How can I write a function to create multiple tables from dataframe columns?

Question

I would like to create contingency tables and run chisq.test() etc. for multiple items in a dataframe.

Various attempts have resulted in 'Error in table(y$x, y$q2) : all arguments must have the same length'.

I think the example below focuses on my central problem, though ultimately I'd write a more complex function. I'd be interested in solutions to my specific function or to my overall approach. Thanks!

my_df <- structure(list(q1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), 
                              .Label = c("Choice1", "Choice2"), 
                              class = "factor"), 
                        q2 = structure(c(1L, 1L, 4L, 5L, 4L, 1L, 1L, 4L), 
                              .Label = c("Agree", "Disagree","N/A or No Opinion", 
                                         "Strongly Agree", "Strongly Disagree"), 
                              class = "factor"),
                        q3 = structure(c(1L, 4L, 1L, 4L, 1L, 4L, 4L, 4L), 
                              .Label = c("Agree", "Disagree","N/A or No Opinion", 
                                    "Strongly Agree", "Strongly Disagree"), 
                              class = "factor")),
                   row.names = c(NA, -8L), 
                   class = c("tbl_df", "tbl", "data.frame"))

my_fn <- function(x, y) {
 table(y$x, y$`q2`)

}

my_fn(names(my_df)[1], my_df)
#Error in table(y$x, y$q2) : all arguments must have the same length

lapply(names(my_df), my_fn, my_df)
#Error in table(y$x, y$q2) : all arguments must have the same length

Duck · Accepted Answer · 2020-09-15 16:31:00Z

Try this. The issue might be connected to the use of $ for variables. In case you want to use names, it is better if you use [[]] so that the strings for names can be understood by the function. Here the code, with slight changes to your function. I added some examples:

#Function
my_fn <- function(x, y) {
  table(y[[x]], y[['q2']])
}
#Code
my_fn('q1', my_df)
lapply(names(my_df),my_fn,y=my_df)

Output:

[[1]]
         
          Agree Disagree N/A or No Opinion Strongly Agree Strongly Disagree
  Choice1     4        0                 0              3                 1
  Choice2     0        0                 0              0                 0

[[2]]
                   
                    Agree Disagree N/A or No Opinion Strongly Agree Strongly Disagree
  Agree                 4        0                 0              0                 0
  Disagree              0        0                 0              0                 0
  N/A or No Opinion     0        0                 0              0                 0
  Strongly Agree        0        0                 0              3                 0
  Strongly Disagree     0        0                 0              0                 1

[[3]]
                   
                    Agree Disagree N/A or No Opinion Strongly Agree Strongly Disagree
  Agree                 1        0                 0              2                 0
  Disagree              0        0                 0              0                 0
  N/A or No Opinion     0        0                 0              0                 0
  Strongly Agree        3        0                 0              1                 1
  Strongly Disagree     0        0                 0              0                 0

akrun · Accepted Answer · 2020-09-15 16:42:20Z

0

We can use count from dplyr. It would get the data in a tibble format

library(dplyr)
library(purrr)
my_fn <- function(data, col1) {data %>% 
          count(!! rlang::sym(col1), q2)}
map(names(my_df), ~ my_fn(my_df, .x))
#[[1]]
# A tibble: 3 x 3
#  q1      q2                    n
#  <fct>   <fct>             <int>
#1 Choice1 Agree                 4
#2 Choice1 Strongly Agree        3
#3 Choice1 Strongly Disagree     1

#[[2]]
# A tibble: 3 x 2
#  q2                    n
#  <fct>             <int>
#1 Agree                 4
#2 Strongly Agree        3
#3 Strongly Disagree     1

#[[3]]
# A tibble: 5 x 3
#  q3             q2                    n
#  <fct>          <fct>             <int>
#1 Agree          Agree                 1
#2 Agree          Strongly Agree        2
#3 Strongly Agree Agree                 3
#4 Strongly Agree Strongly Agree        1
#5 Strongly Agree Strongly Disagree     1

edited Sep 15, 2020 at 16:42

answered Sep 15, 2020 at 16:35

akrun

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1 Comment

nap Over a year ago

Thanks- this solves part of the problem. I'm not sure I can pass the result as an argument for chisq.test() or rcompanion::cramerV() though

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