R - 提取与字符相关的数字

Question

我正在尝试从化学式中提取碳、氢和氧的数量。我之前发现了一些我一直在尝试使用的代码。问题是代码仅在化学式中包含超过 1 个元素时才有效。

V <- DATA # example: CH4O, H2O, C10H18O2
# V is a data.frame

C1 <- as.integer(sub("(?i).*?C:?\\s*(\\d+).*", "\\1", V))
# NA NA 10
H1 <- as.integer(sub("(?i).*?H:?\\s*(\\d+).*", "\\1", V))
# 4 2 18
O1 <- as.integer(sub("(?i).*?O:?\\s*(\\d+).*", "\\1", V))
# NA NA 2

我正在使用

is.na(C1) <- 1

将 NA 更改为 1，然后手动更改 0 值。是否有更有效的代码可以用来获取化学式中元素的正确计数（特别是在值为 0 或 1 并导致 NA 结果的情况下）。如果您需要更多信息或者我是否应该更改某些格式，请告诉我。

编辑： 所需的值是在没有 NA 的情况下获得所有正确的计数，并尽可能手动将值更改为 0。

C1
# 1 0 10
H1
# 4 2 18
O1
# 1 1 2

编辑2： 这是我正在导入的实际数据的示例

Meas. m/z   #   Ion Formula Score   m/z err [mDa]   err [ppm]   mSigma  rdb e¯ Conf Adduct  
84.080700   1   C5H10N  n.a.    84.080776   0.1 0.9 n.a.    2.0 even        
89.060100   1   C4H9O2  n.a.    89.059706   -0.4    -4.4    n.a.    1.0 even        
131.987800  1   C2H4N3P2    n.a.    131.987498  -0.3    -2.3    n.a.    6.0 even        
135.081100  1   C9H11O  n.a.    135.080441  -0.7    -4.9    n.a.    5.0 even        
135.117500  1   C10H15  n.a.    135.116827  -0.7    -5.0    n.a.    4.0 even        
136.061700  1   C5H6N5  n.a.    136.061772  0.1 0.5 n.a.    6.0 even        

在最初的问题中，我刚刚将V 列为来自公式向量，但我实际上拥有的是一个带有其他信息的data.frame，我在执行计算时使用V[,3] 来获取感兴趣的列.

Answer 1

这里有一个替代方案：

vec <- c("CH4O", "H2O", "C10H18O2", "C2H4N3P2")

molecules <- regmatches(vec, gregexpr("\\b[A-Z][a-z]*\\d*", vec))
molecules <- lapply(molecules, function(a) paste0(a, ifelse(grepl("[^0-9]$", a), "1", "")))

atomcounts <- lapply(molecules, function(mol) setNames(as.integer(gsub("\\D", "", mol)), gsub("\\d", "", mol)))

atoms <- unique(unlist(sapply(atomcounts, names)))
atoms <- sapply(atoms, function(atom) sapply(atomcounts, function(a) if (atom %in% names(a)) a[atom] else 0))
rownames(atoms) <- vec
atoms
#           C  H O N P
# CH4O      1  4 1 0 0
# H2O       0  2 1 0 0
# C10H18O2 10 18 2 0 0
# C2H4N3P2  2  4 0 3 2

Answer 2

也许不是我写过的最优雅的代码，但对于给定的化学式，这将返回标记向量中每个元素的计数。中间步骤counts.per.equation 返回每​​个方程中每个元素的计数。

library(stringr)

extract <- str_extract_all(c('CH4O', 'H2O', 'C10H18O2'), '\\D\\d*')

counts.per.equation <- lapply(extract, function(x) {
  elements <- str_extract_all(x, '\\D+')
  counts <- str_extract_all(x, '\\d+', simplify = T)
  counts[counts == ''] <- 1
  counts <- as.numeric(counts)
  names(counts) <- elements
  return(counts)
})

total.counts <- Reduce(function(x, y) {
  names <- union(names(x), names(y))
  counts <- sapply(names, function(z) sum(x[z], y[z], na.rm = T))
  names(counts) <- names
  return(counts)
} , counts.per.equation)

 C  H  O 
11 24  4 

Answer 3

我们可以将base R 方法与strsplit 和xtabs 一起使用

out <- do.call(rbind, Map(cbind,
   lapply(strsplit(gsub("(?<=[A-Z])(?![0-9])", "1", vec,
   perl = TRUE), "(?<=[A-Z])(?=[0-9])|(?<=[0-9])(?=[A-Z])", 
    perl = TRUE), function(x) data.frame(key = x[c(TRUE, FALSE)], 
       value = as.numeric(x[c(FALSE, TRUE)]))), grp = vec))

xtabs(value ~ grp + key, out)
#         key
#grp         C  H  O  N  P
#  CH4O      1  4  1  0  0
#  H2O       0  2  1  0  0
#  C10H18O2 10 18  2  0  0
#  C2H4N3P2  2  4  0  3  2

---

或者tidyverse

library(stringr)
library(dplyr)
library(tidyr)
tibble(col1 = vec,
       col2 = str_replace_all(col1, "(?<=[A-Z])(?![0-9])", "1")) %>%
       separate_rows(col2, sep= "(?<=[A-Z])(?=[0-9])|(?<=[0-9])(?=[A-Z])") %>%
       group_by(col1) %>% 
       summarise(key = list(col2[c(TRUE, FALSE)]),
            val = list(col2[c(FALSE, TRUE)])) %>%
       unnest(c(key, val)) %>% 
   pivot_wider(names_from = key, values_from = val, values_fill = list(val = 0))
# A tibble: 4 x 6
#  col1     C     H     O     N     P    
#  <chr>    <chr> <chr> <chr> <chr> <chr>
#1 C10H18O2 10    18    2     0     0    
#2 C2H4N3P2 2     4     0     3     2    
#3 CH4O     1     4     1     0     0    
#4 H2O      0     2     1     0     0    

数据

vec <- c("CH4O", "H2O", "C10H18O2", "C2H4N3P2")