I am trying to create an edge list from a single character vector. My list to be processed is over 93k elements long, but as an example I will provide a small excerpt.
The chracter strings are part of the ICD10 code hierarchy and the parent child relationships exist within the string. That means that a single string, "A0101", would have a parent of "A010"
It would look like this:
A00
A000
A001
A009
A01
A010
A0100
A0101
A02
A03
etc.
My vector does not contain any other data except the strings but i basically need to convert
dat <- c("A00", "A000", "A001", "A009", "A01", "A010", "A0100", "A0101", "A02")
into an edge list formatted as follows...
# (A00, A000)
# (A00, A001)
# (A00, A009)
# (A01, A010)
# (A010, A0100)
# (A010, A0101)
I am fairly certain there are more efficient ways to accomplish this but this excerpt of code should download the ICD10 CM data from the icd.data package. Use the children detection system from the icd package and then make extensive use of the tidyverse to return an edgelist. I had to get a bit creative to connect the "top" of the hierarchies since they do not include the chapters and sub chapters of ICD10 data as an individual 2 or 1 digit code.
Basically sub-chapters become 2 digit codes, chapters become 1 digit codes, and then there is a root node to connect everything at the top.
library(icd.data)
icd10 <- icd10cm2016
library(icd)
code_children <- lapply(icd10$code, children)
code_vec <- sapply(code_children, paste, collapse = ",")
code_df <- as.data.frame(code_vec, stringsAsFactors = F)
library(dplyr);library(stringr);library(tidyr)
code_df_new <- code_df %>%
mutate(parent = sapply(strsplit(code_vec,","), "[", 1)) %>%
separate(code_vec,
paste("code", 1:max(str_count(code_df$code_vec, ",")), sep ="."),
",",extra = "merge")
library(reshape2)
edgelist <- melt(code_df_new, id = "parent") %>%
filter(!is.na(value)) %>%
select(parent, child = value) %>%
arrange(parent)
edgelist <- subset(edgelist, edgelist$parent != edgelist$child)
edgelist <- subset(edgelist, nchar(edgelist$child) == nchar(edgelist$parent) + 1)
subchaps <- icd10 %>% select(three_digit, sub_chapter, chapter) %>%
mutate(two_digit = substr(three_digit, 1, 2)) %>%
select(parent = two_digit, child = three_digit) %>%
distinct()
chaps <- icd10 %>% select(three_digit, sub_chapter, chapter) %>%
mutate(
two_digit = substr(three_digit, 1, 2),
one_digit = substr(three_digit, 1, 1)) %>%
select(parent = one_digit, child = two_digit) %>%
distinct()
root <- icd10 %>% select(three_digit) %>%
mutate(parent = "root", child = substr(three_digit, 1, 1)) %>%
select(parent, child) %>%
distinct()
edgelist_final <- edgelist %>%
bind_rows(list(chaps, subchaps, root)) %>%
arrange(parent)
If anybody has any tips or methods to improve the efficiency of this code I am all ears. (eyes?)
On the assumption that the length of the node names in ICD10 fully define the order (with shorter ones being parents), here's an approach that connects each node with it's immediate parent, if available.
While I think the logic is legible here, I'd be curious to see what a more streamlined solution would look like.
# Some longer fake data to prove that it works acceptably
# with 93k rows (took a few seconds). These are just
# numbers of different lengths, converted to characters, but they
# should suffice if the assumption about length = order is correct.
set.seed(42)
fake <- runif(93000, 0, 500) %>%
magrittr::raise_to_power(3) %>%
as.integer() %>%
as.character()
# Step 1 - prep
library(dplyr); library(tidyr)
fake_2 <- fake %>%
as_data_frame() %>%
mutate(row = row_number()) %>%
# Step 2 - widen by level and fill in all parent nodes
mutate(level = str_length(value)) %>%
spread(level, value) %>%
fill(everything()) %>%
# Step 3 - Get two highest non-NA nodes
gather(level, code, -row) %>%
arrange(row, level) %>%
filter(!is.na(code)) %>%
group_by(row) %>%
top_n(2, wt = level) %>%
# Step 4 - Spread once more to get pairs
mutate(pos = row_number()) %>%
ungroup() %>%
select(-level) %>%
spread(pos, code)
Output on OP data
# A tibble: 9 x 3
row `1` `2`
<int> <chr> <chr>
1 1 A00 NA
2 2 A00 A000
3 3 A00 A001
4 4 A00 A009
5 5 A01 A009
6 6 A01 A010
7 7 A010 A0100
8 8 A010 A0101
9 9 A010 A0101
Output on 93k fake data
> head(fake, 10)
[1] "55174190" "50801321" "46771275" "6480673"
[5] "20447474" "879955" "4365410" "11434009"
[9] "5002257" "9200296"
> head(fake_2, 10)
# A tibble: 10 x 3
row `1` `2`
<int> <chr> <chr>
1 1 55174190 NA
2 2 50801321 NA
3 3 46771275 NA
4 4 6480673 46771275
5 5 6480673 20447474
6 6 6480673 20447474
7 7 4365410 20447474
8 8 4365410 11434009
9 9 5002257 11434009
10 10 9200296 11434009
Related
I have two dataframes:
set.seed(1)
df1 <- data.frame(k1 = "AFD(1);Acf(2);Vgr7(2);"
,k2 = "ABC(7);BHG(46);TFG(675);")
df2 <- data.frame(site =c("AFD(1);AFD(2);", "Acf(2);", "TFG(677);",
"XX(275);", "ABC(7);", "ABC(9);")
,p1 = rnorm(6, mean = 5, sd = 2)
,p2 = rnorm(6, mean = 6.5, sd = 2))
The first dataframe is in fact a list of often very long strings, made of 'elements". Each "element" is made of a few letters/numbers, followed by a number in brackets, followed by a semicolon. In this example I only put 3 "elements" into each string, but in my real dataframe there are tens to hundreds of them.
> df1
k1 k2
1 AFD(1);Acf(2);Vgr7(2); ABC(7);BHG(46);TFG(675);
The second dataframe shares some of the "elements" with df1. Its first column, called site, contains some (not all) "elements" from the first dataframe, sometimes the "element" forms the whole string, and sometimes is a part of a longer string:
> df2
site p1 p2
1 AFD(1);AFD(2); 4.043700 3.745881
2 Acf(2); 5.835883 5.670011
3 TFG(677); 7.717359 5.711420
4 XX(275); 4.794425 6.381373
5 ABC(7); 5.775343 8.700051
6 ABC(9); 4.892390 8.026351
I would like to filter the whole df2 using df2$site and each k column from df1 (there are many K columns, not all of them contain k in the names).
The easiest way to explain this is to show how the desired output would look like.
> outcome
k site p1 p2
1 k1 AFD(1);AFD(2): 4.043700 3.745881
2 k1 Acf(2); 5.835883 5.670011
3 k2 ABC(7); 5.775343 8.700051
The first column of the outcome dataframe corresponds to the column names in df1. The second column corresponds to the site column of df2 and contains only sites from df1 columns that were found in df2$sites. Other columns are from df2.
I appreciate that this question is made of two separate "problems", one grepping-related and one related to looping through df1 columns. I decided to show the task in its entirety in case there exists a solution that addresses both in one go.
FAILED SOLUTION 1
I can create a string to grep, but for each column separately:
# this replaces the semicolons with "|", but does not escape the brackets.
k1_pattern <- df1 %>%
select(k1) %>%
deframe() %>%
str_replace_all(";","|")
And then I am not sure how to use it. This (below) didn't work, maybe because I didn't escape brackets, but I am struggling with doing it:
k1_result <- df2 %>%
filter(grepl(pattern = k1_pattern, site))
But even if it did work, it would only deal with a single column from df1, and I have many, and would like to perform this operation on all df1 columns at the same time.
FAILED SOLUTION 2
I can create a list of sites to search in df2 from columns in df1:
k1_sites<- df1 %>%
select(k1) %>%
deframe() %>%
strsplit(., "[;]") %>%
unlist()
but the delimiter is lost here, and %in% cannot be used, as the match will sometimes be partial.
library(dplyr)
df2 %>%
mutate(site_list = strsplit(site, ";")) %>%
rowwise() %>%
filter(length(intersect(site_list,
unlist(strsplit(x = paste0(c(t(df1)), collapse=""),
split = ";")))) != 0) %>%
select(-site_list)
#> # A tibble: 3 x 3
#> # Rowwise:
#> site p1 p2
#> <chr> <dbl> <dbl>
#> 1 AFD(1);AFD(2); 3.75 7.47
#> 2 Acf(2); 5.37 7.98
#> 3 ABC(7); 5.66 9.52
Updated answer:
library(dplyr)
library(tidyr)
df1 %>%
rownames_to_column("id") %>%
pivot_longer(-id, names_to = "k", values_to = "site") %>%
separate_rows(site, sep = ";") %>%
filter(site != "") %>%
select(-id) -> df1_k
df2 %>%
tibble::rownames_to_column("id") %>%
separate_rows(site, sep = ";") %>%
full_join(., df1_k, by = c("site")) %>%
group_by(id) %>%
fill(k, .direction = "downup") %>%
filter(!is.na(id) & !is.na(k)) %>%
summarise(k = first(k),
site = paste0(site, collapse = ";"),
p1 = first(p1),
p2 = first(p2), .groups = "drop") %>%
select(-id)
#> # A tibble: 3 x 4
#> k site p1 p2
#> <chr> <chr> <dbl> <dbl>
#> 1 k1 AFD(1);AFD(2); 3.75 7.47
#> 2 k1 Acf(2); 5.37 7.98
#> 3 k2 ABC(7); 5.66 9.52
Here's a way going to a long format for exact matching (so no regex):
library(dplyr)
library(tidyr)
df1_long = df1 |> stack() |>
separate_rows(values, sep = ";") |>
filter(values != "")
df2 |>
mutate(id = row_number()) |>
separate_rows(site, sep = ";") |>
filter(site != "") |>
left_join(df1_long, by = c("site" = "values")) %>%
group_by(id) |>
filter(any(!is.na(ind))) %>%
summarize(
site = paste(site, collapse = ";"),
across(-site, \(x) first(na.omit(x)))
)
# # A tibble: 3 × 5
# id site p1 p2 ind
# <int> <chr> <dbl> <dbl> <fct>
# 1 1 AFD(1);AFD(2) 3.75 7.47 k1
# 2 2 Acf(2) 5.37 7.98 k1
# 3 5 ABC(7) 5.66 9.52 k2
this is not a very good title for the question. I want to sum across certain columns in a data frame for each group, excluding one column for each of my groups. A simple example would be as follows:
df <- tibble(group_name = c("A", "B","C"), mean_A = c(1,2,3), mean_B = c(2,3,4), mean_C=c(3,4,5))
df %>% group_by(group_name) %>% mutate(m1 = sum(across(contains("mean"))))
This creates column m1, which is the sum across mean_a, mean_b, mean_c for each group. What I want to do is exclude mean_a for group a, mean_b for b and mean_c for c. The following does not work though (not surprisingly).
df %>% group_by(group_name) %>% mutate(m1 = sum(across(c(contains("mean") & !contains(group_name)))))
Do you have an idea how I could do this? My original data contains many more groups, so would be hard to do by hand.
Edit: I have tried the following way which solves it in a rudimentary fashion, but something (?grepl maybe) seems to not work great here and I get the wrong result.
df %>% pivot_longer(!group_name) %>% mutate(value2 = case_when(grepl(group_name, name) ~ 0, TRUE ~ value)) %>% group_by(group_name) %>% summarise(m1 = sum(value2))
Edit2: Found out what's wrong with the above, and below works, but still a lot of warnings so I recommend people to follow TarJae's response below
df %>% pivot_longer(!group_name) %>% group_by(group_name) %>% mutate(value2 = case_when(grepl(group_name, name) ~ 0, TRUE ~ value)) %>% group_by(group_name) %>% summarise(m1 = sum(value2))
Here is another option where you can just use group_name directly with the tidyselect helpers:
df %>%
rowwise() %>%
mutate(m1 = rowSums(select(across(starts_with("mean")), -ends_with(group_name)))) %>%
ungroup()
Output
group_name mean_A mean_B mean_C m1
<chr> <dbl> <dbl> <dbl> <dbl>
1 A 1 2 3 5
2 B 2 3 4 6
3 C 3 4 5 7
How it works
The row-wise output of across is a 1-row tibble containing only the variables that start with "mean".
select unselects the subset of the variables from output by across that end with the value from group_name.
At this point you are left with a 1 x 2 tibble, which is then summed using rowSums.
Here is one way how we could do it:
We create a helper column to match column names
We set value of mean column to zeor if column names matches helper name.
Then we use transmute with select to calculate rowSums
Finally we cbind column m1 to df:
library(dplyr)
df %>%
mutate(helper = paste0("mean_", group_name)) %>%
mutate(across(starts_with("mean"), ~ifelse(cur_column()==helper, 0, .))) %>%
transmute(m1 = select(., contains("mean")) %>%
rowSums()) %>%
cbind(df)
m1 group_name mean_a mean_b mean_c
1 5 a 1 2 3
2 6 b 2 3 4
3 7 c 3 4 5
I would like to calculate duration of state using rle() on grouped data. Here is test data frame:
DF <- read.table(text="Time,x,y,sugar,state,ID
0,31,21,0.2,0,L0
1,31,21,0.65,0,L0
2,31,21,1.0,0,L0
3,31,21,1.5,1,L0
4,31,21,1.91,1,L0
5,31,21,2.3,1,L0
6,31,21,2.75,0,L0
7,31,21,3.14,0,L0
8,31,22,3.0,2,L0
9,31,22,3.47,1,L0
10,31,22,3.930,0,L0
0,37,1,0.2,0,L1
1,37,1,0.65,0,L1
2,37,1,1.089,0,L1
3,37,1,1.5198,0,L1
4,36,1,1.4197,2,L1
5,36,1,1.869,0,L1
6,36,1,2.3096,0,L1
7,36,1,2.738,0,L1
8,36,1,3.16,0,L1
9,36,1,3.5703,0,L1
10,36,1,3.970,0,L1
", header = TRUE, sep =",")
I want to know the average length for state == 1, grouped by ID. I have created a function inspired by: https://www.reddit.com/r/rstats/comments/brpzo9/tidyverse_groupby_and_rle/
to calculate the rle average portion:
rle_mean_lengths = function(x, value) {
r = rle(x)
cond = r$values == value
data.frame(count = sum(cond), avg_length = mean(r$lengths[cond]))
}
And then I add in the grouping aspect:
DF %>% group_by(ID) %>% do(rle_mean_lengths(DF$state,1))
However, the values that are generated are incorrect:
ID
count
avg_length
1 L0
2
2
2 L1
2
2
L0 is correct, L1 has no instances of state == 1 so the average should be zero or NA.
I isolated the problem in terms of breaking it down into just summarize:
DF %>% group_by(ID) %>% summarize_at(vars(state),list(name=mean)) # This works but if I use summarize it gives me weird values again.
How do I do the equivalent summarize_at() for do()? Or is there another fix? Thanks
As it is a data.frame column, we may need to unnest afterwards
library(dplyr)
library(tidyr)
DF %>%
group_by(ID) %>%
summarise(new = list(rle_mean_lengths(state, 1)), .groups = "drop") %>%
unnest(new)
Or remove the list and unpack
DF %>%
group_by(ID) %>%
summarise(new = rle_mean_lengths(state, 1), .groups = "drop") %>%
unpack(new)
# A tibble: 2 × 3
ID count avg_length
<chr> <int> <dbl>
1 L0 2 2
2 L1 0 NaN
In the OP's do code, the column that should be extracted should be not from the whole data, but from the data coming fromt the lhs i.e. . (Note that do is kind of deprecated. So it may be better to make use of the summarise with unnest/unpack
DF %>%
group_by(ID) %>%
do(rle_mean_lengths(.$state,1))
# A tibble: 2 × 3
# Groups: ID [2]
ID count avg_length
<chr> <int> <dbl>
1 L0 2 2
2 L1 0 NaN
-- Small edit made to test data. Columns are no longer grouped by round, but instead grouped by Team as is the case in the real dataset.
I have tried a variety of methods for creating multiple new columns, while minimising the repetition in the code. My initial and successful method requires excessive copy and pasting, but I would like to minimise this as much as possible. Below is example data for the problem:
df <- tribble(~R1TeamX, ~R2TeamX,~R3TeamX, ~R1TeamY,~R2TeamY, ~R3TeamY,
10, 11, 12, 15, 19, 20,
11, 13, 14, 25, 18, 15)
This example data is for three rounds with the scores for both team X and team Y. I am looking to create additional columns, finding the difference between the scores of Team X and Y. The real dataset has upwards of 30 rounds.
My initial solution used mutate and works as follows:
df <- df %>%
mutate(R1Diff = R1TeamX - R1TeamY,
R2Diff = R2TeamX - R2TeamY,
R3Diff = R3TeamX - R3TeamY)
While this does the job, it is not scaleable. I have attempted to reduce this down to less code using str_c & mutate, but cannot identify the correct looping method to make this work for several lines of code. Below is my attempt at standardising the code so far:
teamx <- str_c("R", 1:3, "TeamX")
teamy <- str_c("R", 1:3, "TeamY")
round_diff <- str_c("R", 1:3, "Diff")
df <- df %>%
mutate(!!round_diff[1] := UQ(parse_quo(teamx[1], global_env())) - UQ(parse_quo(teamy[1], global_env())),
!!round_diff[2] := UQ(parse_quo(teamx[2], global_env())) - UQ(parse_quo(teamy[2], global_env())),
!!round_diff[3] := UQ(parse_quo(teamx[3], global_env())) - UQ(parse_quo(teamy[3], global_env())))
While additional code is required, this standardises my input to some degree reducing some of the leg work, but I know there must be some way to reduce this into a single line. I have explored mutate_at and for loops to no avail. I suspect this problem could also be tackled with purrr::map, but I do have enough ability in this area to identify the correct approach.
Any help would be greatly appreciated.
Whilst it's possible to do this in dplyr and tidyr functions, remember you still have some useful base R options open to you. This method uses lapply and makes the assumption that your columns are alternating between team X and team Y
seq(length(df)/2) %>%
lapply(function(x) df[[x]] - df[[x + 1]]) %>%
as.data.frame() %>%
setNames(paste0("R", seq(length(df)/2), "Diff")) %>%
cbind(df,.)
#> R1TeamX R1TeamY R2TeamX R2TeamY R3TeamX R3TeamY R1Diff R2Diff R3Diff
#> 1 10 11 12 15 19 20 -1 -1 -3
#> 2 11 13 14 25 18 15 -2 -1 -11
try to do it this way
library(tidyverse)
df %>%
mutate(id = row_number()) %>%
pivot_longer(
-id,
names_to = c("set", ".value"),
names_pattern = "(R\\d+Team)(X|Y)"
) %>%
mutate(Diff = X - Y) %>%
pivot_longer(-c(id, set)) %>%
pivot_wider(id, names_from = c(set, name), values_from = value, names_sep = "")
Here's a solution that I believe is robust with respect to the number of rounds, the number of opponents of Team X and the order in which the results are stored.
First, make the data tidy: remove information about Teams and Rounds from column names.
newDF <- df %>%
mutate(id = row_number()) %>%
pivot_longer(
-id,
names_to = c("Round", "Team"),
names_pattern = "R(\\d+)Team(X|Y)",
values_to="Score"
)
Now calculate the differences in scores
newDF %>%
# Calculate difference in scores
mutate(Team=ifelse(Team == "X", Team, "Opponent")) %>%
pivot_wider(values_from=Score, names_from=Team) %>%
mutate(Diff=X - Opponent) %>%
select(-Opponent) %>%
# Bring in identity of oponent
left_join(
newDF %>%
filter(Team != "X") %>%
select(-Score) %>%
rename(Opposition=Team),
by=c("id", "Round")
)
Giving
# A tibble: 6 x 5
id Round X Diff Opposition
<int> <chr> <dbl> <dbl> <chr>
1 1 1 10 -5 Y
2 1 2 11 -8 Y
3 1 3 12 -8 Y
4 2 1 11 -14 Y
5 2 2 13 -5 Y
6 2 3 14 -1 Y
based on OP's revised input data.
Survey data often contains multiple choice columns with entries separated by commas, for instance:
library("tidyverse")
my_survey <- tibble(
id = 1:5,
question.1 = 1:5,
question.2 = c("Bus", "Bus, Walk, Cycle", "Cycle", "Bus, Cycle", "Walk")
)
It's desirable to have a function multiple_choice_tally that will tally the unique responses for the question:
my_survey %>%
multiple_choice_tally(question = question.2)
### OUTPUT:
# A tibble: 3 x 2
response count
<chr> <int>
1 Bus 3
2 Walk 2
3 Cycle 3
What is the most efficient and flexible way to construct multiple_choice_tally, without any hard coding.
We can use separate_rows from the tidyr package to expand the contents in question.2. Since you are using tidyverse, tidyr has been already loaded with library("tidyverse") and we don't have to load it again. my_survey2 is the final output.
my_survey2 <- my_survey %>%
separate_rows(question.2) %>%
count(question.2) %>%
rename(response = question.2, count = n)
my_survey2
# A tibble: 3 × 2
response count
<chr> <int>
1 Bus 3
2 Cycle 3
3 Walk 2
Update: Design a Function
We can convert the above code into a function as follows.
multiple_choice_tally <- function(survey.data, question){
question <- enquo(question)
survey.data2 <- survey.data %>%
separate_rows(!!question) %>%
count(!!question) %>%
setNames(., c("response", "count"))
return(survey.data2)
}
my_survey %>%
multiple_choice_tally(question = question.2)
# A tibble: 3 x 2
response count
<chr> <int>
1 Bus 3
2 Cycle 3
3 Walk 2
My current solution for this problem is as follows:
multiple_choice_tally <- function(survey.data, question) {
## Require a sym for the RHS of !!response := if_else
question_as_quo <- enquo(question)
question_as_string <- quo_name(question_as_quo)
target_question <- rlang::sym(question_as_string)
## Collate unique responses to the question
unique_responses <- survey.data %>%
select(!!target_question) %>%
na.omit() %>%
.[[1]] %>%
strsplit(",") %>%
unlist() %>%
trimws() %>%
unique()
## Extract responses to question
question_tally <- survey.data %>%
select(!!target_question) %>%
na.omit()
## Iteratively create a column for each unique response
invisible(lapply(unique_responses,
function(response) {
question_tally <<- question_tally %>%
mutate(!!response := if_else(str_detect(!!target_question, response), TRUE, FALSE))
}))
## Gather into tidy form
question_tally %>%
summarise_if(is.logical, funs(sum)) %>%
gather(response, value = count)
}
Which can then be used as follows:
library("tidyverse")
library("rlang")
library("stringr")
my_survey <- tibble(
id = 1:5,
question.1 = 1:5,
question.2 = c("Bus", "Bus, Walk, Cycle", "Cycle", "Bus, Cycle", "Walk")
)
my_survey %>%
multiple_choice_tally(question = question.2)
### OUTPUT:
# A tibble: 3 x 2
response count
<chr> <int>
1 Bus 3
2 Walk 2
3 Cycle 3