Mean columns with almost the same name

Question

I have a data frame containing only one row with named columns. The data frame looks somewhat like this:

  poms_tat1 poms_tat2 poms_tat3      tens1      tens2      tens3 ...
1 0.3708821 0.4915922 0.3958195 -0.1139606 -0.1462545 -0.4411494 ...

I need to calculate the mean of all the columns with similar names. The result should look somewhat like this:

  poms_tat    tens      ...
1 0.4194551  -0.2337881667 ...

My first approach was to use a for loop and a nested while loop to find the indices of the relevant columns and then mean those, but unfortunately I couldn't make it work.

I also found this stackoverflow post which seemed promising but the agrep function seems to match columns in my data frame that should not be matched. I wasn't able to fix that using the max.distance parameter. For example it matches "threat1-3" with "reat1-3". I know those variable names are terrible, but unfortunately that's what I have to work with. What makes this even more complicated is that the number of columns in each category isn't always 3.

I hope I was able to articulate my problem well enough. Thank you.

Edit: Here is a reproducible piece of data:

structure(list(poms_tat1 = 0.370882118644872, poms_tat2 = 0.491592168116328, 
    poms_tat3 = 0.395819547420188, tens1 = -0.113960576459638, 
    tens2 = -0.146254484825426, tens3 = -0.44114940169153, bat_ratio1 = 1, 
    isi1 = 0.0944068640061701, isi2 = 0.597785124823513, isi3 = 0.676617801589949, 
    isi4 = 0.143940321201716, sleepqual = 0.378902118888194, 
    se1 = 0.393610946830482, se2 = 0.0991899501072693, se3 = 0.501745206004254, 
    challenge1 = 0.417855447018672, challenge2 = 0.393610946830482, 
    challenge3 = 0.417855447018672, threat1 = -0.13014390184863, 
    threat2 = -0.34027852368936, threat3 = -0.269679944985297, 
    reat1 = 0.565825152115738, reat2 = 0.571605347479646, reat3 = 0.497468338163091, 
    reat4 = 0.484881137876427, reat5 = 0.494727444918154, selfman1 = 0.389249472080761, 
    selfman2 = 0.40609787800914, selfman3 = 0.418121005003545, 
    selfman4 = 0.467099366496914, selfman5 = 0.205356548067582, 
    selfman6 = 0.464385939554693, selfman7 = 0.379071252751718, 
    eli1 = 0.250872603002127, eli2 = 0, eli3 = 0.265908011739155), row.names = 1L, class = "data.frame")

Answer 1

We could use split.default to split based on the substring of column namesinto a list and then loop over the list with sapply, get the rowMeans in base R

sapply(split.default(df1, sub("\\d+$", "", names(df1))), rowMeans, na.rm = TRUE)

Answer 2

Here's one with tidyr. I only just saw the acceptance of Baraliuh's answer, so my answer here is more for the sake of closure.

library(tidyr)

my_summary <- as.data.frame(sapply(
  X = pivot_longer(
    data = df,
    # Desired columns (all) to summarize.
    cols = everything(),
    # Take each group of columns, which share a common name before different numeric
    # suffixes, and pivot them into multiple rows under a common column by that name.
    names_to = c(
      ".value",
      # Discard anything after the prefix.
      NA
    ),
    # Identify the (optional) numeric suffix.
    names_sep = "\\d*$"
  ),
  # Take the mean of each column; ignore missing values.
  FUN = mean, na.rm = TRUE,
  # Keep as a list, to convert into a data.frame.
  simplify = FALSE
))

In contrast to some alternatives, I do believe my consolidated use of pivoting makes the process cleaner, and the output does achieve the precise format you desire.

   poms_tat       tens bat_ratio       isi sleepqual        se challenge     threat      reat   selfman       eli
1 0.4194313 -0.2337882         1 0.3781875 0.3789021 0.3315154 0.4097739 -0.2467008 0.5229015 0.3899116 0.1722602

With dplyr involved too, the summarization (into a tibble) is even cleaner:

library(tidyr)
library(dplyr)

my_summary <- df %>%
  pivot_longer(
    cols = everything(),
    names_to = c(".value", NA),
    names_sep = "\\d*$"
  ) %>%
  summarize(across(everything(), mean, na.rm = TRUE))

Answer 3

You could either do this by tidyr::pivot_longer, dplyr::mutate, stringr::str_remove, dplyr::group_by, and dplyr::summarise. This would be done like this:

ex_data <- structure(list(poms_tat1 = 0.370882118644872, poms_tat2 = 0.491592168116328, 
               poms_tat3 = 0.395819547420188, tens1 = -0.113960576459638, 
               tens2 = -0.146254484825426, tens3 = -0.44114940169153, bat_ratio1 = 1, 
               isi1 = 0.0944068640061701, isi2 = 0.597785124823513, isi3 = 0.676617801589949, 
               isi4 = 0.143940321201716, sleepqual = 0.378902118888194, 
               se1 = 0.393610946830482, se2 = 0.0991899501072693, se3 = 0.501745206004254, 
               challenge1 = 0.417855447018672, challenge2 = 0.393610946830482, 
               challenge3 = 0.417855447018672, threat1 = -0.13014390184863, 
               threat2 = -0.34027852368936, threat3 = -0.269679944985297, 
               reat1 = 0.565825152115738, reat2 = 0.571605347479646, reat3 = 0.497468338163091, 
               reat4 = 0.484881137876427, reat5 = 0.494727444918154, selfman1 = 0.389249472080761, 
               selfman2 = 0.40609787800914, selfman3 = 0.418121005003545, 
               selfman4 = 0.467099366496914, selfman5 = 0.205356548067582, 
               selfman6 = 0.464385939554693, selfman7 = 0.379071252751718, 
               eli1 = 0.250872603002127, eli2 = 0, eli3 = 0.265908011739155), row.names = 1L, class = "data.frame")
ex_data %>% 
    tidyr::pivot_longer(everything()) %>% 
    dplyr::mutate(
        name = stringr::str_remove(name, '[0-9]$')
    ) %>% 
    dplyr::group_by(name) %>% 
    dplyr::summarise(
        mean = mean(value)
    )
# A tibble: 11 x 2
   name        mean
   <chr>      <dbl>
 1 bat_ratio  1    
 2 challenge  0.410
 3 eli        0.172
 4 isi        0.378
 5 poms_tat   0.419
 6 reat       0.523
 7 se         0.332
 8 selfman    0.390
 9 sleepqual  0.379
10 tens      -0.234
11 threat    -0.247

Alternatively, you could use split.default, together with stringr::str_remove, purrr::map, unlist, purrr::map_dbl, and tibble::enframe as follows:

ex_data %>% 
    split.default(stringr::str_remove(names(.), '[0-9]$')) %>% 
    purrr::map(unlist) %>% 
    purrr::map_dbl(mean) %>% 
    tibble::enframe()
# A tibble: 11 x 2
   name       value
   <chr>      <dbl>
 1 bat_ratio  1    
 2 challenge  0.410
 3 eli        0.172
 4 isi        0.378
 5 poms_tat   0.419
 6 reat       0.523
 7 se         0.332
 8 selfman    0.390
 9 sleepqual  0.379
10 tens      -0.234
11 threat    -0.247

Answer 4

You can also use the following solution:

library(dplyr)
library(tidyr)

df %>%
  pivot_longer(everything()) %>% 
  group_by(grp = sub("\\d+$", "", name)) %>%
  summarise(Avg = mean(value, na.rm = TRUE))

   grp          Avg
   <chr>      <dbl>
 1 bat_ratio  1    
 2 challenge  0.410
 3 eli        0.172
 4 isi        0.378
 5 poms_tat   0.419
 6 reat       0.523
 7 se         0.332
 8 selfman    0.390
 9 sleepqual  0.379
10 tens      -0.234
11 threat    -0.247