R：如何循环从数据框中选择基于名称的变量，并为每个变量创建一个包含第一个列均值的新变量？

Question

I have a dataset containing a number of numeric variables whose names all start with "Ranking". For each of these variables, I want to add another variable to the dataset that contains the column mean of the first variable.

So the data look something like this:

| Ranking_blah | Ranking_bleh | 

| --------     | ----------   |

| 1            | 0            |

| 0            | 1            |

| NA           | 0.5          |

and what I want is:

| Ranking_blah | Ranking_bleh | Ranking_blah_mean | Ranking_bleh_mean |

| --------     | ----------   |----------------   |----------------|

| 1            | 0            | 0                 | 0.5            |

| -1           | 1            | 0                 | 0.5            |

| NA           | 0.5          | 0                 | 0.5    

(I am aware this way the mean variables have the same values in all rows, respectively - I need this because the data will be reshaped later)

What I've tried so far:

#getting a list of all ranking variables I want to create a new mean variable from

ranking_variables = names(data)[grepl("Ranking", names(data))]

#creating a new variable for each base variable in the list and setting it to the mean of the respective base variable

data[paste0(ranking_variables, "_mean")] <- do.call(cbind, lapply(data[ranking_variables], function(x) mean(x, na.rm = TRUE)))

The second part is not working, though, it only yields NA values. What am I doing wrong?

Answer 1

An alternative approach is to use dplyr 's across :

dat |>
    mutate(across(starts_with("Ranking"), ~ mean(., na.rm = TRUE), .names = "{.col}_mean"))

Output:

# A tibble: 3 × 4
  Ranking_blah Ranking_bleh Ranking_blah_mean Ranking_bleh_mean
         <dbl>        <dbl>             <dbl>             <dbl>
1            1          0                   0               0.5
2           -1          1                   0               0.5
3           NA          0.5                 0               0.5

Data:

tibble(Ranking_blah = c(1,-1,NA), Ranking_bleh = c(0,1,0.5))

Answer 2

The across approach is fine, here is another one:

There is less struggle with tidy data, because R makes it easier to compute across rows than across columns.

Tidy data means that every observation has its own row and every variable its own column. Columns are designed to represent variables. I think the "Ranking…" columns are not distinct variables, but different observations of the variable "type". To fix this, we can use tidyr .
See this chapter of R for data science.

library(tidyverse)

data <- data.frame(Ranking_blah = c(1,-1,NA), Ranking_bleh = c(0,1,0.5))
data$id <- c(1:nrow(data))

pivot_longer(data,1:2,names_to = "type") %>%
  group_by(type) %>%
  mutate(mean = mean(value, na.rm = TRUE)) %>%
  ungroup()

# A tibble: 6 × 4
     id type         value  mean
  <int> <chr>        <dbl> <dbl>
1     1 Ranking_blah   1     0  
2     1 Ranking_bleh   0     0.5
3     2 Ranking_blah  -1     0  
4     2 Ranking_bleh   1     0.5
5     3 Ranking_blah  NA     0  
6     3 Ranking_bleh   0.5   0.5

This data is less human readable, but more R friendly.