Description
This shows the output of recoding functions from the package rwf. The functions stack a list of data frames into one data frame, reverse-score a vector (for example reverse-keyed Likert items), turn multiple-response answers into binary dummy columns, and collapse rare factor levels into a single “Other” level. Each section gives the rule the function applies.
Installation instructions for rwf can be found here
The code can be found here
Notation
x_1,\dots,x_n are the values of the input vector and \mathbf{1}[\cdot] is 1 when the condition is true and 0 otherwise. |S| is the number of elements of a set S.
flatten_list
Stacks the elements of a list L=\{L_1,\dots,L_K\} into one data frame. Each
element L_k is converted to a data
frame with n_k rows, the data frames
are placed under each other, and a .id column records the
name of the element each row came from:
\text{rows}=\sum_{k=1}^{K}n_k,\qquad \texttt{.id}_i=\operatorname{name}(L_k)\ \text{ for every row } i \text{ that comes from } L_k
Columns with the same name are matched; a column missing from an
element is filled with NA.
results <- list(sample_a = data.frame(mean = 10.2, sd = 1.1, n = 30),
sample_b = data.frame(mean = 11.8, sd = 1.4, n = 25),
sample_c = data.frame(mean = 9.7, sd = 0.9, n = 41))
flatten_list(results)## .id mean sd n
## 1 sample_a 10.2 1.1 30
## 2 sample_b 11.8 1.4 25
## 3 sample_c 9.7 0.9 41Vectors become one-column data frames:
flatten_list(list(group1 = 1:3, group2 = 4:5))## .id x
## 1 group1 1
## 2 group1 2
## 3 group1 3
## 4 group2 4
## 5 group2 5swap
Reverse-scores a vector by mirroring its distinct observed values. With v_{(1)}<v_{(2)}<\dots<v_{(m)} the m distinct non-missing values in sorted order, each value is replaced by the value at the mirrored position:
x_i=v_{(r)}\ \Rightarrow\ x_i'=v_{(m+1-r)}
When the observed values are all the consecutive integers from a to b, this is the usual reverse-scoring formula:
x_i'=a+b-x_i
Missing values stay missing, and the result has the same class as the input (numeric, integer or character).
## [1] 5 4 3 2 1 1 2 5## [1] TRUEOnly the observed values are mirrored, not the full scale. If a 1–5 item has no answers of 4, the observed values are 1, 2, 3, 5: 1 and 5 swap, 2 and 3 swap, and a+b-x no longer applies:
## [1] 5 3 2 1 1
6 - likert_gap## [1] 5 4 3 1 1Numbers are sorted by value and text alphabetically, so text categories are mirrored in alphabetical order, not in their meaning:
## [1] "low" "high" "mid" "low"dummy_arrange
Turns multiple-response answers into binary dummy columns. Each element i holds one or more responses separated by commas, giving a set S_i (an empty element gives S_i=\varnothing). The columns are all responses given by anyone, sorted as text:
\mathcal{K}=\bigcup_{i=1}^{n}S_i,\qquad D_{ik}=\mathbf{1}\left[k\in S_i\right],\quad i=1,\dots,n,\quad k\in\mathcal{K}
Row sums give the number of responses each person chose, \sum_{k}D_{ik}=|S_i|, and column sums give how often each response was chosen. The function returns the 1s and 0s as text, so convert them to numbers before summing:
answers <- c("Email,Phone", "Phone", "Post,Email,Phone", "", "Email")
dummies <- dummy_arrange(answers)
dummies## Email Phone Post
## 1 1 1 0
## 2 0 1 0
## 3 1 1 1
## 4 0 0 0
## 5 1 0 0
dummies <- data.frame(lapply(dummies, as.numeric))
rowSums(dummies)## [1] 2 1 3 0 1
colSums(dummies)## Email Phone Post
## 3 3 1Single responses work the same way. Because columns are sorted as
text, numeric responses sort as 1, 10,
2:
dummy_arrange(c("1,10", "2", "10,2"))## 1 10 2
## 1 1 1 0
## 2 0 0 1
## 3 0 1 1generate_multiple_response_vector from rwf simulates
multiple-response data; gsub removes the spaces after the
commas:
set.seed(1)
responses <- gsub(" ", "", generate_multiple_response_vector(responses = c("Agree", "Hi", "All"), responded = 1:3, length = 6), fixed = TRUE)
responses## [1] "All" "Hi" "All" "Hi,All,Agree" "Agree,All,Hi" "Hi,All"
dummy_arrange(responses)## Agree All Hi
## 1 0 1 0
## 2 0 0 1
## 3 0 1 0
## 4 1 1 1
## 5 1 1 1
## 6 0 1 1drop_levels
Collapses rare levels of each factor into a single level
"Other" and removes unused levels. With n_\ell=\sum_{i=1}^{n}\mathbf{1}[x_i=\ell] the
frequency of level \ell and c = minimum_frequency:
x_i'=\begin{cases}\text{"Other"} & n_{x_i}\le c\\ x_i & n_{x_i}>c\end{cases}
Levels with no observations (n_\ell=0\le
c) are removed. All factor columns are processed, or only the
columns in factor_index; other columns are left
unchanged.
df <- data.frame(id = 1:12,
city = factor(c(rep("Athens", 6), rep("Patras", 3), rep("Volos", 2), "Chania"),
levels = c("Athens", "Patras", "Volos", "Chania", "Larissa")),
group = factor(rep(c("x", "y"), 6)))
table(df$city)##
## Athens Patras Volos Chania Larissa
## 6 3 2 1 0With c=2, Volos (2) and Chania (1) become “Other” and the empty level Larissa is removed:
table(drop_levels(df, factor_index = 2, minimum_frequency = 2)$city)##
## Athens Patras Other
## 6 3 3The comparison is \le, so with c=3 Patras (exactly 3) is collapsed too:
table(drop_levels(df, factor_index = 2, minimum_frequency = 3)$city)##
## Athens Other
## 6 6Without factor_index, every factor is processed. With
c=6 no level of either factor has more
than 6 cases (Athens and both levels of group have exactly
6), so each factor collapses into a single “Other” level:
result <- drop_levels(df, minimum_frequency = 6)
table(result$city)##
## Other
## 12
table(result$group)##
## Other
## 12