> 7 -0.949 -0.231 0.428 0.290 -0.803 2.15 -1.11 -0.202 -1.21 1
#> 8 1.88 -0.0980 -2.60 -1.86 -0.0258 -0.965 -1.52 -0.539 0.108 0
#> 9 0.221 1.58 -1.46 -0.806 0.749 0.506 1.09 0.523 1.86 0
#> 10 0.0238 -0.389 -0.474 0.512 -0.448 0.178 0.529 1.56 -1.12 1
#> # ... with 2 more variables: var2 <dbl>, var3 <dbl>
Created on 2022-06-08 by the reprex package (v2.0.1)
You could restructure your data along the principles of tidy data (see e.g. https://cran.r-project.org/web/packages/tidyr/vignettes/tidy-data.html).
Here to a long format and using tidyverse:
library(tidyverse)
df <-
df |>
pivot_longer(everything()) |>
separate(name, c("var", "time"), sep = "(?=[0-9])") |>
pivot_wider(id_col = "time",
names_from = "var",
names_prefix = "var_",
values_from = "value",
values_fn = list) |>
unnest(-time) |>
mutate(new_var = ifelse(var_x > 0 & (var_y < 0.5 | var_z < 0.5), 0, 1))
df
You would probably want to keep the data in a long format, but if you want, you can pivot_wider and get back to the format you started with. E.g.
df |>
pivot_wider(values_from = c(starts_with("var_"), "new_var"),
names_from = "time",
values_fn = list) |>
unnest(everything())
As you suggested, a solution using a loop is definitely possible.
# times as unique non-alphabetical parts of column names
times <- unique(gsub('[[:alpha:]]', '', names(df)))
for (time in times) {
# column names for current time
xyz <- paste0(c('x', 'y', 'z'), time)
df[[paste0('var', time)]] <-
ifelse(df[[xyz[1]]]>0 & (df[[xyz[2]]]<.5 | df[[xyz[3]]]<.5), 0, 1)
}
Another way I can think of is transforming the data into a 3D array (observartion × variable × time) so that you can actually do the computation for all variables at once.
times <- unique(gsub('[[:alpha:]]', '', names(df)))
df.arr <- sapply(c('x', 'y', 'z'),
function(var) as.matrix(df[, paste0(var, times)]),
simplify='array')
new.vars <- ifelse(df.arr[, , 1]>0 & (df.arr[, , 2]<0.5 | df.arr[, , 3]<0.5), 0, 1)
colnames(new.vars) <- paste0('var', times)
cbind(df, new.vars)
Here, sapply creates a matrix from columns of measurings for each variable at different times and stacks them into a 3D array.
If you trust (or ensure) correct ordering of columns in the data frame, instead of using sapply you can create the array just by modifying the object's dimensions. I didn't do any benchmarking but i guess this could be the most computationally efficient solution (if it should matter).
df.arr <- as.matrix(df)
dim(df.arr) <- c(dim(df.arr) / c(1, 3), 3)
I have a df that looks something like this like this:
set.seed(42)
ID <- sample(1:30, 100, rep=T)
Trait <- sample(0:1, 100, rep=T)
Year <- sample(1992:1999, 100, rep=T)
df <- cbind(ID, Trait, Year)
df <- as.data.frame(df)
Where ID is an individual organism, trait is a presence/absence of a phenotype and Year is the year an observation was made.
I would like to model if trait is random between individuals, something like this
library(MCMCglmm)
m <- MCMCglmm(Trait ~ ID, random = ~ Year, data = df, family = "categorical")
Now, would like to shuffle the Trait column and run x permutations, to check if my observed mean and CI fall outside of what's expected from random.
I could do this with a for loop, but I'd rather use a tidyverse solution.
I've read that lapply is a bette(?) alternative, but I am struggling to find a specific enough walk-through that I can follow.
I'd appreciate any advice offered here.
Cheers!
Jamie
EDIT October 10th. Cleaned up the code and per comment below added the code to give you back a nice organized tibble\dataframe
### decide how many shuffles you want and name them
### in an orderly fashion for the output
shuffles <- 1:10
names(shuffles) <- paste0("shuffle_", shuffles)
library(MCMCglmm)
library(dplyr)
library(tibble)
library(purrr)
ddd <- purrr::map(shuffles,
~ df %>%
mutate(Trait = sample(Trait)) %>%
MCMCglmm(fixed = Trait ~ ID,
random = ~ Year,
data = .,
family = "categorical",
verbose = FALSE)) %>%
purrr::map( ~ tibble::as_tibble(summary(.x)$solutions, rownames = "model_term")) %>%
dplyr::bind_rows(., .id = 'shuffle')
ddd
#> # A tibble: 20 x 7
#> shuffle model_term post.mean `l-95% CI` `u-95% CI` eff.samp pMCMC
#> <chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 shuffle_1 (Intercept) 112. 6.39 233. 103. 0.016
#> 2 shuffle_1 ID -6.31 -13.5 -0.297 112. 0.014
#> 3 shuffle_2 (Intercept) 24.9 -72.5 133. 778. 0.526
#> 4 shuffle_2 ID -0.327 -6.33 5.33 849. 0.858
#> 5 shuffle_3 (Intercept) 4.39 -77.3 87.4 161. 0.876
#> 6 shuffle_3 ID 1.04 -3.84 5.99 121. 0.662
#> 7 shuffle_4 (Intercept) 7.71 -79.0 107. 418. 0.902
#> 8 shuffle_4 ID 0.899 -4.40 6.57 408. 0.694
#> 9 shuffle_5 (Intercept) 30.4 -62.4 144. 732. 0.51
#> 10 shuffle_5 ID -0.644 -6.61 4.94 970. 0.866
#> 11 shuffle_6 (Intercept) -45.5 -148. 42.7 208. 0.302
#> 12 shuffle_6 ID 4.73 -0.211 11.6 89.1 0.058
#> 13 shuffle_7 (Intercept) -16.2 -133. 85.9 108. 0.696
#> 14 shuffle_7 ID 2.47 -2.42 10.3 47.8 0.304
#> 15 shuffle_8 (Intercept) 0.568 0.549 0.581 6.60 0.001
#> 16 shuffle_8 ID -0.0185 -0.0197 -0.0168 2.96 0.001
#> 17 shuffle_9 (Intercept) -6.95 -112. 92.2 452. 0.886
#> 18 shuffle_9 ID 2.07 -3.30 8.95 370. 0.476
#> 19 shuffle_10 (Intercept) 43.8 -57.0 159. 775. 0.396
#> 20 shuffle_10 ID -1.36 -7.44 5.08 901. 0.62
Your original data
set.seed(42)
ID <- sample(1:30, 100, rep=T)
Trait <- sample(0:1, 100, rep=T)
Year <- sample(1992:1999, 100, rep=T)
df <- cbind(ID, Trait, Year)
df <- as.data.frame(df)
I am trying to implement analyses across a posterior of matrices. What I start with is a tibble of k^2 columns, where k is the dimensions of the matrix. The ith row forms the matrix of the ith iteration.
So, for example for a 3x3 matrix, this is:
set.seed(12)
n <- 1000
z1z1 <- rnorm(n, 5, 1)
z2z2 <- rnorm(n, 5, 1)
z3z3 <- rnorm(n, 5, 1)
z1z2 <- rnorm(n, 0, 1)
z1z3 <- rnorm(n, 0, 1)
z2z3 <- rnorm(n, 0, 1)
post3 <- as_tibble(matrix(c(z1z1, z1z2, z1z3,
z1z2, z2z2, z2z3,
z1z3, z2z3, z3z3),
ncol = 9))
post3
Giving:
# A tibble: 1,000 x 9
V1 V2 V3 V4 V5 V6 V7 V8 V9
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 3.52 -0.618 2.96 -0.618 2.48 -0.634 2.96 -0.634 5.98
2 6.58 -0.827 0.0909 -0.827 5.52 -1.84 0.0909 -1.84 6.20
3 4.04 1.48 -1.66 1.48 6.58 0.166 -1.66 0.166 5.58
4 4.08 -1.01 0.809 -1.01 5.49 0.607 0.809 0.607 6.55
5 3.00 0.582 -0.485 0.582 6.20 0.0765 -0.485 0.0765 6.38
6 4.73 0.718 1.97 0.718 4.00 -0.147 1.97 -0.147 4.35
7 4.68 -0.372 0.572 -0.372 4.65 -1.68 0.572 -1.68 3.83
8 4.37 -0.809 0.883 -0.809 3.96 0.985 0.883 0.985 4.97
9 4.89 0.405 0.686 0.405 6.02 0.252 0.686 0.252 6.29
10 5.43 0.124 0.199 0.124 5.75 0.354 0.199 0.354 4.20
# ... with 990 more rows
Where this is the matrix in the first iteration:
k <- sqrt(length(post3))
matrix(post3[1,], nrow = k)
[,1] [,2] [,3]
[1,] 3.519432 -0.618137 2.962622
[2,] -0.618137 2.479522 -0.6338298
[3,] 2.962622 -0.6338298 5.977552
I am then working along this posterior to calculate the dominance of the first eigenvector:
post3 %>%
rowwise %>%
mutate(
pre_eig = list(eigen(matrix(c(V1, V2, V3, V4, V5, V6, V7, V8, V9), nrow = k))),
dom = pre_eig[[1]][1] / sum(pre_eig[[1]][1:k])) %>%
select('dom')
Giving:
# A tibble: 1,000 x 1
dom
<dbl>
1 0.676
2 0.437
3 0.462
4 0.427
5 0.414
6 0.504
7 0.474
8 0.429
9 0.394
10 0.383
# ... with 990 more rows
What I would like to do is make this script versatile so that it can take posteriors for any value of k. The issue I am having is in how to define the matrix without having to hand write all the column names - when applying this to 2000x2000 matrices I don't want to write out V1, V2, V3... V4000000!
I tried a few things (including ...eigen(matrix(c(paste0('V', 1:(k^2))), nrow = k)))..., which I think is not working because it wants V1, V2... rather than "V1", "V2"...) and I all out of ideas. How do I get it to automatically take the column names from the posterior tibble?
I would then be able to use the exact same piece of script for example on post3 <- as_tibble(matrix(c(z1z1, z1z2, z1z2, z2z2), ncol = 4))...
You can avoid naming all the columns explicitly if you gather each row's values into key-value pairs:
library(tidyr)
post3 %>%
# add row ID (so that results can be sorted back into original order)
mutate(row.id = seq(1, n())) %>%
# convert each row to long format, with values sorted from 1st to k^2th column
gather(position, value, -row.id) %>%
mutate(position = as.numeric(gsub("^V", "", position))) %>%
arrange(row.id, position) %>%
select(-position) %>%
# group by row ID & calculate
group_by(row.id) %>%
summarise(pre_eig = list(eigen(matrix(value, nrow = k))[["values"]]),
dom = pre_eig[[1]][1] / sum(pre_eig[[1]][1:k])) %>%
ungroup() %>%
# sort results in original order
arrange(row.id) %>%
select(dom)
The results should be the same as before:
# A tibble: 1,000 x 1
dom
<dbl>
1 0.676
2 0.437
3 0.462
4 0.427
5 0.414
6 0.504
7 0.474
8 0.429
9 0.394
10 0.383
# ... with 990 more rows
I have several models fit to predict an outcome y = x1 + x2 + .....+x22. That's a fair number of predictors and a fair number of models. My customers want to know what's the marginal impact of each X on the estimated y. The models may include splines and interaction terms. I can do this, but it's cumbersome and requires loops or a lot of copy paste, which is slow or error prone. Can I do this better by writing my function differently and/or using purrr or an *apply function? Reproducible example is below. Ideally, I could write one function and apply it to longdata.
## create my fake data.
library(tidyverse)
library (rms)
ltrans<- function(l1){
newvar <- exp(l1)/(exp(l1)+1)
return(newvar)
}
set.seed(123)
mystates <- c("AL","AR","TN")
mydf <- data.frame(idno = seq(1:1500),state = rep(mystates,500))
mydf$x1[mydf$state=='AL'] <- rnorm(500,50,7)
mydf$x1[mydf$state=='AR'] <- rnorm(500,55,8)
mydf$x1[mydf$state=='TN'] <- rnorm(500,48,10)
mydf$x2 <- sample(1:5,500, replace = T)
mydf$x3 <- (abs(rnorm(1500,10,20)))^2
mydf$outcome <- as.numeric(cut2(sample(1:100,1500,replace = T),95))-1
dd<- datadist(mydf)
options(datadist = 'dd')
m1 <- lrm(outcome ~ x1 + x2+ rcs(x3,3), data = mydf)
dothemath <- function(x1 = x1ref,x2 = x2ref,x3 = x3ref) {
ltrans(-2.1802256-0.01114239*x1+0.050319692*x2-0.00079289232* x3+
7.6508189e-10*pmax(x3-7.4686271,0)^3-9.0897627e-10*pmax(x3- 217.97865,0)^3+
1.4389439e-10*pmax(x3-1337.2538,0)^3)}
x1ref <- 51.4
x2ref <- 3
x3ref <- 217.9
dothemath() ## 0.0591
mydf$referent <- dothemath()
mydf$thisobs <- dothemath(x1 = mydf$x1, x2 = mydf$x2, x3 = mydf$x3)
mydf$predicted <- predict(m1,mydf,type = "fitted.ind") ## yes, matches.
mydf$x1_marginaleffect <- dothemath(x1= mydf$x1)/mydf$referent
mydf$x2_marginaleffect <- dothemath(x2 = mydf$x2)/mydf$referent
mydf$x3_marginaleffect <- dothemath(x3 = mydf$x3)/mydf$referent
## can I do this with long data?
longdata <- mydf %>%
select(idno,state,referent,thisobs,x1,x2,x3) %>%
gather(varname,value,x1:x3)
##longdata$marginaleffect <- dothemath(longdata$varname = longdata$value) ## no, this does not work.
## I need to communicate to the function which variable it is evaluating.
longdata$marginaleffect[longdata$varname=="x1"] <- dothemath(x1 = longdata$value[longdata$varname=="x1"])/
longdata$referent[longdata$varname=="x1"]
longdata$marginaleffect[longdata$varname=="x2"] <- dothemath(x2 = longdata$value[longdata$varname=="x2"])/
longdata$referent[longdata$varname=="x2"]
longdata$marginaleffect[longdata$varname=="x3"] <- dothemath(x3 = longdata$value[longdata$varname=="x3"])/
longdata$referent[longdata$varname=="x3"]
testing<- inner_join(longdata[longdata$varname=="x1",c(1,7)],mydf[,c(1,10)])
head(testing) ## yes, both methods work.
Mostly you're just talking about a grouped mutate, with the caveat that dothemath is built such that you need to specify the variable name, which can be done by using do.call or purrr::invoke to call it on a named list of parameters:
longdata <- longdata %>%
group_by(varname) %>%
mutate(marginaleffect = invoke(dothemath, setNames(list(value), varname[1])) / referent)
longdata
#> # A tibble: 4,500 x 7
#> # Groups: varname [3]
#> idno state referent thisobs varname value marginaleffect
#> <int> <fct> <dbl> <dbl> <chr> <dbl> <dbl>
#> 1 1 AL 0.0591 0.0688 x1 46.1 1.06
#> 2 2 AR 0.0591 0.0516 x1 50.2 1.01
#> 3 3 TN 0.0591 0.0727 x1 38.0 1.15
#> 4 4 AL 0.0591 0.0667 x1 48.4 1.03
#> 5 5 AR 0.0591 0.0515 x1 47.1 1.05
#> 6 6 TN 0.0591 0.0484 x1 37.6 1.15
#> 7 7 AL 0.0591 0.0519 x1 60.9 0.905
#> 8 8 AR 0.0591 0.0531 x1 63.2 0.883
#> 9 9 TN 0.0591 0.0780 x1 47.8 1.04
#> 10 10 AL 0.0591 0.0575 x1 50.5 1.01
#> # ... with 4,490 more rows
# the first values look similar
inner_join(longdata[longdata$varname == "x1", c(1,7)], mydf[,c(1,10)])
#> Joining, by = "idno"
#> # A tibble: 1,500 x 3
#> idno marginaleffect x1_marginaleffect
#> <int> <dbl> <dbl>
#> 1 1 1.06 1.06
#> 2 2 1.01 1.01
#> 3 3 1.15 1.15
#> 4 4 1.03 1.03
#> 5 5 1.05 1.05
#> 6 6 1.15 1.15
#> 7 7 0.905 0.905
#> 8 8 0.883 0.883
#> 9 9 1.04 1.04
#> 10 10 1.01 1.01
#> # ... with 1,490 more rows
# check everything is the same
mydf %>%
gather(varname, marginaleffect, x1_marginaleffect:x3_marginaleffect) %>%
select(idno, varname, marginaleffect) %>%
mutate(varname = substr(varname, 1, 2)) %>%
all_equal(select(longdata, idno, varname, marginaleffect))
#> [1] TRUE
It may be easier to reconfigure dothemath to take an additional parameter of the variable name so as to avoid the gymnastics.