R-将每个字母因子的字母数字字符观测值拆分为列，并为每个观测值分配数值

Question

I am not quite sure how to best word the title for what I want to do.

I have a data frame that looks like this:

 ID = c(1, 2, 3, 4, 5, 6, 7)
 observation = c("a2", NA, "b3", "c5", NA, "b", "a3")
 df <- data.frame(cbind(ID, observation))

 df

  ID observation
1  1          a2
2  2        <NA>
3  3          b3
4  4          c5
5  5        <NA>
6  6           b
7  7          a3

My desired output is a data frame that splits observations by numbers and letters, with a new column for each unique letter where each row contains the associated observation number for that letter.

The desired output should look like this:

desired_df <- data.frame(cbind(ID, a = c(2, NA, 0, 0, 0 , 0, 3), 
                                   b = c(0, NA, 3, 0, 0, 0, 0),
                                   c = c(0, NA, 0, 5, 0, 0, 0)))
desired_df

  ID  a  b  c
1  1  2  0  0
2  2 NA NA NA
3  3  0  3  0
4  4  0  0  5
5  5  0 NA NA
6  6  0  0  0
7  7  3  0  0

I've tried approaching this by splitting observations into letters and numbers with a regular expression and saving the result into a new column:

library(stringr)
char <- unlist(str_replace_all(observation, "[[:digit:]]", ""))
num <- unlist(str_extract(observation, "[[:digit:]]"))
df_new <- cbind(ID, char, num)
df_new

  ID char  num
1  1    a    2
2  2 <NA> <NA>
3  3    b    3
4  4    c    5
5  5 <NA> <NA>
6  6    b <NA>
7  7    a    3

Then tried converting char to a factors to a binary form based on the answer to this SO Question

df_new <- data.frame(cbind(df, sapply(levels(as.factor((char))), 
function(x) as.integer(x == char))))

  ID char  num  a  b  c
1  1    a    2  1  0  0
2  2 <NA> <NA> NA NA NA
3  3    b    3  0  1  0
4  4    c    5  0  0  1
5  5 <NA> <NA> NA NA NA
6  6    b <NA>  0  1  0
7  7    a    3  1  0  0

I then tried to replace each 1 observation with the the corresponding value in df_new1$num for that row, based on the answer to this SO question :

df_new2 <- data.frame(with(df_new1, ifelse(df_new1 == 1, df_new1$num, 0)))

df_new2
  ID char num  a  b  c
1  1    0   0  1  0  0
2  0   NA  NA NA NA NA
3  0    0   0  0  2  0
4  0    0   0  0  0  3
5  0   NA  NA NA NA NA
6  0    0  NA  0 NA  0
7  0    0   0  2  0  0

Which outputs the wrong result. I've been struggling to figure this out. I am OK with all non 1 values being replaced with 0 as long as the values in columns a, b, c are correct.

I'm not sure if splitting letters and numbers into separate columns, and trying to replace binary observations for letters as factors is even the best approach for trying to solve my original problem and am open to any approach that works.

My real data frame is generated by a script that extracts patterns from .txt files, where the alphanumeric observations vary from file to file. I need something that will work for any unique letters that get assigned to the char column.

I appreciate any advice or help on figuring this out as I am a novice to R. I'm still getting familiar with SO etiquette and would appreciate any comments on how to improve the question and/or reproducible example.

Answer 1

You can use extract from tidyr to split observation into var and value column, then use spread to reshape the table. Note that <NA> is now its own column because of the NA values in ID == 2 . A select gets rid of that column:

library(dplyr)
library(tidyr)

df %>%
  extract(observation, c("var", "value"), regex = "([a-z])?(\\d)?") %>%
  spread(var, value) %>%
  select(-`<NA>`)

Result:

  ID    a    b    c
1  1    2 <NA> <NA>
2  2 <NA> <NA> <NA>
3  3 <NA>    3 <NA>
4  4 <NA> <NA>    5
5  5 <NA> <NA> <NA>
6  6    3 <NA> <NA>

Answer 2

Since you mentioned that non-digit values can be 0 or NA

library(tidyverse)
df %>%
  nest(-ID) %>%
  mutate(data = map(data, ~data.frame(key = gsub("\\d", "", unlist(.x)), val = gsub("\\D", "", unlist(.x))))) %>%
  unnest() %>%
  spread(key, val, fill = 0) %>%
  select(-ncol(.)) %>%
  replace(.=="", 0)

  # ID    a     b     c    
  # <fct> <chr> <chr> <chr>
# 1 1     2     0     0    
# 2 2     0     0     0    
# 3 3     0     3     0    
# 4 4     0     0     5    
# 5 5     0     0     0    
# 6 6     3     0     0    
# There were 14 warnings (use warnings() to see them)