I'm trying to pass vectors, each with a different number of NA values, through to a map() function but it's returning an error.
I have a tibble of N numeric columns and 1 categorical column. I want to compare the distributions for each of the numeric columns against the other split by the values of the categorical column. I use overlapping::overlap() to calculate the overlap of the distributions, and i feed the numeric columns into a map_dfr function for the iteration. For example:
require(overlapping)
require(dplyr)
require(purrr)
set.seed( 1 )
n <- 100
G1 <- sample( 0:30, size = n, replace = TRUE )
G2 <- sample( 0:30, size = n, replace = TRUE, prob = dbinom( 0:30, 31, .55 ))
G3 <- sample( 0:30, size = n, replace = TRUE, prob = dbinom( 0:30, 41, .65 ))
Data <- data.frame(y = G1, x = G2, z = G3, group = rep(c("G1","G2", "G3"), each = n), class = rep(c("C1","C2", "C3"), each = 1)) %>% as_tibble()
Data
overlap_fcn <- function(.x) {
## construct list of vectors
dist_list <- list(
"C1" = Data %>%
filter(class == 'C1', !is.na(.x)) %>%
pull(.x),
"C2" = Data %>%
filter(class == 'C2', !is.na(.x)) %>%
pull(.x),
"C3" = Data %>%
filter(class == 'C3', !is.na(.x)) %>%
pull(.x)
)
## calculate distribution overlaps
return(
enframe(
overlapping::overlap(dist_list)$OV*100
) %>%
mutate(value = paste0(round(value, 2), "%"),
class = .x) %>%
rename(comparison = name, overlap = value) %>%
relocate(class)
)
}
overlap_table <- purrr::map_dfr(
.x = c('y', 'x', "z"),
.f = ~overlap_fcn(.x))
overlap_table
The above works as intended. However, in practice I have different amounts of missingess in each of x, y, and z. I try to account for this with the filter on !is.na(.x) but it's not working. For example:
Data$x[1:3] <- NA
Data$y[10:20] <- NA
Data$z[100:150] <- NA
overlap_table <- purrr::map_dfr(
.x = c('x', 'y', "z"),
.f = ~overlap_fcn(.x))
returns this error:
Error in density.default(x[[j]], n = nbins, ...): 'x' contains missing values
Error in density.default(x[[j]], n = nbins, ...): 'x' contains missing values
Traceback:
1. purrr::map_dfr(.x = c("x", "y", "z"), .f = ~overlap_fcn(.x))
2. map(.x, .f, ...)
3. .f(.x[[i]], ...)
4. overlap_fcn(.x)
5. enframe(overlapping::overlap(dist_list)$OV * 100) %>% mutate(value = paste0(round(value,
. 2), "%"), class = .x) %>% rename(comparison = name, overlap = value) %>%
. relocate(class) # at line 25-33 of file <text>
6. relocate(., class)
7. rename(., comparison = name, overlap = value)
8. mutate(., value = paste0(round(value, 2), "%"), class = .x)
9. enframe(overlapping::overlap(dist_list)$OV * 100)
10. overlapping::overlap(dist_list)
11. density(x[[j]], n = nbins, ...)
12. density.default(x[[j]], n = nbins, ...)
13. stop("'x' contains missing values")
Can anyone help me out here please? I'm sure it's something super obvious i'm missing; i just can't see what!
Here, the .x is character class. We may need to convert to symbol and evaluate (!!)
overlap_fcn <- function(.x) {
## construct list of vectors
dist_list <- list(
"C1" = Data %>%
filter(class == 'C1', !is.na(!! rlang::sym(.x))) %>%
pull(.x),
"C2" = Data %>%
filter(class == 'C2', !is.na(!! rlang::sym(.x))) %>%
pull(.x),
"C3" = Data %>%
filter(class == 'C3', !is.na(!! rlang::sym(.x))) %>%
pull(.x)
)
## calculate distribution overlaps
return(
enframe(
overlapping::overlap(dist_list)$OV*100
) %>%
mutate(value = paste0(round(value, 2), "%"),
class = .x) %>%
rename(comparison = name, overlap = value) %>%
relocate(class)
)
}
-testing after creating the NAs in Data
> purrr::map_dfr(
+ .x = c('x', 'y', "z"),
+ .f = ~overlap_fcn(.x))
# A tibble: 9 × 3
class comparison overlap
<chr> <chr> <chr>
1 x C1-C2 98.61%
2 x C1-C3 97.46%
3 x C2-C3 97.5%
4 y C1-C2 95.47%
5 y C1-C3 96.22%
6 y C2-C3 97.14%
7 z C1-C2 90.17%
8 z C1-C3 94.9%
9 z C2-C3 89.24%
Related
I have a data frame that contains nested list based on ID. I am trying to apply a function to the nested list within this data frame, but I am running into this error:
Error in make_track(tbl = x, .x = x, .y = y, .t = date, uid = ID, crs = sp::CRS("+init=epsg:32612")) : Non existent columns from tbl were requested.
Here is my reproducible example. I was wondering what the best way to apply a function to a nested list might be, and how I can go about fixing this error. Do I have to do a double lapply to fix this problem?
set.seed(12345)
library(lubridate)
library(dplyr)
library(amt)
f = function(data){
data %>% mutate(
new = floor_date(data$date, "10 days"),
new = if_else(day(new) == 31, new - days(10), new)
) %>%
group_split(new)
}
nested <- tibble(
ID = rep(c("A","B","C","D", "E"), 100),
date = rep_len(seq(dmy("01-01-2010"), dmy("31-12-2013"), by = "days"), 500),
x = runif(length(date), min = 60000, max = 80000),
y = runif(length(date), min = 800000, max = 900000)
) %>% group_by(ID) %>%
nest() %>%
mutate(data = map(data, f))
track_list <- lapply(nested, function (x){
make_track(tbl = x, .x = x, .y = y, .t = date,
uid = ID,
# lat/long: 4326 (lat/long, WGS84 datum).
# utm: crs = sp::CRS("+init=epsg:32612"))
crs = sp::CRS("+init=epsg:32612"))
})
The issue is that the data is nested, so we need to do one more level inside to pick up the data. Also, the make_track requires all columns to be in the same data object, so we need to create the corresponding uid from the 'ID' column of nested object
library(purrr)
library(dplyr)
library(amt)
out <- map2_dfr(nested$ID, nested$data, function(z, lst1)
map_dfr(lst1, ~ {
dat <- .x %>%
mutate(ID = z)
make_track(tbl = dat, .x = x, .y = y, .t = date, uid = ID,
crs = sp::CRS("+init=epsg:32612"))
}))
-output
> out
# A tibble: 500 x 4
x_ y_ t_ uid
<dbl> <dbl> <date> <chr>
1 74418. 820935. 2010-01-01 A
2 63327. 885896. 2010-01-06 A
3 60691. 873949. 2010-01-11 A
4 69250. 868411. 2010-01-16 A
5 69075. 876142. 2010-01-21 A
6 67797. 829892. 2010-01-26 A
7 75860. 843542. 2010-01-31 A
8 67233. 882318. 2010-02-05 A
9 75644. 826283. 2010-02-10 A
10 66424. 853789. 2010-02-15 A
# … with 490 more rows
If we want the output as a nested list, use remove the _dfr
out <- map2(nested$ID, nested$data, function(z, lst1)
map(lst1, ~ {
dat <- .x %>%
mutate(ID = z)
make_track(tbl = dat, .x = x, .y = y, .t = date, uid = ID,
crs = sp::CRS("+init=epsg:32612"))
}))
I have a list of 2D matrices. Each matrix is filled using the function fillMatrices. This function adds a number of individuals to each day 0 in a matrix and updates the columns a_M, b_M and c_M. The numbers of individuals come from an initial matrix ind. The code works but it is slow when the number of matrices within the list is large. For example with n = 10000:
user system elapsed
3.73 0.83 4.55
If possible, I would like to reduce the elapsed time to <= 1 sec and increase the n to 720000 matrices. So, I am looking for way to optimize only the section 3. Here is the code:
###############################################
###############################################
## Section 3
## Run the function "fillMatrices"
indexTime <- 1
dt_t_1 <- do.call(rbind, lapply(list_matrices, function(x) x[1,]))
dt_t <- fillMatrices(dt_t_1 = dt_t_1, species = c("a_M", "b_M", "c_M"), maxDuration = 5, matrixColumns = col_mat)
## Fill the matrices within the list
system.time(for(i in 1:n){
list_matrices[[i]][indexTime + 1,] <- dt_t[,i]
})
## test <- list_matrices[[1]]
The code of the section 1 is used to initialize the matrices and the function fillMatrices can be found in the section 2. In my example, the function is used to fill matrices for one species. In reality, the function is used for 3 species (i.e., is applied three times) by changing the argument species = c("a_M", "b_M", "c_M"). How can I speed up my code? Any advice would be much appreciated.
Here are the codes of sections 1 and 2:
rm(list=ls(all=TRUE))
library(ff)
library(dplyr)
set.seed(12345)
## Define the number of individuals
n <- 10000
###############################################
###############################################
## Section 1
## Build the list of 2D matrices
v_date <- as.vector(outer(c(paste(seq(0, 1, by = 1), "day", sep="_"), paste(seq(2, 5, by = 1), "days", sep="_")), c("a_M", "b_M", "c_M"), paste, sep="|"))
col_mat <- c("year", "day", "time", "ID", "died", v_date)
list_matrices <- list()
for(i in 1:n){
print(i)
list_matrices[[i]] <- ff(-999, dim=c(3650, length(col_mat)), dimnames=list(NULL, col_mat), vmode="double", overwrite = TRUE)
}
## test <- list_matrices[[1]]
## dim(list_matrices[[1]])
## Fill the first row of each matrix
for(i in 1:n){
print(i)
list_matrices[[i]][1,] <- c(1, 1, 1, i-1, 0, rep(0, length(v_date)))
}
## test <- list_matrices[[2]]
## Build the matrix "individual"
ind <- as.matrix(data.frame(year = rep(1, n), day = rep(1, n), time = rep(1, n), died = rep(0, n), ID = (seq(1, n, 1))- 1, a_M = sample(1:10, n, replace = T), b_M = sample(1:10, n, replace = T), c_M = sample(1:10, n, replace = T)))
## print(ind)
###############################################
###############################################
## Section 2
## Function to convert a data frame into a matrix
convertDFToMat <- function(x){
mat <- as.matrix(x[,-1])
ifelse(is(x[,1], "data.frame"), rownames(mat) <- pull(x[,1]), rownames(mat) <- x[,1])
## Convert character matrix into numeric matrix
mat <- apply(mat, 2, as.numeric)
return(mat)
}
## Define the function that is used to fill the matrices within the list
fillMatrices <- function(dt_t_1, species, maxDuration, matrixColumns){
## Format data
dt <- as.data.frame(dt_t_1) %>%
reshape::melt(id = c("ID")) %>%
arrange(ID) %>%
dplyr::mutate_all(as.character)
## summary(dt)
## Break out the variable "variable" into different columns, with one row for each individual-day
dt_reshape_filter_1 <- dt %>%
dplyr::filter(!variable %in% c("year", "day", "time", "ID", "died")) %>%
dplyr::mutate(day = variable %>% gsub(pattern = "\\_.*", replacement = "", x = .), col = variable %>% gsub(pattern = ".*\\|", replacement = "", x = .)) %>%
dplyr::select(-variable) %>%
tidyr::spread(col, value) %>%
dplyr::mutate_all(as.numeric) %>%
dplyr::arrange(ID, day)
## summary(dt_reshape_filter_1)
## Apply requested transformations and build the data frame
dt_transform <- dt_reshape_filter_1 %>%
dplyr::rename_at(vars(species), ~ c("a", "b", "c")) %>%
dplyr::mutate(day = day + 1) %>%
dplyr::filter(day < maxDuration + 1) %>%
dplyr::bind_rows(tibble(ID = ind[,c("ID")], day = 0, a = ind[,c("a_M")], b = ind[,c("b_M")])) %>%
dplyr::mutate(c = a + b) %>%
dplyr::rename_at(vars("a", "b", "c"), ~ species) %>%
dplyr::arrange(ID, day)
## summary(dt_transform)
## Take different columns of the data frame and gather them into a single column
dt_gather <- dt_transform %>%
tidyr::gather(variable, value, species) %>%
dplyr::mutate(day = if_else(day > 1, paste0(day, "_days"), paste0(day, "_day"))) %>%
tidyr::unite(variable, c("day", "variable"), sep = "|") %>%
dplyr::rename(var2 = ID) %>%
dplyr::mutate_all(as.character)
## summary(dt_gather)
## Add the other columns in the data frame and convert the resulting data frame into a matrix
dt_reshape_filter_2 <- dt %>%
dplyr::rename(var2 = ID) %>%
dplyr::filter(variable %in% c("year", "day", "time", "ID", "died")) %>%
tidyr::spread(variable, value) %>%
dplyr::arrange(as.numeric(var2)) %>%
dplyr::mutate(year = ind[,c("year")],
day = ind[,c("day")],
time = ind[,c("time")],
ID = ind[,c("ID")],
died = ind[,c("died")]) %>%
tidyr::gather(variable, value, c(year, day, time, ID, died)) %>%
dplyr::arrange(as.numeric(var2)) %>%
dplyr::mutate_all(as.character)
## summary(dt_reshape_filter_2)
## Build the output matrix
dt_bind <- bind_rows(dt_reshape_filter_2, dt_gather) %>%
tidyr::spread(var2, value) %>%
dplyr::arrange(match(variable, matrixColumns)) %>%
dplyr::select("variable", as.character(ind[,c("ID")]))
## summary(dt_bind)
dt_mat <- convertDFToMat(dt_bind)
## summary(dt_mat)
return(dt_mat)
}
Making a 3D array instead of a 2D list of matrices gives you more options
library(ff)
library(dplyr)
set.seed(12345)
## Define the number of individuals
n <- 10000L
n_row <- 3650L
#array way:
v_date <- as.vector(outer(c(paste(seq(0, 1, by = 1), "day", sep="_"), paste(seq(2, 5, by = 1), "days", sep="_")), c("a_M", "b_M", "c_M"), paste, sep="|"))
col_mat <- c("year", "day", "time", "ID", "died", v_date)
arr1 <- ff(-999L, dim = c(n_row, length(col_mat), n), dimnames = list(NULL, col_mat, NULL))
## Fill the first row of each matrix slice
arr1[1, , ] <- c(1L, 1L, 1L, NA, 0L, rep(0L, length(v_date)))
arr1[1, 4, ] <- seq_len(n)-1L
## Build the matrix "individual"
ind <- as.matrix(data.frame(year = rep(1L, n), day = rep(1L, n), time = rep(1L, n), died = rep(0L, n), ID = (seq(1L, n, 1L))- 1L, a_M = sample(1L:10L, n, replace = T), b_M = sample(1L:10L, n, replace = T), c_M = sample(1L:10L, n, replace = T)))
##fill the matrix
indexTime <- 1L
dt_t <- fillMatrices(dt_t_1 = t(arr1[1, ,]), species = c("a_M", "b_M", "c_M"), maxDuration = 5, matrixColumns = col_mat)
## reassign
system.time(
arr1[indexTime + 1, ,] <- dt_t
)
user system elapsed
0.05 0.70 0.7
# for comparison
#> system.time(for(i in 1:n){
#+ list_matrices[[i]][indexTime + 1,] <- dt_t[,i]
#+ })
# user system elapsed
# 4.75 1.08 5.90
As far as I can tell, it's giving me the same results as your original approach but does so a lot faster.
I am working with the iris dataset, and manipulating it as follows to get a species, feature1, feature2, value data frame:
gatherpairs <- function(data, ...,
xkey = '.xkey', xvalue = '.xvalue',
ykey = '.ykey', yvalue = '.yvalue',
na.rm = FALSE, convert = FALSE, factor_key = FALSE) {
vars <- quos(...)
xkey <- enquo(xkey)
xvalue <- enquo(xvalue)
ykey <- enquo(ykey)
yvalue <- enquo(yvalue)
data %>% {
cbind(gather(., key = !!xkey, value = !!xvalue, !!!vars,
na.rm = na.rm, convert = convert, factor_key = factor_key),
select(., !!!vars))
} %>% gather(., key = !!ykey, value = !!yvalue, !!!vars,
na.rm = na.rm, convert = convert, factor_key = factor_key)%>%
filter(!(.xkey == .ykey)) %>%
mutate(var = apply(.[, c(".xkey", ".ykey")], 1, function(x) paste(sort(x), collapse = ""))) %>%
arrange(var)
}
test = iris %>%
gatherpairs(sapply(colnames(iris[, -ncol(iris)]), eval))
This was taken from https://stackoverflow.com/a/47731111/8315659
What this does is give me that data frame with all combinations of feature1 and feature2, but I want to remove duplicates where it is just the reverse being shown. For example, Petal.Length vs Petal.Width is the same as Petal.Width vs Petal.Length. But if there are two rows with identical values for Petal.Length vs Petal.Width, I do not want to drop that row. Therefore, just dropping rows where all values are identical except that .xkey and .ykey are reversed is what I would want to do. Essentially, this is just to recreate the bottom triangle of the ggplot matrix shown in the above linked answer.
How can this be done?
Jack
I think this could be accomplished using the first part of the source code, which performs a single gathering operation. Using the iris example, this will produce 600 rows of output, one for each of the 150 rows x 4 columns in iris.
gatherpairs <- function(data, ...,
xkey = '.xkey', xvalue = '.xvalue',
ykey = '.ykey', yvalue = '.yvalue',
na.rm = FALSE, convert = FALSE, factor_key = FALSE) {
vars <- quos(...)
xkey <- enquo(xkey)
xvalue <- enquo(xvalue)
ykey <- enquo(ykey)
yvalue <- enquo(yvalue)
data %>% {
cbind(gather(., key = !!xkey, value = !!xvalue, !!!vars,
na.rm = na.rm, convert = convert, factor_key = factor_key),
select(., !!!vars))
} # %>% gather(., key = !!ykey, value = !!yvalue, !!!vars,
# na.rm = na.rm, convert = convert, factor_key = factor_key)%>%
# filter(!(.xkey == .ykey)) %>%
# mutate(var = apply(.[, c(".xkey", ".ykey")], 1, function(x) paste(sort(x), collapse = ""))) %>%
# arrange(var)
}
My data is below
grp <- paste('group', sample(1:3, 100, replace = T))
x <- rnorm(100, 100)
y <- rnorm(100, 10)
df <- data.frame(grp = grp, x =x , y =y , stringsAsFactors = F)
lag_size <- c(10, 4, 9)
Now when I try to use
df %>% group_by(grp) %>% mutate_all(lag, n = lag_size) %>% arrange(grp)
it gives an error
Error in mutate_impl(.data, dots) :
Expecting a single value:
whereas this works fine
df %>% group_by(grp) %>% mutate_all(lag, n = 10) %>% arrange(grp)
If we need to do the lag based on the 'grp' i.e. to lag the corresponding 'grp' with the value specified in 'lag_size'
library(tidyverse)
res <- map2(split(df[2:3], df$grp) , lag_size, ~.x %>%
mutate_all(lag, n = .y)) %>%
bind_rows(., .id = 'grp')
We can check the lag in 'grp' by the position of the first non-NA element
res %>%
group_by(grp) %>%
summarise(n = which(!is.na(x))[1]-1)
# A tibble: 3 x 2
# grp n
# <chr> <dbl>
#1 group 1 10
#2 group 2 4
#3 group 3 9
Here is a data structure that I have:
structure(list(UDD_beta = c(1.17136554204268, 0.939587997289016
), UDD_pval = c(0, 0), UDD_R.sq = c(0.749044972637797, 0.516943886705951
), SSX_beta = c(1.05356804780772, 0.927948300464624), SSX_pval = c(0,
0), SSX_R.sq = c(0.60226298037862, 0.629111666509209), SPP_beta = c(0.675765151939885,
0.516425218613404), SPP_pval = c(0, 0), SPP_R.sq = c(0.479849538274406,
0.378266618442121), EEE_beta = c(0.690521022226874, 0.639380962824289
), EEE_pval = c(0, 0), EEE_R.sq = c(0.585610742768951, 0.676073352909597
)), .Names = c("UDD_beta", "UDD_pval", "UDD_R.sq", "SSX_beta",
"SSX_pval", "SSX_R.sq", "SPP_beta", "SPP_pval", "SPP_R.sq",
"EEE_beta", "EEE_pval", "EEE_R.sq"), row.names = c("DDK", "DDL"
), class = "data.frame")
I want to take R.sq columns and for each row find the max and the column name of the max value. Then take corresponding beta. Expected output:
Name Value
DDK UDD 1.17136554204268
DDL EEE 0.690521022226874
Sorry, the second expected value should be 0.639380962824289.
We could use max.col. Subset the columns of interest i.e. columns that have 'R.sq' using the grep, then get the column index of max value with max.col. Use that to get the column names and also the values that correspond to a particular row (row/column indexing)
i1 <- grep("R.sq", names(df1))
i2 <- max.col(df1[i1], "first")
i3 <- grep("beta", names(df1))
res <- data.frame(Names = sub("_.*", "", names(df1)[i1][i2]),
Value = df1[i3][cbind(1:nrow(df1), i2)])
row.names(res) <- row.names(df1)
sub_data <- data[grep("R.sq", colnames(data))]
colnames(sub_data) <- gsub("_R.sq", "", colnames(sub_data))
sub_data$Name <- NA
sub_data$Value <- NA
for (i in 1:nrow(sub_data)){
sub_data$Name[i] <- names(sub_data[i,])[which.max(apply(sub_data[i,], 2, max))]
sub_data$Value[i] <- max(data[grep(paste0(sub_data$Name[i], "_beta"), colnames(data))], na.rm=T)
}
sub_data[c("Name", "Value")]
# Name Value
#DDK UDD 1.171366
#DDL EEE 0.690521
You can use a tidyverse approach via gathering your df to long and filtering both R.sq vars and max value, i.e.
library(tidyverse)
df %>%
rownames_to_column('ID') %>%
gather(var, val, -ID) %>%
filter(grepl('R.sq|beta', var)) %>%
group_by(ID) %>%
mutate(max1=as.integer(val == max(val[grepl('R.sq', var)]))) %>%
group_by(ID, grp = sub('_.*', '', var)) %>%
filter(!all(max1 == 0) & grepl('beta', var)) %>%
ungroup() %>% select(-c(max1, grp))
which gives,
# A tibble: 2 x 3
ID var val
<chr> <chr> <dbl>
1 DDK UDD_beta 1.171366
2 DDL EEE_beta 0.639381
# Need ID for all possible betas and Rsq
ID <- gsub("_R.sq", "", grep("_R.sq$", names(INPUT), value = TRUE))
dummy <- function(x) {
# Find out which Rsq is largest
i <- ID[which.max(x[paste0(ID, "_R.sq")])]
# Extract beta for largest Rsq
data.frame(Name = i, Value = x[paste0(i, "_beta")])
}
do.call("rbind", apply(INPUT, 1, dummy))