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Calculate empirical joint type 1/type 2 response probabilities

Source: R/epred_draws_metad.R
joint_probabilities.Rd

Given a dataset data, determine the probability of each combination of type 1 and type 2 responses, optionally conditional on stimulus.

Usage

joint_probabilities(
  data,
  ...,
  .stimulus = "stimulus",
  .response = "response",
  .confidence = "confidence",
  .joint_response = "joint_response",
  K = NULL,
  by_stimulus = TRUE
)

Arguments

data

The data frame to aggregate

...

Grouping columns in data. These columns will be converted to factors.

.stimulus

The name of "stimulus" column

.response

The name of "response" column

.confidence

The name of "confidence" column

.joint_response

The name of "joint_response" column

K

The number of confidence levels in data. If NULL, this is estimated from data using the maximum value of either the confidence column or joint response column.

by_stimulus

If TRUE (default), calculate type 2 response probabilities conditional on stimulus.

Value

A tibble with columns:

  • ...: the grouping columns in data

  • {.stimulus} (if by_stimulus=TRUE): the stimulus

  • {.response}: the type 1 response

  • {.confidence}: the type 2 response

  • {.joint_response}: the joint type 1/type 2 response

  • n: the number of rows in data with the corresponding stimulus (if by_stimulus=TRUE), response, confidence, and joint_response

  • p: the proportion of rows in data with the corresponding response (per stimulus if by_stimulus=TRUE)

See also

epred_draws(), epred_rvars()

Examples

# calculate type 2 response probabilities by stimulus
joint_probabilities(example_data())
#> `hmetad` has inferred that there are K=4 confidence levels in the data. If this is incorrect, please set this manually using the argument `K=<K>`
#> # A tibble: 16 × 6
#> # Groups:   stimulus [2]
#>    stimulus response confidence joint_response     n     p
#>       <int>    <int>      <int>          <int> <int> <dbl>
#>  1        0        0          4              1    71 0.142
#>  2        0        0          3              2    94 0.188
#>  3        0        0          2              3   101 0.202
#>  4        0        0          1              4    86 0.172
#>  5        0        1          1              5    74 0.148
#>  6        0        1          2              6    44 0.088
#>  7        0        1          3              7    24 0.048
#>  8        0        1          4              8     6 0.012
#>  9        1        0          4              1    10 0.02 
#> 10        1        0          3              2    26 0.052
#> 11        1        0          2              3    46 0.092
#> 12        1        0          1              4    75 0.15 
#> 13        1        1          1              5    86 0.172
#> 14        1        1          2              6   104 0.208
#> 15        1        1          3              7    75 0.15 
#> 16        1        1          4              8    78 0.156

# calculate type 2 response probabilities by condition, averaging over stimuli
joint_probabilities(sim_metad_condition(), condition, by_stimulus = FALSE)
#> `hmetad` has inferred that there are K=4 confidence levels in the data. If this is incorrect, please set this manually using the argument `K=<K>`
#> # A tibble: 16 × 6
#> # Groups:   condition [2]
#>    condition response confidence joint_response     n     p
#>        <int>    <int>      <int>          <int> <int> <dbl>
#>  1         1        0          4              1    13  0.13
#>  2         1        0          3              2    13  0.13
#>  3         1        0          2              3     8  0.08
#>  4         1        0          1              4    16  0.16
#>  5         1        1          1              5    16  0.16
#>  6         1        1          2              6    11  0.11
#>  7         1        1          3              7    11  0.11
#>  8         1        1          4              8    12  0.12
#>  9         2        0          4              1     8  0.08
#> 10         2        0          3              2     9  0.09
#> 11         2        0          2              3    22  0.22
#> 12         2        0          1              4    18  0.18
#> 13         2        1          1              5    18  0.18
#> 14         2        1          2              6     9  0.09
#> 15         2        1          3              7     7  0.07
#> 16         2        1          4              8     9  0.09