R 解析字符串值中的层次结构

我试图从单个字符向量创建一个边列表。我要处理的列表长度超过93k个元素，但作为示例，我将提供一个小摘录

chracter字符串是ICD10代码层次结构的一部分，并且该字符串中存在父子关系。这意味着单个字符串A0101的父级为A010

它看起来是这样的：

A00 A000 A001 A009 A01 A010 A0100 A0101 A02 A03 等 我的向量不包含除字符串以外的任何其他数据，但我基本上需要转换

dat <- c("A00", "A000", "A001", "A009", "A01", "A010", "A0100", "A0101", "A02")

---

假设ICD10中节点名称的长度完全定义了顺序，较短的节点是父节点，这里有一种方法可以将每个节点与其直接父节点（如果可用）连接起来

虽然我认为这里的逻辑清晰可辨，但我很想知道一个更精简的解决方案会是什么样子

# 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)

输出操作数据

# 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

93k伪数据输出

> 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

---

我相当肯定有更有效的方法来实现这一点，但这段代码应该从icd.data包下载icd10cm数据。使用icd软件包中的儿童检测系统，然后广泛使用tidyverse返回边缘列表。我必须有点创造性地连接层次结构的顶部，因为它们不包括ICD10数据的章节和子章节作为单独的2位或1位代码

基本上，子章节变成2位代码，章节变成1位代码，然后有一个根节点连接顶部的所有内容

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)

---

如果有人有任何技巧或方法来提高这段代码的效率，我洗耳恭听。眼睛？

Hi Trip，将列表转换为边缘列表的逻辑是什么？起初我认为A00-A000是第一条边，A001-A009是第二条边，A01-A010是第三条边，但示例输出看起来不同。谢谢：我的想法可能有误，但我正在尝试创建一种分层数据格式，然后可以使用它来测量树上各个节点之间的距离。这些值代表ICD10诊断值，我试图测量诊断代码集的相似性。哦，这真的很有趣。我想下面的人明白了！不幸的是，对字符长度的假设是不够的。较短的代码是附加了chracter的代码的父节点，但根字符串必须匹配。。。换句话说，A0101是A010的孩子，而不是A011，否则代码运行得很好，但这可能是我没有正确解释ICD10的层次结构的错误。我对这项工作的目标很感兴趣。也许我们可以连接？Hi@JackWasey…目标是利用此方法计算icd本体中的距离。这里有一些关于根据患者的诊断包计算患者相似性的已发表论文。这很有趣，但我在这方面寻找了没有使用共病的出版物。请给我指一些好吗？

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)